Using supeerposition, determine the current through the inductance X L for the network of Fig. 19.105.

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Answer 1

Textbook Question

Chapter 19, Problem 1P

Using supeerposition, determine the current through the inductance X_L for the network of Fig. 19.105.

Chapter 19, Problem 1P, Using supeerposition, determine the current through the inductance XL for the network of Fig.

Expert Solution & Answer

To determine

The current flowing through the inductance $XL$ of the given circuit.

Answer to Problem 1P

The current through the inductor is $IL=6.1∠−32.11° A$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 1

Calculation:

To apply superposition theorem, first consider the effect of voltage source $E1$ and replace the source $E2$ by a short circuit.

The required diagram is shown in Figure 2.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 2

Let the voltage at node 1 is $V$ . Apply KCL at node 1.

$V−30∠30°3+Vj8+V−j6=0(13+1 j8+1 −j6)V=30∠30°3(0.333+j0.0417)=10∠30°V=29.77∠22.87° V$

The current through the inductor is given by

$IL1=Vj8$

Substitute $29.77∠22.87°$ for $V$ in the above equation.

$IL1=29.77∠22.87°j8=3.72∠−67.12° A$

Now consider the effect of source voltage $E2$ and replace the source voltage $E1$ by short circuit.

The required diagram is shown in Figure 3.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 3

Apply the KCL at node 2.

$V3+Vj8+V−60∠10°−j6=0(13+1 j8+1 −j6)V=60∠10°−j6(0.333+j0.0417)=10∠100°V=29.77∠92.87° V$

The current through the inductor is given by,

$IL2=29.77∠92.87°j8=3.72∠2.87° A$

Therefore, according to super position theorem the current through the inductor is given by

$IL=IL1+IL2$

Substitute $3.72∠−67.12° A$ for $IL1$ and $3.72∠2.87° A$ for $IL2$ in the above equation.

$IL=3.87∠−67.12°+3.87∠92.87°IL=6.1∠−32.11° A$

Conclusion:

Therefore, the current through the inductor is $I=6.1∠−32.11° A$ .

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Answer 2

Textbook Question

Chapter 19, Problem 1P

Using supeerposition, determine the current through the inductance X_L for the network of Fig. 19.105.

Chapter 19, Problem 1P, Using supeerposition, determine the current through the inductance XL for the network of Fig.

Expert Solution & Answer

To determine

The current flowing through the inductance $XL$ of the given circuit.

Answer to Problem 1P

The current through the inductor is $IL=6.1∠−32.11° A$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 1

Calculation:

To apply superposition theorem, first consider the effect of voltage source $E1$ and replace the source $E2$ by a short circuit.

The required diagram is shown in Figure 2.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 2

Let the voltage at node 1 is $V$ . Apply KCL at node 1.

$V−30∠30°3+Vj8+V−j6=0(13+1 j8+1 −j6)V=30∠30°3(0.333+j0.0417)=10∠30°V=29.77∠22.87° V$

The current through the inductor is given by

$IL1=Vj8$

Substitute $29.77∠22.87°$ for $V$ in the above equation.

$IL1=29.77∠22.87°j8=3.72∠−67.12° A$

Now consider the effect of source voltage $E2$ and replace the source voltage $E1$ by short circuit.

The required diagram is shown in Figure 3.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 3

Apply the KCL at node 2.

$V3+Vj8+V−60∠10°−j6=0(13+1 j8+1 −j6)V=60∠10°−j6(0.333+j0.0417)=10∠100°V=29.77∠92.87° V$

The current through the inductor is given by,

$IL2=29.77∠92.87°j8=3.72∠2.87° A$

Therefore, according to super position theorem the current through the inductor is given by

$IL=IL1+IL2$

Substitute $3.72∠−67.12° A$ for $IL1$ and $3.72∠2.87° A$ for $IL2$ in the above equation.

$IL=3.87∠−67.12°+3.87∠92.87°IL=6.1∠−32.11° A$

Conclusion:

Therefore, the current through the inductor is $I=6.1∠−32.11° A$ .

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Answer 3

Textbook Question

Chapter 19, Problem 1P

Using supeerposition, determine the current through the inductance X_L for the network of Fig. 19.105.

Chapter 19, Problem 1P, Using supeerposition, determine the current through the inductance XL for the network of Fig.

Expert Solution & Answer

To determine

The current flowing through the inductance $XL$ of the given circuit.

Answer to Problem 1P

The current through the inductor is $IL=6.1∠−32.11° A$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 1

Calculation:

To apply superposition theorem, first consider the effect of voltage source $E1$ and replace the source $E2$ by a short circuit.

The required diagram is shown in Figure 2.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 2

Let the voltage at node 1 is $V$ . Apply KCL at node 1.

$V−30∠30°3+Vj8+V−j6=0(13+1 j8+1 −j6)V=30∠30°3(0.333+j0.0417)=10∠30°V=29.77∠22.87° V$

The current through the inductor is given by

$IL1=Vj8$

Substitute $29.77∠22.87°$ for $V$ in the above equation.

$IL1=29.77∠22.87°j8=3.72∠−67.12° A$

Now consider the effect of source voltage $E2$ and replace the source voltage $E1$ by short circuit.

The required diagram is shown in Figure 3.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 3

Apply the KCL at node 2.

$V3+Vj8+V−60∠10°−j6=0(13+1 j8+1 −j6)V=60∠10°−j6(0.333+j0.0417)=10∠100°V=29.77∠92.87° V$

The current through the inductor is given by,

$IL2=29.77∠92.87°j8=3.72∠2.87° A$

Therefore, according to super position theorem the current through the inductor is given by

$IL=IL1+IL2$

Substitute $3.72∠−67.12° A$ for $IL1$ and $3.72∠2.87° A$ for $IL2$ in the above equation.

$IL=3.87∠−67.12°+3.87∠92.87°IL=6.1∠−32.11° A$

Conclusion:

Therefore, the current through the inductor is $I=6.1∠−32.11° A$ .

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Answer 4

Textbook Question

Chapter 19, Problem 1P

Using supeerposition, determine the current through the inductance X_L for the network of Fig. 19.105.

Chapter 19, Problem 1P, Using supeerposition, determine the current through the inductance XL for the network of Fig.

Expert Solution & Answer

To determine

The current flowing through the inductance $XL$ of the given circuit.

Answer to Problem 1P

The current through the inductor is $IL=6.1∠−32.11° A$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 1

Calculation:

To apply superposition theorem, first consider the effect of voltage source $E1$ and replace the source $E2$ by a short circuit.

The required diagram is shown in Figure 2.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 2

Let the voltage at node 1 is $V$ . Apply KCL at node 1.

$V−30∠30°3+Vj8+V−j6=0(13+1 j8+1 −j6)V=30∠30°3(0.333+j0.0417)=10∠30°V=29.77∠22.87° V$

The current through the inductor is given by

$IL1=Vj8$

Substitute $29.77∠22.87°$ for $V$ in the above equation.

$IL1=29.77∠22.87°j8=3.72∠−67.12° A$

Now consider the effect of source voltage $E2$ and replace the source voltage $E1$ by short circuit.

The required diagram is shown in Figure 3.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 3

Apply the KCL at node 2.

$V3+Vj8+V−60∠10°−j6=0(13+1 j8+1 −j6)V=60∠10°−j6(0.333+j0.0417)=10∠100°V=29.77∠92.87° V$

The current through the inductor is given by,

$IL2=29.77∠92.87°j8=3.72∠2.87° A$

Therefore, according to super position theorem the current through the inductor is given by

$IL=IL1+IL2$

Substitute $3.72∠−67.12° A$ for $IL1$ and $3.72∠2.87° A$ for $IL2$ in the above equation.

$IL=3.87∠−67.12°+3.87∠92.87°IL=6.1∠−32.11° A$

Conclusion:

Therefore, the current through the inductor is $I=6.1∠−32.11° A$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The current flowing through the inductance $XL$ of the given circuit.

Answer to Problem 1P

The current through the inductor is $IL=6.1∠−32.11° A$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 1

Calculation:

To apply superposition theorem, first consider the effect of voltage source $E1$ and replace the source $E2$ by a short circuit.

The required diagram is shown in Figure 2.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 2

Let the voltage at node 1 is $V$ . Apply KCL at node 1.

$V−30∠30°3+Vj8+V−j6=0(13+1 j8+1 −j6)V=30∠30°3(0.333+j0.0417)=10∠30°V=29.77∠22.87° V$

The current through the inductor is given by

$IL1=Vj8$

Substitute $29.77∠22.87°$ for $V$ in the above equation.

$IL1=29.77∠22.87°j8=3.72∠−67.12° A$

Now consider the effect of source voltage $E2$ and replace the source voltage $E1$ by short circuit.

The required diagram is shown in Figure 3.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 3

Apply the KCL at node 2.

$V3+Vj8+V−60∠10°−j6=0(13+1 j8+1 −j6)V=60∠10°−j6(0.333+j0.0417)=10∠100°V=29.77∠92.87° V$

The current through the inductor is given by,

$IL2=29.77∠92.87°j8=3.72∠2.87° A$

Therefore, according to super position theorem the current through the inductor is given by

$IL=IL1+IL2$

Substitute $3.72∠−67.12° A$ for $IL1$ and $3.72∠2.87° A$ for $IL2$ in the above equation.

$IL=3.87∠−67.12°+3.87∠92.87°IL=6.1∠−32.11° A$

Conclusion:

Therefore, the current through the inductor is $I=6.1∠−32.11° A$ .

Answer 8

Expert Solution & Answer

To determine

The current flowing through the inductance $XL$ of the given circuit.

Answer to Problem 1P

The current through the inductor is $IL=6.1∠−32.11° A$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 1

Calculation:

To apply superposition theorem, first consider the effect of voltage source $E1$ and replace the source $E2$ by a short circuit.

The required diagram is shown in Figure 2.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 2

Let the voltage at node 1 is $V$ . Apply KCL at node 1.

$V−30∠30°3+Vj8+V−j6=0(13+1 j8+1 −j6)V=30∠30°3(0.333+j0.0417)=10∠30°V=29.77∠22.87° V$

The current through the inductor is given by

$IL1=Vj8$

Substitute $29.77∠22.87°$ for $V$ in the above equation.

$IL1=29.77∠22.87°j8=3.72∠−67.12° A$

Now consider the effect of source voltage $E2$ and replace the source voltage $E1$ by short circuit.

The required diagram is shown in Figure 3.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 3

Apply the KCL at node 2.

$V3+Vj8+V−60∠10°−j6=0(13+1 j8+1 −j6)V=60∠10°−j6(0.333+j0.0417)=10∠100°V=29.77∠92.87° V$

The current through the inductor is given by,

$IL2=29.77∠92.87°j8=3.72∠2.87° A$

Therefore, according to super position theorem the current through the inductor is given by

$IL=IL1+IL2$

Substitute $3.72∠−67.12° A$ for $IL1$ and $3.72∠2.87° A$ for $IL2$ in the above equation.

$IL=3.87∠−67.12°+3.87∠92.87°IL=6.1∠−32.11° A$

Conclusion:

Therefore, the current through the inductor is $I=6.1∠−32.11° A$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The current flowing through the inductance $XL$ of the given circuit.

Answer 12

Answer to Problem 1P

The current through the inductor is $IL=6.1∠−32.11° A$ .

Answer 13

Explanation of Solution

Given:

The given circuit is shown in Figure 1.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 1

Calculation:

To apply superposition theorem, first consider the effect of voltage source $E1$ and replace the source $E2$ by a short circuit.

The required diagram is shown in Figure 2.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 2

Let the voltage at node 1 is $V$ . Apply KCL at node 1.

$V−30∠30°3+Vj8+V−j6=0(13+1 j8+1 −j6)V=30∠30°3(0.333+j0.0417)=10∠30°V=29.77∠22.87° V$

The current through the inductor is given by

$IL1=Vj8$

Substitute $29.77∠22.87°$ for $V$ in the above equation.

$IL1=29.77∠22.87°j8=3.72∠−67.12° A$

Now consider the effect of source voltage $E2$ and replace the source voltage $E1$ by short circuit.

The required diagram is shown in Figure 3.

Introductory Circuit Analysis (13th Edition), Chapter 19, Problem 1P , additional homework tip 3

Apply the KCL at node 2.

$V3+Vj8+V−60∠10°−j6=0(13+1 j8+1 −j6)V=60∠10°−j6(0.333+j0.0417)=10∠100°V=29.77∠92.87° V$

The current through the inductor is given by,

$IL2=29.77∠92.87°j8=3.72∠2.87° A$

Therefore, according to super position theorem the current through the inductor is given by

$IL=IL1+IL2$

Substitute $3.72∠−67.12° A$ for $IL1$ and $3.72∠2.87° A$ for $IL2$ in the above equation.

$IL=3.87∠−67.12°+3.87∠92.87°IL=6.1∠−32.11° A$

Conclusion:

Therefore, the current through the inductor is $I=6.1∠−32.11° A$ .

Answer 14

Using supeerposition, determine the current through the inductance X L for the network of Fig. 19.105.

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Answer to Problem 1P

Explanation of Solution

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