Determine the Norton equivalent of the circuit depicted in Fig. 5.74 as seen looking into the two open terminals. ( b ) Compute power dissipated in a 5 Ω resistor connected in parallel with the existing 5 Ω resistor. ( c ) Compute the current flowing through a short circuit connecting the two terminals.

Question

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Answer 1

Textbook Question

Chapter 5, Problem 33E

Determine the Norton equivalent of the circuit depicted in Fig. 5.74 as seen looking into the two open terminals. (b) Compute power dissipated in a 5 Ω resistor connected in parallel with the existing 5 Ω resistor. (c) Compute the current flowing through a short circuit connecting the two terminals.

Chapter 5, Problem 33E, Determine the Norton equivalent of the circuit depicted in Fig. 5.74 as seen looking into the two

(a)

Expert Solution

To determine

Find the Norton equivalent of the circuit as seen looking into the two open terminals.

Answer to Problem 33E

The Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

Explanation of Solution

Formula used:

The expression for voltage is as follows.

$v=iR$ (1)

Here,

$v$ is the voltage,

$i$ is the current and

$R$ is value of resistance.

The expression for series combination of resistance is as follows.

$R=(R1+R2)$ (2)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

The expression for parallel combination of resistance is as follows.

$1R=(1R1+1R2)$ (3)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

Calculation:

The circuit diagram is redrawn as shown in Figure 1.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 1

Refer to redrawn Figure 1,

The expression for KVL in super mesh $BACEFDB$ is as follows.

$v1+i1R1+i2R2+(i2−i3)R3=0$ (4)

Here,

$v1$ is voltage in loop $BACDB$ ,

$R1$ , $R2$ and $R3$ are the resistances in the circuit,

$i1$ is the current flowing in loop $BACDB$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $2 V$ for $v1$ , $2 Ω$ for $R1$ , $5 Ω$ for $R2$ and $1 Ω$ for $R3$ in equation (4).

$2 V+i1(2 Ω)+i2(5 Ω)+(i2−i3)(1 Ω)=02 V+i1(2 Ω)+i2(5 Ω)+(1 Ω)i2−(1 Ω)i3=0$

$2 V+i1(2 Ω)+i2(6 Ω)−(1 Ω)i3=0$ (5)

The expression for KVL in mesh $FEGHF$ is as follows.

$(i3−i2)R3+i3R4+v2=0$ (6)

Here,

$R3$ and $R4$ are the resistances in the circuit,

$v2$ is voltage in loop $FEGHF$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $1 Ω$ for $R3$ , $3 Ω$ for $R4$ and $4 V$ for $v2$ in the equation (6).

$(i3−i2)(1 Ω)+i3(3 Ω)+4 V=0(1 Ω)i3−(1 Ω)i2+i3(3 Ω)+4 V=0$

$(4 Ω)i3−(1 Ω)i2+4 V=0$ (7)

The expression for current $i$ is as follows.

$i4=i2−i1$ (8)

Here,

$i4$ is current through independent current source,

$i2$ is the current flowing in loop $DCEFD$ and

$i1$ is the current flowing in loop $BACDB$ .

Substitute $2 A$ for $i4$ in equation (8).

$2 A=i2−i1$ (9)

Rearrange equations(5),(7) and (9).

$2i1+6i2−i3=−20i1−i2+4i3=−4−i1+i2+0i3=2$

The equations so formed can be written in matrix form as,

$(26−10−14−110)(i1i2i3)=(−2−42)$

Therefore, by Cramer’s rule,

The determinant of coefficient matrix is as follows.

$Δ=|26−10−14−110|=−31$

The 1^st determinant is as follows.

$Δ1=|−26−1−4−14210|=58$

The 2^nd determinant is as follows,

$Δ2=|2−2−10−44−120|=−4$

The 3rd determinant is as follows.

$Δ3=|26−20−1−4−112|=30$

Simplify for $i1$ .

$i1=Δ1Δ=58−31=−1.87 A$

Simplify for $i2$ .

$i2=Δ2Δ=−4−31=0.129 A=129 mA {∵ 1 A=103 mA}$

Simplify for $i3$ .

$i3=Δ3Δ=30−31=−0.9677 A$

Substitute $129 mA$ for $i$ and $5 Ω$ for $R$ in equation (1).

$v=(129 mA)×(5 Ω)=(129×10−3 A)×(5 Ω) {∵1 mA=10−3 A}=645×10−3 V =645 mV {∵10−3 V=1 mV}$

So, the Thevenin voltage is $645 mV$ .

The circuit diagram is redrawn as shown in Figure 2.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 2

As $R3$ and $R4$ are in parallel therefore from equation (3).

$1R5=(1R3+1R4)1R5=(11 Ω+13 Ω)1R5=10.75 Ω$

Rearrange for $R5$ ,

$R5=0.75 Ω$

The circuit diagram is redrawn as shown in Figure 3.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 3

As $R1$ and $R5$ are in series therefore from equation (2).

$R6=R1+R5=2 Ω+0.75 Ω=2.75 Ω$

The circuit diagram is redrawn as shown in Figure 4.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 4

Refer to redrawn Figure 4,

As $R2$ and $R6$ are in parallel therefore from equation (3),

$1Rth=(1R2+1R6)1Rth=(15 Ω+12.75 Ω)1Rth=11.774 Ω$

Rearrange for $Rth$ ,

$Rth=1.774 Ω$

So, the Thevenin equivalent resistance across the branch $XY$ is $1.774 Ω$ .

Thevenin equivalent resistance is same as the Norton equivalent resistance,

Hence,

$RN=1.774 Ω$

The circuit diagram is redrawn as shown in Figure 5,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 5

Refer to redrawn Figure 5,

Norton current is the current in the load resistor when load resistance is replaced by a short circuit.

The expression for the Norton current flowing in the circuit is as follows,

$IN=VthRth$ (10)

Here,

$IN$ is the Norton current,

$Vth$ is the Thevenin voltage and

$Rth$ is the Thevenin equivalent resistance.

Substitute $1.774 Ω$ for $Rth$ and $645 mV$ for $Vth$ in equation (10),

$IN=(645 mV)(1.774 Ω)=363.58 mA$

Conclusion:

Thus the Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

(b)

Expert Solution

To determine

Find the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor.

Answer to Problem 33E

The power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

Explanation of Solution

Given Data:

The load resistance is $5 Ω$ .

Formula used:

The expression for the power dissipated in a resistor is as follows,

$p=(iL)2RL$ (`11)

Here,

$p$ is the power dissipated in a resistor,

$RL$ is the load resistance and

$iL$ is the current flowing through the load.

Calculation:

The circuit is drawn as shown in Figure 6,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 6

Refer to redrawn Figure 6,

The simplified Norton equivalent of the circuit is drawn as shown in Figure 7,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 7

Refer to redrawn Figure 7,

The expression for the current flowing through the load is as follows,

$iL=IN(RNRN+RL)$ (12)

Substitute $363.58 mA$ for $IN$ , $1.774 Ω$ for $RN$ and $5 Ω$ for $RL$ in equation (12),

$iL=(363.58 mA)(1.774 Ω(1.774 Ω)+(5 Ω))=(363.58 mA)(0.26188)=95.22 mA$

Substitute $95.22 mA$ for $iL$ and $5 Ω$ for $RL$ in equation (11),

$p=(95.22 mA)2(5 Ω)=(95.22×10−3 A)2(5 Ω) {∵1 mA=10−3 A}=(9066.023×10−6)(5) W=45330.115×10−6W$

$p=45.33 mW {∵10−6 A=10−3 mA}$

Conclusion:

Thus, the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

(c)

Expert Solution

To determine

Find the current flowing through a short circuit connecting the two terminals.

Answer to Problem 33E

The current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

Explanation of Solution

Calculation:

The current flowing through a short circuit connecting the two terminals is Norton current which is $363.58 mA$ as calculated in sub-part (a).

So, the current flowing through a short circuit connecting the two terminals is $363.58 mA$

Conclusion:

Thus, the current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

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Answer 2

Textbook Question

Chapter 5, Problem 33E

Determine the Norton equivalent of the circuit depicted in Fig. 5.74 as seen looking into the two open terminals. (b) Compute power dissipated in a 5 Ω resistor connected in parallel with the existing 5 Ω resistor. (c) Compute the current flowing through a short circuit connecting the two terminals.

Chapter 5, Problem 33E, Determine the Norton equivalent of the circuit depicted in Fig. 5.74 as seen looking into the two

(a)

Expert Solution

To determine

Find the Norton equivalent of the circuit as seen looking into the two open terminals.

Answer to Problem 33E

The Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

Explanation of Solution

Formula used:

The expression for voltage is as follows.

$v=iR$ (1)

Here,

$v$ is the voltage,

$i$ is the current and

$R$ is value of resistance.

The expression for series combination of resistance is as follows.

$R=(R1+R2)$ (2)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

The expression for parallel combination of resistance is as follows.

$1R=(1R1+1R2)$ (3)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

Calculation:

The circuit diagram is redrawn as shown in Figure 1.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 1

Refer to redrawn Figure 1,

The expression for KVL in super mesh $BACEFDB$ is as follows.

$v1+i1R1+i2R2+(i2−i3)R3=0$ (4)

Here,

$v1$ is voltage in loop $BACDB$ ,

$R1$ , $R2$ and $R3$ are the resistances in the circuit,

$i1$ is the current flowing in loop $BACDB$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $2 V$ for $v1$ , $2 Ω$ for $R1$ , $5 Ω$ for $R2$ and $1 Ω$ for $R3$ in equation (4).

$2 V+i1(2 Ω)+i2(5 Ω)+(i2−i3)(1 Ω)=02 V+i1(2 Ω)+i2(5 Ω)+(1 Ω)i2−(1 Ω)i3=0$

$2 V+i1(2 Ω)+i2(6 Ω)−(1 Ω)i3=0$ (5)

The expression for KVL in mesh $FEGHF$ is as follows.

$(i3−i2)R3+i3R4+v2=0$ (6)

Here,

$R3$ and $R4$ are the resistances in the circuit,

$v2$ is voltage in loop $FEGHF$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $1 Ω$ for $R3$ , $3 Ω$ for $R4$ and $4 V$ for $v2$ in the equation (6).

$(i3−i2)(1 Ω)+i3(3 Ω)+4 V=0(1 Ω)i3−(1 Ω)i2+i3(3 Ω)+4 V=0$

$(4 Ω)i3−(1 Ω)i2+4 V=0$ (7)

The expression for current $i$ is as follows.

$i4=i2−i1$ (8)

Here,

$i4$ is current through independent current source,

$i2$ is the current flowing in loop $DCEFD$ and

$i1$ is the current flowing in loop $BACDB$ .

Substitute $2 A$ for $i4$ in equation (8).

$2 A=i2−i1$ (9)

Rearrange equations(5),(7) and (9).

$2i1+6i2−i3=−20i1−i2+4i3=−4−i1+i2+0i3=2$

The equations so formed can be written in matrix form as,

$(26−10−14−110)(i1i2i3)=(−2−42)$

Therefore, by Cramer’s rule,

The determinant of coefficient matrix is as follows.

$Δ=|26−10−14−110|=−31$

The 1^st determinant is as follows.

$Δ1=|−26−1−4−14210|=58$

The 2^nd determinant is as follows,

$Δ2=|2−2−10−44−120|=−4$

The 3rd determinant is as follows.

$Δ3=|26−20−1−4−112|=30$

Simplify for $i1$ .

$i1=Δ1Δ=58−31=−1.87 A$

Simplify for $i2$ .

$i2=Δ2Δ=−4−31=0.129 A=129 mA {∵ 1 A=103 mA}$

Simplify for $i3$ .

$i3=Δ3Δ=30−31=−0.9677 A$

Substitute $129 mA$ for $i$ and $5 Ω$ for $R$ in equation (1).

$v=(129 mA)×(5 Ω)=(129×10−3 A)×(5 Ω) {∵1 mA=10−3 A}=645×10−3 V =645 mV {∵10−3 V=1 mV}$

So, the Thevenin voltage is $645 mV$ .

The circuit diagram is redrawn as shown in Figure 2.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 2

As $R3$ and $R4$ are in parallel therefore from equation (3).

$1R5=(1R3+1R4)1R5=(11 Ω+13 Ω)1R5=10.75 Ω$

Rearrange for $R5$ ,

$R5=0.75 Ω$

The circuit diagram is redrawn as shown in Figure 3.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 3

As $R1$ and $R5$ are in series therefore from equation (2).

$R6=R1+R5=2 Ω+0.75 Ω=2.75 Ω$

The circuit diagram is redrawn as shown in Figure 4.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 4

Refer to redrawn Figure 4,

As $R2$ and $R6$ are in parallel therefore from equation (3),

$1Rth=(1R2+1R6)1Rth=(15 Ω+12.75 Ω)1Rth=11.774 Ω$

Rearrange for $Rth$ ,

$Rth=1.774 Ω$

So, the Thevenin equivalent resistance across the branch $XY$ is $1.774 Ω$ .

Thevenin equivalent resistance is same as the Norton equivalent resistance,

Hence,

$RN=1.774 Ω$

The circuit diagram is redrawn as shown in Figure 5,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 5

Refer to redrawn Figure 5,

Norton current is the current in the load resistor when load resistance is replaced by a short circuit.

The expression for the Norton current flowing in the circuit is as follows,

$IN=VthRth$ (10)

Here,

$IN$ is the Norton current,

$Vth$ is the Thevenin voltage and

$Rth$ is the Thevenin equivalent resistance.

Substitute $1.774 Ω$ for $Rth$ and $645 mV$ for $Vth$ in equation (10),

$IN=(645 mV)(1.774 Ω)=363.58 mA$

Conclusion:

Thus the Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

(b)

Expert Solution

To determine

Find the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor.

Answer to Problem 33E

The power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

Explanation of Solution

Given Data:

The load resistance is $5 Ω$ .

Formula used:

The expression for the power dissipated in a resistor is as follows,

$p=(iL)2RL$ (`11)

Here,

$p$ is the power dissipated in a resistor,

$RL$ is the load resistance and

$iL$ is the current flowing through the load.

Calculation:

The circuit is drawn as shown in Figure 6,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 6

Refer to redrawn Figure 6,

The simplified Norton equivalent of the circuit is drawn as shown in Figure 7,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 7

Refer to redrawn Figure 7,

The expression for the current flowing through the load is as follows,

$iL=IN(RNRN+RL)$ (12)

Substitute $363.58 mA$ for $IN$ , $1.774 Ω$ for $RN$ and $5 Ω$ for $RL$ in equation (12),

$iL=(363.58 mA)(1.774 Ω(1.774 Ω)+(5 Ω))=(363.58 mA)(0.26188)=95.22 mA$

Substitute $95.22 mA$ for $iL$ and $5 Ω$ for $RL$ in equation (11),

$p=(95.22 mA)2(5 Ω)=(95.22×10−3 A)2(5 Ω) {∵1 mA=10−3 A}=(9066.023×10−6)(5) W=45330.115×10−6W$

$p=45.33 mW {∵10−6 A=10−3 mA}$

Conclusion:

Thus, the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

(c)

Expert Solution

To determine

Find the current flowing through a short circuit connecting the two terminals.

Answer to Problem 33E

The current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

Explanation of Solution

Calculation:

The current flowing through a short circuit connecting the two terminals is Norton current which is $363.58 mA$ as calculated in sub-part (a).

So, the current flowing through a short circuit connecting the two terminals is $363.58 mA$

Conclusion:

Thus, the current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

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Answer 3

Textbook Question

Chapter 5, Problem 33E

Determine the Norton equivalent of the circuit depicted in Fig. 5.74 as seen looking into the two open terminals. (b) Compute power dissipated in a 5 Ω resistor connected in parallel with the existing 5 Ω resistor. (c) Compute the current flowing through a short circuit connecting the two terminals.

Chapter 5, Problem 33E, Determine the Norton equivalent of the circuit depicted in Fig. 5.74 as seen looking into the two

(a)

Expert Solution

To determine

Find the Norton equivalent of the circuit as seen looking into the two open terminals.

Answer to Problem 33E

The Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

Explanation of Solution

Formula used:

The expression for voltage is as follows.

$v=iR$ (1)

Here,

$v$ is the voltage,

$i$ is the current and

$R$ is value of resistance.

The expression for series combination of resistance is as follows.

$R=(R1+R2)$ (2)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

The expression for parallel combination of resistance is as follows.

$1R=(1R1+1R2)$ (3)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

Calculation:

The circuit diagram is redrawn as shown in Figure 1.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 1

Refer to redrawn Figure 1,

The expression for KVL in super mesh $BACEFDB$ is as follows.

$v1+i1R1+i2R2+(i2−i3)R3=0$ (4)

Here,

$v1$ is voltage in loop $BACDB$ ,

$R1$ , $R2$ and $R3$ are the resistances in the circuit,

$i1$ is the current flowing in loop $BACDB$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $2 V$ for $v1$ , $2 Ω$ for $R1$ , $5 Ω$ for $R2$ and $1 Ω$ for $R3$ in equation (4).

$2 V+i1(2 Ω)+i2(5 Ω)+(i2−i3)(1 Ω)=02 V+i1(2 Ω)+i2(5 Ω)+(1 Ω)i2−(1 Ω)i3=0$

$2 V+i1(2 Ω)+i2(6 Ω)−(1 Ω)i3=0$ (5)

The expression for KVL in mesh $FEGHF$ is as follows.

$(i3−i2)R3+i3R4+v2=0$ (6)

Here,

$R3$ and $R4$ are the resistances in the circuit,

$v2$ is voltage in loop $FEGHF$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $1 Ω$ for $R3$ , $3 Ω$ for $R4$ and $4 V$ for $v2$ in the equation (6).

$(i3−i2)(1 Ω)+i3(3 Ω)+4 V=0(1 Ω)i3−(1 Ω)i2+i3(3 Ω)+4 V=0$

$(4 Ω)i3−(1 Ω)i2+4 V=0$ (7)

The expression for current $i$ is as follows.

$i4=i2−i1$ (8)

Here,

$i4$ is current through independent current source,

$i2$ is the current flowing in loop $DCEFD$ and

$i1$ is the current flowing in loop $BACDB$ .

Substitute $2 A$ for $i4$ in equation (8).

$2 A=i2−i1$ (9)

Rearrange equations(5),(7) and (9).

$2i1+6i2−i3=−20i1−i2+4i3=−4−i1+i2+0i3=2$

The equations so formed can be written in matrix form as,

$(26−10−14−110)(i1i2i3)=(−2−42)$

Therefore, by Cramer’s rule,

The determinant of coefficient matrix is as follows.

$Δ=|26−10−14−110|=−31$

The 1^st determinant is as follows.

$Δ1=|−26−1−4−14210|=58$

The 2^nd determinant is as follows,

$Δ2=|2−2−10−44−120|=−4$

The 3rd determinant is as follows.

$Δ3=|26−20−1−4−112|=30$

Simplify for $i1$ .

$i1=Δ1Δ=58−31=−1.87 A$

Simplify for $i2$ .

$i2=Δ2Δ=−4−31=0.129 A=129 mA {∵ 1 A=103 mA}$

Simplify for $i3$ .

$i3=Δ3Δ=30−31=−0.9677 A$

Substitute $129 mA$ for $i$ and $5 Ω$ for $R$ in equation (1).

$v=(129 mA)×(5 Ω)=(129×10−3 A)×(5 Ω) {∵1 mA=10−3 A}=645×10−3 V =645 mV {∵10−3 V=1 mV}$

So, the Thevenin voltage is $645 mV$ .

The circuit diagram is redrawn as shown in Figure 2.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 2

As $R3$ and $R4$ are in parallel therefore from equation (3).

$1R5=(1R3+1R4)1R5=(11 Ω+13 Ω)1R5=10.75 Ω$

Rearrange for $R5$ ,

$R5=0.75 Ω$

The circuit diagram is redrawn as shown in Figure 3.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 3

As $R1$ and $R5$ are in series therefore from equation (2).

$R6=R1+R5=2 Ω+0.75 Ω=2.75 Ω$

The circuit diagram is redrawn as shown in Figure 4.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 4

Refer to redrawn Figure 4,

As $R2$ and $R6$ are in parallel therefore from equation (3),

$1Rth=(1R2+1R6)1Rth=(15 Ω+12.75 Ω)1Rth=11.774 Ω$

Rearrange for $Rth$ ,

$Rth=1.774 Ω$

So, the Thevenin equivalent resistance across the branch $XY$ is $1.774 Ω$ .

Thevenin equivalent resistance is same as the Norton equivalent resistance,

Hence,

$RN=1.774 Ω$

The circuit diagram is redrawn as shown in Figure 5,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 5

Refer to redrawn Figure 5,

Norton current is the current in the load resistor when load resistance is replaced by a short circuit.

The expression for the Norton current flowing in the circuit is as follows,

$IN=VthRth$ (10)

Here,

$IN$ is the Norton current,

$Vth$ is the Thevenin voltage and

$Rth$ is the Thevenin equivalent resistance.

Substitute $1.774 Ω$ for $Rth$ and $645 mV$ for $Vth$ in equation (10),

$IN=(645 mV)(1.774 Ω)=363.58 mA$

Conclusion:

Thus the Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

(b)

Expert Solution

To determine

Find the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor.

Answer to Problem 33E

The power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

Explanation of Solution

Given Data:

The load resistance is $5 Ω$ .

Formula used:

The expression for the power dissipated in a resistor is as follows,

$p=(iL)2RL$ (`11)

Here,

$p$ is the power dissipated in a resistor,

$RL$ is the load resistance and

$iL$ is the current flowing through the load.

Calculation:

The circuit is drawn as shown in Figure 6,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 6

Refer to redrawn Figure 6,

The simplified Norton equivalent of the circuit is drawn as shown in Figure 7,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 7

Refer to redrawn Figure 7,

The expression for the current flowing through the load is as follows,

$iL=IN(RNRN+RL)$ (12)

Substitute $363.58 mA$ for $IN$ , $1.774 Ω$ for $RN$ and $5 Ω$ for $RL$ in equation (12),

$iL=(363.58 mA)(1.774 Ω(1.774 Ω)+(5 Ω))=(363.58 mA)(0.26188)=95.22 mA$

Substitute $95.22 mA$ for $iL$ and $5 Ω$ for $RL$ in equation (11),

$p=(95.22 mA)2(5 Ω)=(95.22×10−3 A)2(5 Ω) {∵1 mA=10−3 A}=(9066.023×10−6)(5) W=45330.115×10−6W$

$p=45.33 mW {∵10−6 A=10−3 mA}$

Conclusion:

Thus, the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

(c)

Expert Solution

To determine

Find the current flowing through a short circuit connecting the two terminals.

Answer to Problem 33E

The current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

Explanation of Solution

Calculation:

The current flowing through a short circuit connecting the two terminals is Norton current which is $363.58 mA$ as calculated in sub-part (a).

So, the current flowing through a short circuit connecting the two terminals is $363.58 mA$

Conclusion:

Thus, the current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

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Answer 4

Textbook Question

Chapter 5, Problem 33E

Determine the Norton equivalent of the circuit depicted in Fig. 5.74 as seen looking into the two open terminals. (b) Compute power dissipated in a 5 Ω resistor connected in parallel with the existing 5 Ω resistor. (c) Compute the current flowing through a short circuit connecting the two terminals.

Chapter 5, Problem 33E, Determine the Norton equivalent of the circuit depicted in Fig. 5.74 as seen looking into the two

(a)

Expert Solution

To determine

Find the Norton equivalent of the circuit as seen looking into the two open terminals.

Answer to Problem 33E

The Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

Explanation of Solution

Formula used:

The expression for voltage is as follows.

$v=iR$ (1)

Here,

$v$ is the voltage,

$i$ is the current and

$R$ is value of resistance.

The expression for series combination of resistance is as follows.

$R=(R1+R2)$ (2)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

The expression for parallel combination of resistance is as follows.

$1R=(1R1+1R2)$ (3)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

Calculation:

The circuit diagram is redrawn as shown in Figure 1.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 1

Refer to redrawn Figure 1,

The expression for KVL in super mesh $BACEFDB$ is as follows.

$v1+i1R1+i2R2+(i2−i3)R3=0$ (4)

Here,

$v1$ is voltage in loop $BACDB$ ,

$R1$ , $R2$ and $R3$ are the resistances in the circuit,

$i1$ is the current flowing in loop $BACDB$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $2 V$ for $v1$ , $2 Ω$ for $R1$ , $5 Ω$ for $R2$ and $1 Ω$ for $R3$ in equation (4).

$2 V+i1(2 Ω)+i2(5 Ω)+(i2−i3)(1 Ω)=02 V+i1(2 Ω)+i2(5 Ω)+(1 Ω)i2−(1 Ω)i3=0$

$2 V+i1(2 Ω)+i2(6 Ω)−(1 Ω)i3=0$ (5)

The expression for KVL in mesh $FEGHF$ is as follows.

$(i3−i2)R3+i3R4+v2=0$ (6)

Here,

$R3$ and $R4$ are the resistances in the circuit,

$v2$ is voltage in loop $FEGHF$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $1 Ω$ for $R3$ , $3 Ω$ for $R4$ and $4 V$ for $v2$ in the equation (6).

$(i3−i2)(1 Ω)+i3(3 Ω)+4 V=0(1 Ω)i3−(1 Ω)i2+i3(3 Ω)+4 V=0$

$(4 Ω)i3−(1 Ω)i2+4 V=0$ (7)

The expression for current $i$ is as follows.

$i4=i2−i1$ (8)

Here,

$i4$ is current through independent current source,

$i2$ is the current flowing in loop $DCEFD$ and

$i1$ is the current flowing in loop $BACDB$ .

Substitute $2 A$ for $i4$ in equation (8).

$2 A=i2−i1$ (9)

Rearrange equations(5),(7) and (9).

$2i1+6i2−i3=−20i1−i2+4i3=−4−i1+i2+0i3=2$

The equations so formed can be written in matrix form as,

$(26−10−14−110)(i1i2i3)=(−2−42)$

Therefore, by Cramer’s rule,

The determinant of coefficient matrix is as follows.

$Δ=|26−10−14−110|=−31$

The 1^st determinant is as follows.

$Δ1=|−26−1−4−14210|=58$

The 2^nd determinant is as follows,

$Δ2=|2−2−10−44−120|=−4$

The 3rd determinant is as follows.

$Δ3=|26−20−1−4−112|=30$

Simplify for $i1$ .

$i1=Δ1Δ=58−31=−1.87 A$

Simplify for $i2$ .

$i2=Δ2Δ=−4−31=0.129 A=129 mA {∵ 1 A=103 mA}$

Simplify for $i3$ .

$i3=Δ3Δ=30−31=−0.9677 A$

Substitute $129 mA$ for $i$ and $5 Ω$ for $R$ in equation (1).

$v=(129 mA)×(5 Ω)=(129×10−3 A)×(5 Ω) {∵1 mA=10−3 A}=645×10−3 V =645 mV {∵10−3 V=1 mV}$

So, the Thevenin voltage is $645 mV$ .

The circuit diagram is redrawn as shown in Figure 2.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 2

As $R3$ and $R4$ are in parallel therefore from equation (3).

$1R5=(1R3+1R4)1R5=(11 Ω+13 Ω)1R5=10.75 Ω$

Rearrange for $R5$ ,

$R5=0.75 Ω$

The circuit diagram is redrawn as shown in Figure 3.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 3

As $R1$ and $R5$ are in series therefore from equation (2).

$R6=R1+R5=2 Ω+0.75 Ω=2.75 Ω$

The circuit diagram is redrawn as shown in Figure 4.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 4

Refer to redrawn Figure 4,

As $R2$ and $R6$ are in parallel therefore from equation (3),

$1Rth=(1R2+1R6)1Rth=(15 Ω+12.75 Ω)1Rth=11.774 Ω$

Rearrange for $Rth$ ,

$Rth=1.774 Ω$

So, the Thevenin equivalent resistance across the branch $XY$ is $1.774 Ω$ .

Thevenin equivalent resistance is same as the Norton equivalent resistance,

Hence,

$RN=1.774 Ω$

The circuit diagram is redrawn as shown in Figure 5,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 5

Refer to redrawn Figure 5,

Norton current is the current in the load resistor when load resistance is replaced by a short circuit.

The expression for the Norton current flowing in the circuit is as follows,

$IN=VthRth$ (10)

Here,

$IN$ is the Norton current,

$Vth$ is the Thevenin voltage and

$Rth$ is the Thevenin equivalent resistance.

Substitute $1.774 Ω$ for $Rth$ and $645 mV$ for $Vth$ in equation (10),

$IN=(645 mV)(1.774 Ω)=363.58 mA$

Conclusion:

Thus the Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

(b)

Expert Solution

To determine

Find the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor.

Answer to Problem 33E

The power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

Explanation of Solution

Given Data:

The load resistance is $5 Ω$ .

Formula used:

The expression for the power dissipated in a resistor is as follows,

$p=(iL)2RL$ (`11)

Here,

$p$ is the power dissipated in a resistor,

$RL$ is the load resistance and

$iL$ is the current flowing through the load.

Calculation:

The circuit is drawn as shown in Figure 6,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 6

Refer to redrawn Figure 6,

The simplified Norton equivalent of the circuit is drawn as shown in Figure 7,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 7

Refer to redrawn Figure 7,

The expression for the current flowing through the load is as follows,

$iL=IN(RNRN+RL)$ (12)

Substitute $363.58 mA$ for $IN$ , $1.774 Ω$ for $RN$ and $5 Ω$ for $RL$ in equation (12),

$iL=(363.58 mA)(1.774 Ω(1.774 Ω)+(5 Ω))=(363.58 mA)(0.26188)=95.22 mA$

Substitute $95.22 mA$ for $iL$ and $5 Ω$ for $RL$ in equation (11),

$p=(95.22 mA)2(5 Ω)=(95.22×10−3 A)2(5 Ω) {∵1 mA=10−3 A}=(9066.023×10−6)(5) W=45330.115×10−6W$

$p=45.33 mW {∵10−6 A=10−3 mA}$

Conclusion:

Thus, the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

(c)

Expert Solution

To determine

Find the current flowing through a short circuit connecting the two terminals.

Answer to Problem 33E

The current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

Explanation of Solution

Calculation:

The current flowing through a short circuit connecting the two terminals is Norton current which is $363.58 mA$ as calculated in sub-part (a).

So, the current flowing through a short circuit connecting the two terminals is $363.58 mA$

Conclusion:

Thus, the current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

Find the Norton equivalent of the circuit as seen looking into the two open terminals.

Answer to Problem 33E

The Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

Explanation of Solution

Formula used:

The expression for voltage is as follows.

$v=iR$ (1)

Here,

$v$ is the voltage,

$i$ is the current and

$R$ is value of resistance.

The expression for series combination of resistance is as follows.

$R=(R1+R2)$ (2)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

The expression for parallel combination of resistance is as follows.

$1R=(1R1+1R2)$ (3)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

Calculation:

The circuit diagram is redrawn as shown in Figure 1.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 1

Refer to redrawn Figure 1,

The expression for KVL in super mesh $BACEFDB$ is as follows.

$v1+i1R1+i2R2+(i2−i3)R3=0$ (4)

Here,

$v1$ is voltage in loop $BACDB$ ,

$R1$ , $R2$ and $R3$ are the resistances in the circuit,

$i1$ is the current flowing in loop $BACDB$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $2 V$ for $v1$ , $2 Ω$ for $R1$ , $5 Ω$ for $R2$ and $1 Ω$ for $R3$ in equation (4).

$2 V+i1(2 Ω)+i2(5 Ω)+(i2−i3)(1 Ω)=02 V+i1(2 Ω)+i2(5 Ω)+(1 Ω)i2−(1 Ω)i3=0$

$2 V+i1(2 Ω)+i2(6 Ω)−(1 Ω)i3=0$ (5)

The expression for KVL in mesh $FEGHF$ is as follows.

$(i3−i2)R3+i3R4+v2=0$ (6)

Here,

$R3$ and $R4$ are the resistances in the circuit,

$v2$ is voltage in loop $FEGHF$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $1 Ω$ for $R3$ , $3 Ω$ for $R4$ and $4 V$ for $v2$ in the equation (6).

$(i3−i2)(1 Ω)+i3(3 Ω)+4 V=0(1 Ω)i3−(1 Ω)i2+i3(3 Ω)+4 V=0$

$(4 Ω)i3−(1 Ω)i2+4 V=0$ (7)

The expression for current $i$ is as follows.

$i4=i2−i1$ (8)

Here,

$i4$ is current through independent current source,

$i2$ is the current flowing in loop $DCEFD$ and

$i1$ is the current flowing in loop $BACDB$ .

Substitute $2 A$ for $i4$ in equation (8).

$2 A=i2−i1$ (9)

Rearrange equations(5),(7) and (9).

$2i1+6i2−i3=−20i1−i2+4i3=−4−i1+i2+0i3=2$

The equations so formed can be written in matrix form as,

$(26−10−14−110)(i1i2i3)=(−2−42)$

Therefore, by Cramer’s rule,

The determinant of coefficient matrix is as follows.

$Δ=|26−10−14−110|=−31$

The 1^st determinant is as follows.

$Δ1=|−26−1−4−14210|=58$

The 2^nd determinant is as follows,

$Δ2=|2−2−10−44−120|=−4$

The 3rd determinant is as follows.

$Δ3=|26−20−1−4−112|=30$

Simplify for $i1$ .

$i1=Δ1Δ=58−31=−1.87 A$

Simplify for $i2$ .

$i2=Δ2Δ=−4−31=0.129 A=129 mA {∵ 1 A=103 mA}$

Simplify for $i3$ .

$i3=Δ3Δ=30−31=−0.9677 A$

Substitute $129 mA$ for $i$ and $5 Ω$ for $R$ in equation (1).

$v=(129 mA)×(5 Ω)=(129×10−3 A)×(5 Ω) {∵1 mA=10−3 A}=645×10−3 V =645 mV {∵10−3 V=1 mV}$

So, the Thevenin voltage is $645 mV$ .

The circuit diagram is redrawn as shown in Figure 2.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 2

As $R3$ and $R4$ are in parallel therefore from equation (3).

$1R5=(1R3+1R4)1R5=(11 Ω+13 Ω)1R5=10.75 Ω$

Rearrange for $R5$ ,

$R5=0.75 Ω$

The circuit diagram is redrawn as shown in Figure 3.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 3

As $R1$ and $R5$ are in series therefore from equation (2).

$R6=R1+R5=2 Ω+0.75 Ω=2.75 Ω$

The circuit diagram is redrawn as shown in Figure 4.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 4

Refer to redrawn Figure 4,

As $R2$ and $R6$ are in parallel therefore from equation (3),

$1Rth=(1R2+1R6)1Rth=(15 Ω+12.75 Ω)1Rth=11.774 Ω$

Rearrange for $Rth$ ,

$Rth=1.774 Ω$

So, the Thevenin equivalent resistance across the branch $XY$ is $1.774 Ω$ .

Thevenin equivalent resistance is same as the Norton equivalent resistance,

Hence,

$RN=1.774 Ω$

The circuit diagram is redrawn as shown in Figure 5,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 5

Refer to redrawn Figure 5,

Norton current is the current in the load resistor when load resistance is replaced by a short circuit.

The expression for the Norton current flowing in the circuit is as follows,

$IN=VthRth$ (10)

Here,

$IN$ is the Norton current,

$Vth$ is the Thevenin voltage and

$Rth$ is the Thevenin equivalent resistance.

Substitute $1.774 Ω$ for $Rth$ and $645 mV$ for $Vth$ in equation (10),

$IN=(645 mV)(1.774 Ω)=363.58 mA$

Conclusion:

Thus the Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

(b)

Expert Solution

To determine

Find the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor.

Answer to Problem 33E

The power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

Explanation of Solution

Given Data:

The load resistance is $5 Ω$ .

Formula used:

The expression for the power dissipated in a resistor is as follows,

$p=(iL)2RL$ (`11)

Here,

$p$ is the power dissipated in a resistor,

$RL$ is the load resistance and

$iL$ is the current flowing through the load.

Calculation:

The circuit is drawn as shown in Figure 6,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 6

Refer to redrawn Figure 6,

The simplified Norton equivalent of the circuit is drawn as shown in Figure 7,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 7

Refer to redrawn Figure 7,

The expression for the current flowing through the load is as follows,

$iL=IN(RNRN+RL)$ (12)

Substitute $363.58 mA$ for $IN$ , $1.774 Ω$ for $RN$ and $5 Ω$ for $RL$ in equation (12),

$iL=(363.58 mA)(1.774 Ω(1.774 Ω)+(5 Ω))=(363.58 mA)(0.26188)=95.22 mA$

Substitute $95.22 mA$ for $iL$ and $5 Ω$ for $RL$ in equation (11),

$p=(95.22 mA)2(5 Ω)=(95.22×10−3 A)2(5 Ω) {∵1 mA=10−3 A}=(9066.023×10−6)(5) W=45330.115×10−6W$

$p=45.33 mW {∵10−6 A=10−3 mA}$

Conclusion:

Thus, the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

(c)

Expert Solution

To determine

Find the current flowing through a short circuit connecting the two terminals.

Answer to Problem 33E

The current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

Explanation of Solution

Calculation:

The current flowing through a short circuit connecting the two terminals is Norton current which is $363.58 mA$ as calculated in sub-part (a).

So, the current flowing through a short circuit connecting the two terminals is $363.58 mA$

Conclusion:

Thus, the current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

Answer 8

(a)

Expert Solution

To determine

Find the Norton equivalent of the circuit as seen looking into the two open terminals.

Answer to Problem 33E

The Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

Explanation of Solution

Formula used:

The expression for voltage is as follows.

$v=iR$ (1)

Here,

$v$ is the voltage,

$i$ is the current and

$R$ is value of resistance.

The expression for series combination of resistance is as follows.

$R=(R1+R2)$ (2)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

The expression for parallel combination of resistance is as follows.

$1R=(1R1+1R2)$ (3)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

Calculation:

The circuit diagram is redrawn as shown in Figure 1.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 1

Refer to redrawn Figure 1,

The expression for KVL in super mesh $BACEFDB$ is as follows.

$v1+i1R1+i2R2+(i2−i3)R3=0$ (4)

Here,

$v1$ is voltage in loop $BACDB$ ,

$R1$ , $R2$ and $R3$ are the resistances in the circuit,

$i1$ is the current flowing in loop $BACDB$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $2 V$ for $v1$ , $2 Ω$ for $R1$ , $5 Ω$ for $R2$ and $1 Ω$ for $R3$ in equation (4).

$2 V+i1(2 Ω)+i2(5 Ω)+(i2−i3)(1 Ω)=02 V+i1(2 Ω)+i2(5 Ω)+(1 Ω)i2−(1 Ω)i3=0$

$2 V+i1(2 Ω)+i2(6 Ω)−(1 Ω)i3=0$ (5)

The expression for KVL in mesh $FEGHF$ is as follows.

$(i3−i2)R3+i3R4+v2=0$ (6)

Here,

$R3$ and $R4$ are the resistances in the circuit,

$v2$ is voltage in loop $FEGHF$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $1 Ω$ for $R3$ , $3 Ω$ for $R4$ and $4 V$ for $v2$ in the equation (6).

$(i3−i2)(1 Ω)+i3(3 Ω)+4 V=0(1 Ω)i3−(1 Ω)i2+i3(3 Ω)+4 V=0$

$(4 Ω)i3−(1 Ω)i2+4 V=0$ (7)

The expression for current $i$ is as follows.

$i4=i2−i1$ (8)

Here,

$i4$ is current through independent current source,

$i2$ is the current flowing in loop $DCEFD$ and

$i1$ is the current flowing in loop $BACDB$ .

Substitute $2 A$ for $i4$ in equation (8).

$2 A=i2−i1$ (9)

Rearrange equations(5),(7) and (9).

$2i1+6i2−i3=−20i1−i2+4i3=−4−i1+i2+0i3=2$

The equations so formed can be written in matrix form as,

$(26−10−14−110)(i1i2i3)=(−2−42)$

Therefore, by Cramer’s rule,

The determinant of coefficient matrix is as follows.

$Δ=|26−10−14−110|=−31$

The 1^st determinant is as follows.

$Δ1=|−26−1−4−14210|=58$

The 2^nd determinant is as follows,

$Δ2=|2−2−10−44−120|=−4$

The 3rd determinant is as follows.

$Δ3=|26−20−1−4−112|=30$

Simplify for $i1$ .

$i1=Δ1Δ=58−31=−1.87 A$

Simplify for $i2$ .

$i2=Δ2Δ=−4−31=0.129 A=129 mA {∵ 1 A=103 mA}$

Simplify for $i3$ .

$i3=Δ3Δ=30−31=−0.9677 A$

Substitute $129 mA$ for $i$ and $5 Ω$ for $R$ in equation (1).

$v=(129 mA)×(5 Ω)=(129×10−3 A)×(5 Ω) {∵1 mA=10−3 A}=645×10−3 V =645 mV {∵10−3 V=1 mV}$

So, the Thevenin voltage is $645 mV$ .

The circuit diagram is redrawn as shown in Figure 2.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 2

As $R3$ and $R4$ are in parallel therefore from equation (3).

$1R5=(1R3+1R4)1R5=(11 Ω+13 Ω)1R5=10.75 Ω$

Rearrange for $R5$ ,

$R5=0.75 Ω$

The circuit diagram is redrawn as shown in Figure 3.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 3

As $R1$ and $R5$ are in series therefore from equation (2).

$R6=R1+R5=2 Ω+0.75 Ω=2.75 Ω$

The circuit diagram is redrawn as shown in Figure 4.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 4

Refer to redrawn Figure 4,

As $R2$ and $R6$ are in parallel therefore from equation (3),

$1Rth=(1R2+1R6)1Rth=(15 Ω+12.75 Ω)1Rth=11.774 Ω$

Rearrange for $Rth$ ,

$Rth=1.774 Ω$

So, the Thevenin equivalent resistance across the branch $XY$ is $1.774 Ω$ .

Thevenin equivalent resistance is same as the Norton equivalent resistance,

Hence,

$RN=1.774 Ω$

The circuit diagram is redrawn as shown in Figure 5,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 5

Refer to redrawn Figure 5,

Norton current is the current in the load resistor when load resistance is replaced by a short circuit.

The expression for the Norton current flowing in the circuit is as follows,

$IN=VthRth$ (10)

Here,

$IN$ is the Norton current,

$Vth$ is the Thevenin voltage and

$Rth$ is the Thevenin equivalent resistance.

Substitute $1.774 Ω$ for $Rth$ and $645 mV$ for $Vth$ in equation (10),

$IN=(645 mV)(1.774 Ω)=363.58 mA$

Conclusion:

Thus the Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Find the Norton equivalent of the circuit as seen looking into the two open terminals.

Answer 13

Answer to Problem 33E

The Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

Answer 14

Explanation of Solution

Formula used:

The expression for voltage is as follows.

$v=iR$ (1)

Here,

$v$ is the voltage,

$i$ is the current and

$R$ is value of resistance.

The expression for series combination of resistance is as follows.

$R=(R1+R2)$ (2)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

The expression for parallel combination of resistance is as follows.

$1R=(1R1+1R2)$ (3)

Here,

$R$ is the total resistances and

$R1$ and $R2$ are the resistances in the circuit.

Calculation:

The circuit diagram is redrawn as shown in Figure 1.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 1

Refer to redrawn Figure 1,

The expression for KVL in super mesh $BACEFDB$ is as follows.

$v1+i1R1+i2R2+(i2−i3)R3=0$ (4)

Here,

$v1$ is voltage in loop $BACDB$ ,

$R1$ , $R2$ and $R3$ are the resistances in the circuit,

$i1$ is the current flowing in loop $BACDB$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $2 V$ for $v1$ , $2 Ω$ for $R1$ , $5 Ω$ for $R2$ and $1 Ω$ for $R3$ in equation (4).

$2 V+i1(2 Ω)+i2(5 Ω)+(i2−i3)(1 Ω)=02 V+i1(2 Ω)+i2(5 Ω)+(1 Ω)i2−(1 Ω)i3=0$

$2 V+i1(2 Ω)+i2(6 Ω)−(1 Ω)i3=0$ (5)

The expression for KVL in mesh $FEGHF$ is as follows.

$(i3−i2)R3+i3R4+v2=0$ (6)

Here,

$R3$ and $R4$ are the resistances in the circuit,

$v2$ is voltage in loop $FEGHF$ ,

$i2$ is the current flowing in loop $DCEFD$ and

$i3$ is the current flowing in loop $FEGHF$ .

Substitute $1 Ω$ for $R3$ , $3 Ω$ for $R4$ and $4 V$ for $v2$ in the equation (6).

$(i3−i2)(1 Ω)+i3(3 Ω)+4 V=0(1 Ω)i3−(1 Ω)i2+i3(3 Ω)+4 V=0$

$(4 Ω)i3−(1 Ω)i2+4 V=0$ (7)

The expression for current $i$ is as follows.

$i4=i2−i1$ (8)

Here,

$i4$ is current through independent current source,

$i2$ is the current flowing in loop $DCEFD$ and

$i1$ is the current flowing in loop $BACDB$ .

Substitute $2 A$ for $i4$ in equation (8).

$2 A=i2−i1$ (9)

Rearrange equations(5),(7) and (9).

$2i1+6i2−i3=−20i1−i2+4i3=−4−i1+i2+0i3=2$

The equations so formed can be written in matrix form as,

$(26−10−14−110)(i1i2i3)=(−2−42)$

Therefore, by Cramer’s rule,

The determinant of coefficient matrix is as follows.

$Δ=|26−10−14−110|=−31$

The 1^st determinant is as follows.

$Δ1=|−26−1−4−14210|=58$

The 2^nd determinant is as follows,

$Δ2=|2−2−10−44−120|=−4$

The 3rd determinant is as follows.

$Δ3=|26−20−1−4−112|=30$

Simplify for $i1$ .

$i1=Δ1Δ=58−31=−1.87 A$

Simplify for $i2$ .

$i2=Δ2Δ=−4−31=0.129 A=129 mA {∵ 1 A=103 mA}$

Simplify for $i3$ .

$i3=Δ3Δ=30−31=−0.9677 A$

Substitute $129 mA$ for $i$ and $5 Ω$ for $R$ in equation (1).

$v=(129 mA)×(5 Ω)=(129×10−3 A)×(5 Ω) {∵1 mA=10−3 A}=645×10−3 V =645 mV {∵10−3 V=1 mV}$

So, the Thevenin voltage is $645 mV$ .

The circuit diagram is redrawn as shown in Figure 2.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 2

As $R3$ and $R4$ are in parallel therefore from equation (3).

$1R5=(1R3+1R4)1R5=(11 Ω+13 Ω)1R5=10.75 Ω$

Rearrange for $R5$ ,

$R5=0.75 Ω$

The circuit diagram is redrawn as shown in Figure 3.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 3

As $R1$ and $R5$ are in series therefore from equation (2).

$R6=R1+R5=2 Ω+0.75 Ω=2.75 Ω$

The circuit diagram is redrawn as shown in Figure 4.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 4

Refer to redrawn Figure 4,

As $R2$ and $R6$ are in parallel therefore from equation (3),

$1Rth=(1R2+1R6)1Rth=(15 Ω+12.75 Ω)1Rth=11.774 Ω$

Rearrange for $Rth$ ,

$Rth=1.774 Ω$

So, the Thevenin equivalent resistance across the branch $XY$ is $1.774 Ω$ .

Thevenin equivalent resistance is same as the Norton equivalent resistance,

Hence,

$RN=1.774 Ω$

The circuit diagram is redrawn as shown in Figure 5,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 5

Refer to redrawn Figure 5,

Norton current is the current in the load resistor when load resistance is replaced by a short circuit.

The expression for the Norton current flowing in the circuit is as follows,

$IN=VthRth$ (10)

Here,

$IN$ is the Norton current,

$Vth$ is the Thevenin voltage and

$Rth$ is the Thevenin equivalent resistance.

Substitute $1.774 Ω$ for $Rth$ and $645 mV$ for $Vth$ in equation (10),

$IN=(645 mV)(1.774 Ω)=363.58 mA$

Conclusion:

Thus the Norton current is $363.58 mA$ and Norton resistance is $1.774 Ω$ .

Answer 15

(b)

Expert Solution

To determine

Find the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor.

Answer to Problem 33E

The power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

Explanation of Solution

Given Data:

The load resistance is $5 Ω$ .

Formula used:

The expression for the power dissipated in a resistor is as follows,

$p=(iL)2RL$ (`11)

Here,

$p$ is the power dissipated in a resistor,

$RL$ is the load resistance and

$iL$ is the current flowing through the load.

Calculation:

The circuit is drawn as shown in Figure 6,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 6

Refer to redrawn Figure 6,

The simplified Norton equivalent of the circuit is drawn as shown in Figure 7,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 7

Refer to redrawn Figure 7,

The expression for the current flowing through the load is as follows,

$iL=IN(RNRN+RL)$ (12)

Substitute $363.58 mA$ for $IN$ , $1.774 Ω$ for $RN$ and $5 Ω$ for $RL$ in equation (12),

$iL=(363.58 mA)(1.774 Ω(1.774 Ω)+(5 Ω))=(363.58 mA)(0.26188)=95.22 mA$

Substitute $95.22 mA$ for $iL$ and $5 Ω$ for $RL$ in equation (11),

$p=(95.22 mA)2(5 Ω)=(95.22×10−3 A)2(5 Ω) {∵1 mA=10−3 A}=(9066.023×10−6)(5) W=45330.115×10−6W$

$p=45.33 mW {∵10−6 A=10−3 mA}$

Conclusion:

Thus, the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Find the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor.

Answer 20

Answer to Problem 33E

The power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

Answer 21

Explanation of Solution

Given Data:

The load resistance is $5 Ω$ .

Formula used:

The expression for the power dissipated in a resistor is as follows,

$p=(iL)2RL$ (`11)

Here,

$p$ is the power dissipated in a resistor,

$RL$ is the load resistance and

$iL$ is the current flowing through the load.

Calculation:

The circuit is drawn as shown in Figure 6,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 6

Refer to redrawn Figure 6,

The simplified Norton equivalent of the circuit is drawn as shown in Figure 7,

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 5, Problem 33E , additional homework tip 7

Refer to redrawn Figure 7,

The expression for the current flowing through the load is as follows,

$iL=IN(RNRN+RL)$ (12)

Substitute $363.58 mA$ for $IN$ , $1.774 Ω$ for $RN$ and $5 Ω$ for $RL$ in equation (12),

$iL=(363.58 mA)(1.774 Ω(1.774 Ω)+(5 Ω))=(363.58 mA)(0.26188)=95.22 mA$

Substitute $95.22 mA$ for $iL$ and $5 Ω$ for $RL$ in equation (11),

$p=(95.22 mA)2(5 Ω)=(95.22×10−3 A)2(5 Ω) {∵1 mA=10−3 A}=(9066.023×10−6)(5) W=45330.115×10−6W$

$p=45.33 mW {∵10−6 A=10−3 mA}$

Conclusion:

Thus, the power dissipated in a $5 Ω$ resistor connected in parallel with the existing $5 Ω$ resistor is $45.33 mW$ .

Answer 22

(c)

Expert Solution

To determine

Find the current flowing through a short circuit connecting the two terminals.

Answer to Problem 33E

The current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

Explanation of Solution

Calculation:

The current flowing through a short circuit connecting the two terminals is Norton current which is $363.58 mA$ as calculated in sub-part (a).

So, the current flowing through a short circuit connecting the two terminals is $363.58 mA$

Conclusion:

Thus, the current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

Find the current flowing through a short circuit connecting the two terminals.

Answer 27

Answer to Problem 33E

The current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

Answer 28

Explanation of Solution

Calculation:

The current flowing through a short circuit connecting the two terminals is Norton current which is $363.58 mA$ as calculated in sub-part (a).

So, the current flowing through a short circuit connecting the two terminals is $363.58 mA$

Conclusion:

Thus, the current flowing through a short circuit connecting the two terminals is $363.58 mA$ .

Answer 29

Ch. 5.1 - For the circuit of Fig. 5.4, use superposition to...Ch. 5.2 - For the circuit of Fig. 5.7, use superposition to...Ch. 5.2 - For the circuit of Fig. 5.18, compute the current...Ch. 5.2 - For the circuit of Fig. 5.20, compute the voltage...Ch. 5.3 - Using repeated source transformations, determine...Ch. 5.3 - Use Thvenins theorem to find the current through...Ch. 5.3 - Determine the Thvenin and Norton equivalents of...Ch. 5.3 - Find the Thvenin equivalent for the network of...Ch. 5.3 - Find the Thvenin equivalent for the network of...Ch. 5.4 - Consider the circuit of Fig. 5.43. FIGURE 5.43...

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