The voltage waveform in Fig. 6.46 is applied across a 55- μ F capacitor. Draw the current waveform through it. Figure 6.46 For Prob. 6.6.

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Answer 1

Textbook Question

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Chapter 6, Problem 6P

The voltage waveform in Fig. 6.46 is applied across a 55-μF capacitor. Draw the current waveform through it.

Chapter 6, Problem 6P, The voltage waveform in Fig. 6.46 is applied across a 55-F capacitor. Draw the current waveform

Figure 6.46

For Prob. 6.6.

Expert Solution & Answer

To determine

Find the current waveform.

Explanation of Solution

Given data:

The value of the capacitor $(C)$ is $55 μF$ .

Formula used:

Write the expression to calculate the straight line equation for two points $(x1,y1)$ and $(x2,y2)$ .

$(y−y1)=y2−y1x2−x1(x−x1)$ (1)

Refer to Figure 6.46 in the textbook.

From the given graph, substitute $t$ for $x$ and $v(t)$ for $y$ in equation (1).

$(v(t)−y1)=y2−y1x2−x1(t−x1)$ (2)

Write the expression to calculate the current through the inductor.

$i(t)=Cdv(t)dt$ (3)

Here,

$C$ is the value of the capacitor, and

$dv(t)dt$ is the rate of change of voltage with time.

Calculation:

The given voltage waveform is redrawn as Figure 1.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 1

Refer to Figure 1, split up the time period as five divisions such as $0 ms<t<2 ms$ , $2 ms<t<4 ms$ , $4 ms<t<8 ms$ , $8 ms<t<10 ms$ and $10 ms<t<12 ms$ to find the respective voltage value.

Case (i): $0 ms<t<2 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(0 ms,0 V)$ and $(2 ms,10 V)$ .

Substitute $0 ms$ for $x1$ , $0 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=10 V−0 V2 ms−0 ms(t−0 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=10 V2 ms(t)=102×10−3t V {∵1 m=10−3}=(5000 t) V$

Case (ii): $2 ms<t<4 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(4 ms,10 V)$ and $(2 ms,10 V)$ .

Substitute $4 ms$ for $x1$ , $10 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−10 V=10 V−10 V2 ms−4 ms(t−4 ms)v(t)−10 V=0 V−2 ms(t−4 ms)v(t)−10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=10 V$

Case (iii): $4 ms<t<8 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(4 ms,10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $4 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−(−10 V))=10 V−(−10 V)4 ms−8 ms(t−8 ms)(v(t)+10 V)=10 V+10 V−4 ms(t−8 ms)(v(t)+10 V)=−20 V4 ms(t−8 ms)$

Simplify the equation to find $v(t)$ .

$v(t)= (−20 V4 ms(t−8 ms)−10 V)=(−(204×10−3)(t−(8×10−3))−10) V {∵1 m=10−3}=(−5000 t+40−10) V=(−5000 t+30) V$

Case (iv): $8 ms<t<10 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(10 ms,−10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−(−10 V)=(−10 V)−(−10 V)10 ms−8 ms(t−8 ms)v(t)+10 V=−10 V+10 V2 ms(t−8 ms)v(t)+10 V=0 V2 ms(t−8 ms)v(t)+10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V$

Case (v): $10 ms<t<12 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(12 ms,0 V)$ and $(10 ms,−10 V)$ .

Substitute $12 ms$ for $x1$ , $0 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=−10 V−0 V10 ms−12 ms(t−12 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V−2 ms(t−12 ms)=(102×10−3(t−(12×10−3))) V {∵1 m=10−3}=(5000 t−60) V$

Therefore, the voltage function of the signal in Figure 1 is,

$v(t)={(5000 t) V, 0 ms<t<2 ms10 V, 2 ms<t<4 ms(−5000 t+30) V, 4 ms<t<8 ms−10 V, 8 ms<t<10 ms(5000 t−60) V, 10 ms<t<12 ms$

For $0 ms<t<2 ms$ :

Substitute $55 μF$ for $C$ and $5000 t V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t V)=(55×10−6)(5000)ddt(t) V⋅F {∵1 μ=10−6}=(55×10−6)(5000)(1) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

For $2 ms<t<4 ms$ :

Substitute $55 μF$ for $C$ and $10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(10 V)=(55×10−6)ddt(10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $4 ms<t<8 ms$ :

Substitute $55 μF$ for $C$ and $(−5000 t+30) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−5000 t+30) V=(55×10−6)ddt(−5000 t+30) V⋅F {∵1 μ=10−6}=(55×10−6)(−5000 (1)+0) V⋅Fs=−0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=−0.275×103×10−3 A=−275 mA {∵1 m=103}$

For $8 ms<t<10 ms$ :

Substitute $55 μF$ for $C$ and $−10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−10 V)=(55×10−6)ddt(−10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $10 ms<t<12 ms$ :

Substitute $55 μF$ for $C$ and $(5000 t−60) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t−60) V=(55×10−6)ddt(5000 t−60) V⋅F {∵1 μ=10−6}=(55×10−6)(5000 (1)−0) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

Therefore, the current function of the signal in Figure 1 is,

$i(t)={275 mA, 0 ms<t<2 ms0 A, 2 ms<t<4 ms−275 mA, 4 ms<t<8 ms0 A, 8 ms<t<10 ms275 mA, 10 ms<t<12 ms$

From the current expression $i(t)$ , the current waveform is drawn in Figure 2.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 2

Conclusion:

Thus, the current equation is found and its respective waveform is drawn.

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Answer 2

Textbook Question

100%

Chapter 6, Problem 6P

The voltage waveform in Fig. 6.46 is applied across a 55-μF capacitor. Draw the current waveform through it.

Chapter 6, Problem 6P, The voltage waveform in Fig. 6.46 is applied across a 55-F capacitor. Draw the current waveform

Figure 6.46

For Prob. 6.6.

Expert Solution & Answer

To determine

Find the current waveform.

Explanation of Solution

Given data:

The value of the capacitor $(C)$ is $55 μF$ .

Formula used:

Write the expression to calculate the straight line equation for two points $(x1,y1)$ and $(x2,y2)$ .

$(y−y1)=y2−y1x2−x1(x−x1)$ (1)

Refer to Figure 6.46 in the textbook.

From the given graph, substitute $t$ for $x$ and $v(t)$ for $y$ in equation (1).

$(v(t)−y1)=y2−y1x2−x1(t−x1)$ (2)

Write the expression to calculate the current through the inductor.

$i(t)=Cdv(t)dt$ (3)

Here,

$C$ is the value of the capacitor, and

$dv(t)dt$ is the rate of change of voltage with time.

Calculation:

The given voltage waveform is redrawn as Figure 1.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 1

Refer to Figure 1, split up the time period as five divisions such as $0 ms<t<2 ms$ , $2 ms<t<4 ms$ , $4 ms<t<8 ms$ , $8 ms<t<10 ms$ and $10 ms<t<12 ms$ to find the respective voltage value.

Case (i): $0 ms<t<2 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(0 ms,0 V)$ and $(2 ms,10 V)$ .

Substitute $0 ms$ for $x1$ , $0 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=10 V−0 V2 ms−0 ms(t−0 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=10 V2 ms(t)=102×10−3t V {∵1 m=10−3}=(5000 t) V$

Case (ii): $2 ms<t<4 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(4 ms,10 V)$ and $(2 ms,10 V)$ .

Substitute $4 ms$ for $x1$ , $10 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−10 V=10 V−10 V2 ms−4 ms(t−4 ms)v(t)−10 V=0 V−2 ms(t−4 ms)v(t)−10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=10 V$

Case (iii): $4 ms<t<8 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(4 ms,10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $4 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−(−10 V))=10 V−(−10 V)4 ms−8 ms(t−8 ms)(v(t)+10 V)=10 V+10 V−4 ms(t−8 ms)(v(t)+10 V)=−20 V4 ms(t−8 ms)$

Simplify the equation to find $v(t)$ .

$v(t)= (−20 V4 ms(t−8 ms)−10 V)=(−(204×10−3)(t−(8×10−3))−10) V {∵1 m=10−3}=(−5000 t+40−10) V=(−5000 t+30) V$

Case (iv): $8 ms<t<10 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(10 ms,−10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−(−10 V)=(−10 V)−(−10 V)10 ms−8 ms(t−8 ms)v(t)+10 V=−10 V+10 V2 ms(t−8 ms)v(t)+10 V=0 V2 ms(t−8 ms)v(t)+10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V$

Case (v): $10 ms<t<12 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(12 ms,0 V)$ and $(10 ms,−10 V)$ .

Substitute $12 ms$ for $x1$ , $0 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=−10 V−0 V10 ms−12 ms(t−12 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V−2 ms(t−12 ms)=(102×10−3(t−(12×10−3))) V {∵1 m=10−3}=(5000 t−60) V$

Therefore, the voltage function of the signal in Figure 1 is,

$v(t)={(5000 t) V, 0 ms<t<2 ms10 V, 2 ms<t<4 ms(−5000 t+30) V, 4 ms<t<8 ms−10 V, 8 ms<t<10 ms(5000 t−60) V, 10 ms<t<12 ms$

For $0 ms<t<2 ms$ :

Substitute $55 μF$ for $C$ and $5000 t V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t V)=(55×10−6)(5000)ddt(t) V⋅F {∵1 μ=10−6}=(55×10−6)(5000)(1) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

For $2 ms<t<4 ms$ :

Substitute $55 μF$ for $C$ and $10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(10 V)=(55×10−6)ddt(10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $4 ms<t<8 ms$ :

Substitute $55 μF$ for $C$ and $(−5000 t+30) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−5000 t+30) V=(55×10−6)ddt(−5000 t+30) V⋅F {∵1 μ=10−6}=(55×10−6)(−5000 (1)+0) V⋅Fs=−0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=−0.275×103×10−3 A=−275 mA {∵1 m=103}$

For $8 ms<t<10 ms$ :

Substitute $55 μF$ for $C$ and $−10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−10 V)=(55×10−6)ddt(−10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $10 ms<t<12 ms$ :

Substitute $55 μF$ for $C$ and $(5000 t−60) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t−60) V=(55×10−6)ddt(5000 t−60) V⋅F {∵1 μ=10−6}=(55×10−6)(5000 (1)−0) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

Therefore, the current function of the signal in Figure 1 is,

$i(t)={275 mA, 0 ms<t<2 ms0 A, 2 ms<t<4 ms−275 mA, 4 ms<t<8 ms0 A, 8 ms<t<10 ms275 mA, 10 ms<t<12 ms$

From the current expression $i(t)$ , the current waveform is drawn in Figure 2.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 2

Conclusion:

Thus, the current equation is found and its respective waveform is drawn.

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Answer 3

Textbook Question

100%

Chapter 6, Problem 6P

The voltage waveform in Fig. 6.46 is applied across a 55-μF capacitor. Draw the current waveform through it.

Chapter 6, Problem 6P, The voltage waveform in Fig. 6.46 is applied across a 55-F capacitor. Draw the current waveform

Figure 6.46

For Prob. 6.6.

Expert Solution & Answer

To determine

Find the current waveform.

Explanation of Solution

Given data:

The value of the capacitor $(C)$ is $55 μF$ .

Formula used:

Write the expression to calculate the straight line equation for two points $(x1,y1)$ and $(x2,y2)$ .

$(y−y1)=y2−y1x2−x1(x−x1)$ (1)

Refer to Figure 6.46 in the textbook.

From the given graph, substitute $t$ for $x$ and $v(t)$ for $y$ in equation (1).

$(v(t)−y1)=y2−y1x2−x1(t−x1)$ (2)

Write the expression to calculate the current through the inductor.

$i(t)=Cdv(t)dt$ (3)

Here,

$C$ is the value of the capacitor, and

$dv(t)dt$ is the rate of change of voltage with time.

Calculation:

The given voltage waveform is redrawn as Figure 1.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 1

Refer to Figure 1, split up the time period as five divisions such as $0 ms<t<2 ms$ , $2 ms<t<4 ms$ , $4 ms<t<8 ms$ , $8 ms<t<10 ms$ and $10 ms<t<12 ms$ to find the respective voltage value.

Case (i): $0 ms<t<2 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(0 ms,0 V)$ and $(2 ms,10 V)$ .

Substitute $0 ms$ for $x1$ , $0 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=10 V−0 V2 ms−0 ms(t−0 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=10 V2 ms(t)=102×10−3t V {∵1 m=10−3}=(5000 t) V$

Case (ii): $2 ms<t<4 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(4 ms,10 V)$ and $(2 ms,10 V)$ .

Substitute $4 ms$ for $x1$ , $10 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−10 V=10 V−10 V2 ms−4 ms(t−4 ms)v(t)−10 V=0 V−2 ms(t−4 ms)v(t)−10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=10 V$

Case (iii): $4 ms<t<8 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(4 ms,10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $4 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−(−10 V))=10 V−(−10 V)4 ms−8 ms(t−8 ms)(v(t)+10 V)=10 V+10 V−4 ms(t−8 ms)(v(t)+10 V)=−20 V4 ms(t−8 ms)$

Simplify the equation to find $v(t)$ .

$v(t)= (−20 V4 ms(t−8 ms)−10 V)=(−(204×10−3)(t−(8×10−3))−10) V {∵1 m=10−3}=(−5000 t+40−10) V=(−5000 t+30) V$

Case (iv): $8 ms<t<10 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(10 ms,−10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−(−10 V)=(−10 V)−(−10 V)10 ms−8 ms(t−8 ms)v(t)+10 V=−10 V+10 V2 ms(t−8 ms)v(t)+10 V=0 V2 ms(t−8 ms)v(t)+10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V$

Case (v): $10 ms<t<12 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(12 ms,0 V)$ and $(10 ms,−10 V)$ .

Substitute $12 ms$ for $x1$ , $0 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=−10 V−0 V10 ms−12 ms(t−12 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V−2 ms(t−12 ms)=(102×10−3(t−(12×10−3))) V {∵1 m=10−3}=(5000 t−60) V$

Therefore, the voltage function of the signal in Figure 1 is,

$v(t)={(5000 t) V, 0 ms<t<2 ms10 V, 2 ms<t<4 ms(−5000 t+30) V, 4 ms<t<8 ms−10 V, 8 ms<t<10 ms(5000 t−60) V, 10 ms<t<12 ms$

For $0 ms<t<2 ms$ :

Substitute $55 μF$ for $C$ and $5000 t V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t V)=(55×10−6)(5000)ddt(t) V⋅F {∵1 μ=10−6}=(55×10−6)(5000)(1) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

For $2 ms<t<4 ms$ :

Substitute $55 μF$ for $C$ and $10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(10 V)=(55×10−6)ddt(10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $4 ms<t<8 ms$ :

Substitute $55 μF$ for $C$ and $(−5000 t+30) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−5000 t+30) V=(55×10−6)ddt(−5000 t+30) V⋅F {∵1 μ=10−6}=(55×10−6)(−5000 (1)+0) V⋅Fs=−0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=−0.275×103×10−3 A=−275 mA {∵1 m=103}$

For $8 ms<t<10 ms$ :

Substitute $55 μF$ for $C$ and $−10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−10 V)=(55×10−6)ddt(−10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $10 ms<t<12 ms$ :

Substitute $55 μF$ for $C$ and $(5000 t−60) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t−60) V=(55×10−6)ddt(5000 t−60) V⋅F {∵1 μ=10−6}=(55×10−6)(5000 (1)−0) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

Therefore, the current function of the signal in Figure 1 is,

$i(t)={275 mA, 0 ms<t<2 ms0 A, 2 ms<t<4 ms−275 mA, 4 ms<t<8 ms0 A, 8 ms<t<10 ms275 mA, 10 ms<t<12 ms$

From the current expression $i(t)$ , the current waveform is drawn in Figure 2.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 2

Conclusion:

Thus, the current equation is found and its respective waveform is drawn.

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Answer 4

Textbook Question

100%

Chapter 6, Problem 6P

The voltage waveform in Fig. 6.46 is applied across a 55-μF capacitor. Draw the current waveform through it.

Chapter 6, Problem 6P, The voltage waveform in Fig. 6.46 is applied across a 55-F capacitor. Draw the current waveform

Figure 6.46

For Prob. 6.6.

Expert Solution & Answer

To determine

Find the current waveform.

Explanation of Solution

Given data:

The value of the capacitor $(C)$ is $55 μF$ .

Formula used:

Write the expression to calculate the straight line equation for two points $(x1,y1)$ and $(x2,y2)$ .

$(y−y1)=y2−y1x2−x1(x−x1)$ (1)

Refer to Figure 6.46 in the textbook.

From the given graph, substitute $t$ for $x$ and $v(t)$ for $y$ in equation (1).

$(v(t)−y1)=y2−y1x2−x1(t−x1)$ (2)

Write the expression to calculate the current through the inductor.

$i(t)=Cdv(t)dt$ (3)

Here,

$C$ is the value of the capacitor, and

$dv(t)dt$ is the rate of change of voltage with time.

Calculation:

The given voltage waveform is redrawn as Figure 1.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 1

Refer to Figure 1, split up the time period as five divisions such as $0 ms<t<2 ms$ , $2 ms<t<4 ms$ , $4 ms<t<8 ms$ , $8 ms<t<10 ms$ and $10 ms<t<12 ms$ to find the respective voltage value.

Case (i): $0 ms<t<2 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(0 ms,0 V)$ and $(2 ms,10 V)$ .

Substitute $0 ms$ for $x1$ , $0 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=10 V−0 V2 ms−0 ms(t−0 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=10 V2 ms(t)=102×10−3t V {∵1 m=10−3}=(5000 t) V$

Case (ii): $2 ms<t<4 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(4 ms,10 V)$ and $(2 ms,10 V)$ .

Substitute $4 ms$ for $x1$ , $10 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−10 V=10 V−10 V2 ms−4 ms(t−4 ms)v(t)−10 V=0 V−2 ms(t−4 ms)v(t)−10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=10 V$

Case (iii): $4 ms<t<8 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(4 ms,10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $4 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−(−10 V))=10 V−(−10 V)4 ms−8 ms(t−8 ms)(v(t)+10 V)=10 V+10 V−4 ms(t−8 ms)(v(t)+10 V)=−20 V4 ms(t−8 ms)$

Simplify the equation to find $v(t)$ .

$v(t)= (−20 V4 ms(t−8 ms)−10 V)=(−(204×10−3)(t−(8×10−3))−10) V {∵1 m=10−3}=(−5000 t+40−10) V=(−5000 t+30) V$

Case (iv): $8 ms<t<10 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(10 ms,−10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−(−10 V)=(−10 V)−(−10 V)10 ms−8 ms(t−8 ms)v(t)+10 V=−10 V+10 V2 ms(t−8 ms)v(t)+10 V=0 V2 ms(t−8 ms)v(t)+10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V$

Case (v): $10 ms<t<12 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(12 ms,0 V)$ and $(10 ms,−10 V)$ .

Substitute $12 ms$ for $x1$ , $0 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=−10 V−0 V10 ms−12 ms(t−12 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V−2 ms(t−12 ms)=(102×10−3(t−(12×10−3))) V {∵1 m=10−3}=(5000 t−60) V$

Therefore, the voltage function of the signal in Figure 1 is,

$v(t)={(5000 t) V, 0 ms<t<2 ms10 V, 2 ms<t<4 ms(−5000 t+30) V, 4 ms<t<8 ms−10 V, 8 ms<t<10 ms(5000 t−60) V, 10 ms<t<12 ms$

For $0 ms<t<2 ms$ :

Substitute $55 μF$ for $C$ and $5000 t V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t V)=(55×10−6)(5000)ddt(t) V⋅F {∵1 μ=10−6}=(55×10−6)(5000)(1) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

For $2 ms<t<4 ms$ :

Substitute $55 μF$ for $C$ and $10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(10 V)=(55×10−6)ddt(10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $4 ms<t<8 ms$ :

Substitute $55 μF$ for $C$ and $(−5000 t+30) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−5000 t+30) V=(55×10−6)ddt(−5000 t+30) V⋅F {∵1 μ=10−6}=(55×10−6)(−5000 (1)+0) V⋅Fs=−0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=−0.275×103×10−3 A=−275 mA {∵1 m=103}$

For $8 ms<t<10 ms$ :

Substitute $55 μF$ for $C$ and $−10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−10 V)=(55×10−6)ddt(−10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $10 ms<t<12 ms$ :

Substitute $55 μF$ for $C$ and $(5000 t−60) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t−60) V=(55×10−6)ddt(5000 t−60) V⋅F {∵1 μ=10−6}=(55×10−6)(5000 (1)−0) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

Therefore, the current function of the signal in Figure 1 is,

$i(t)={275 mA, 0 ms<t<2 ms0 A, 2 ms<t<4 ms−275 mA, 4 ms<t<8 ms0 A, 8 ms<t<10 ms275 mA, 10 ms<t<12 ms$

From the current expression $i(t)$ , the current waveform is drawn in Figure 2.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 2

Conclusion:

Thus, the current equation is found and its respective waveform is drawn.

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Answer 5

Textbook Question

100%

Answer 6

Textbook Question

100%

Answer 7

Expert Solution & Answer

To determine

Find the current waveform.

Explanation of Solution

Given data:

The value of the capacitor $(C)$ is $55 μF$ .

Formula used:

Write the expression to calculate the straight line equation for two points $(x1,y1)$ and $(x2,y2)$ .

$(y−y1)=y2−y1x2−x1(x−x1)$ (1)

Refer to Figure 6.46 in the textbook.

From the given graph, substitute $t$ for $x$ and $v(t)$ for $y$ in equation (1).

$(v(t)−y1)=y2−y1x2−x1(t−x1)$ (2)

Write the expression to calculate the current through the inductor.

$i(t)=Cdv(t)dt$ (3)

Here,

$C$ is the value of the capacitor, and

$dv(t)dt$ is the rate of change of voltage with time.

Calculation:

The given voltage waveform is redrawn as Figure 1.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 1

Refer to Figure 1, split up the time period as five divisions such as $0 ms<t<2 ms$ , $2 ms<t<4 ms$ , $4 ms<t<8 ms$ , $8 ms<t<10 ms$ and $10 ms<t<12 ms$ to find the respective voltage value.

Case (i): $0 ms<t<2 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(0 ms,0 V)$ and $(2 ms,10 V)$ .

Substitute $0 ms$ for $x1$ , $0 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=10 V−0 V2 ms−0 ms(t−0 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=10 V2 ms(t)=102×10−3t V {∵1 m=10−3}=(5000 t) V$

Case (ii): $2 ms<t<4 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(4 ms,10 V)$ and $(2 ms,10 V)$ .

Substitute $4 ms$ for $x1$ , $10 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−10 V=10 V−10 V2 ms−4 ms(t−4 ms)v(t)−10 V=0 V−2 ms(t−4 ms)v(t)−10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=10 V$

Case (iii): $4 ms<t<8 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(4 ms,10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $4 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−(−10 V))=10 V−(−10 V)4 ms−8 ms(t−8 ms)(v(t)+10 V)=10 V+10 V−4 ms(t−8 ms)(v(t)+10 V)=−20 V4 ms(t−8 ms)$

Simplify the equation to find $v(t)$ .

$v(t)= (−20 V4 ms(t−8 ms)−10 V)=(−(204×10−3)(t−(8×10−3))−10) V {∵1 m=10−3}=(−5000 t+40−10) V=(−5000 t+30) V$

Case (iv): $8 ms<t<10 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(10 ms,−10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−(−10 V)=(−10 V)−(−10 V)10 ms−8 ms(t−8 ms)v(t)+10 V=−10 V+10 V2 ms(t−8 ms)v(t)+10 V=0 V2 ms(t−8 ms)v(t)+10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V$

Case (v): $10 ms<t<12 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(12 ms,0 V)$ and $(10 ms,−10 V)$ .

Substitute $12 ms$ for $x1$ , $0 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=−10 V−0 V10 ms−12 ms(t−12 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V−2 ms(t−12 ms)=(102×10−3(t−(12×10−3))) V {∵1 m=10−3}=(5000 t−60) V$

Therefore, the voltage function of the signal in Figure 1 is,

$v(t)={(5000 t) V, 0 ms<t<2 ms10 V, 2 ms<t<4 ms(−5000 t+30) V, 4 ms<t<8 ms−10 V, 8 ms<t<10 ms(5000 t−60) V, 10 ms<t<12 ms$

For $0 ms<t<2 ms$ :

Substitute $55 μF$ for $C$ and $5000 t V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t V)=(55×10−6)(5000)ddt(t) V⋅F {∵1 μ=10−6}=(55×10−6)(5000)(1) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

For $2 ms<t<4 ms$ :

Substitute $55 μF$ for $C$ and $10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(10 V)=(55×10−6)ddt(10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $4 ms<t<8 ms$ :

Substitute $55 μF$ for $C$ and $(−5000 t+30) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−5000 t+30) V=(55×10−6)ddt(−5000 t+30) V⋅F {∵1 μ=10−6}=(55×10−6)(−5000 (1)+0) V⋅Fs=−0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=−0.275×103×10−3 A=−275 mA {∵1 m=103}$

For $8 ms<t<10 ms$ :

Substitute $55 μF$ for $C$ and $−10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−10 V)=(55×10−6)ddt(−10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $10 ms<t<12 ms$ :

Substitute $55 μF$ for $C$ and $(5000 t−60) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t−60) V=(55×10−6)ddt(5000 t−60) V⋅F {∵1 μ=10−6}=(55×10−6)(5000 (1)−0) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

Therefore, the current function of the signal in Figure 1 is,

$i(t)={275 mA, 0 ms<t<2 ms0 A, 2 ms<t<4 ms−275 mA, 4 ms<t<8 ms0 A, 8 ms<t<10 ms275 mA, 10 ms<t<12 ms$

From the current expression $i(t)$ , the current waveform is drawn in Figure 2.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 2

Conclusion:

Thus, the current equation is found and its respective waveform is drawn.

Answer 8

Expert Solution & Answer

To determine

Find the current waveform.

Explanation of Solution

Given data:

The value of the capacitor $(C)$ is $55 μF$ .

Formula used:

Write the expression to calculate the straight line equation for two points $(x1,y1)$ and $(x2,y2)$ .

$(y−y1)=y2−y1x2−x1(x−x1)$ (1)

Refer to Figure 6.46 in the textbook.

From the given graph, substitute $t$ for $x$ and $v(t)$ for $y$ in equation (1).

$(v(t)−y1)=y2−y1x2−x1(t−x1)$ (2)

Write the expression to calculate the current through the inductor.

$i(t)=Cdv(t)dt$ (3)

Here,

$C$ is the value of the capacitor, and

$dv(t)dt$ is the rate of change of voltage with time.

Calculation:

The given voltage waveform is redrawn as Figure 1.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 1

Refer to Figure 1, split up the time period as five divisions such as $0 ms<t<2 ms$ , $2 ms<t<4 ms$ , $4 ms<t<8 ms$ , $8 ms<t<10 ms$ and $10 ms<t<12 ms$ to find the respective voltage value.

Case (i): $0 ms<t<2 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(0 ms,0 V)$ and $(2 ms,10 V)$ .

Substitute $0 ms$ for $x1$ , $0 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=10 V−0 V2 ms−0 ms(t−0 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=10 V2 ms(t)=102×10−3t V {∵1 m=10−3}=(5000 t) V$

Case (ii): $2 ms<t<4 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(4 ms,10 V)$ and $(2 ms,10 V)$ .

Substitute $4 ms$ for $x1$ , $10 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−10 V=10 V−10 V2 ms−4 ms(t−4 ms)v(t)−10 V=0 V−2 ms(t−4 ms)v(t)−10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=10 V$

Case (iii): $4 ms<t<8 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(4 ms,10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $4 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−(−10 V))=10 V−(−10 V)4 ms−8 ms(t−8 ms)(v(t)+10 V)=10 V+10 V−4 ms(t−8 ms)(v(t)+10 V)=−20 V4 ms(t−8 ms)$

Simplify the equation to find $v(t)$ .

$v(t)= (−20 V4 ms(t−8 ms)−10 V)=(−(204×10−3)(t−(8×10−3))−10) V {∵1 m=10−3}=(−5000 t+40−10) V=(−5000 t+30) V$

Case (iv): $8 ms<t<10 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(10 ms,−10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−(−10 V)=(−10 V)−(−10 V)10 ms−8 ms(t−8 ms)v(t)+10 V=−10 V+10 V2 ms(t−8 ms)v(t)+10 V=0 V2 ms(t−8 ms)v(t)+10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V$

Case (v): $10 ms<t<12 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(12 ms,0 V)$ and $(10 ms,−10 V)$ .

Substitute $12 ms$ for $x1$ , $0 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=−10 V−0 V10 ms−12 ms(t−12 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V−2 ms(t−12 ms)=(102×10−3(t−(12×10−3))) V {∵1 m=10−3}=(5000 t−60) V$

Therefore, the voltage function of the signal in Figure 1 is,

$v(t)={(5000 t) V, 0 ms<t<2 ms10 V, 2 ms<t<4 ms(−5000 t+30) V, 4 ms<t<8 ms−10 V, 8 ms<t<10 ms(5000 t−60) V, 10 ms<t<12 ms$

For $0 ms<t<2 ms$ :

Substitute $55 μF$ for $C$ and $5000 t V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t V)=(55×10−6)(5000)ddt(t) V⋅F {∵1 μ=10−6}=(55×10−6)(5000)(1) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

For $2 ms<t<4 ms$ :

Substitute $55 μF$ for $C$ and $10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(10 V)=(55×10−6)ddt(10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $4 ms<t<8 ms$ :

Substitute $55 μF$ for $C$ and $(−5000 t+30) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−5000 t+30) V=(55×10−6)ddt(−5000 t+30) V⋅F {∵1 μ=10−6}=(55×10−6)(−5000 (1)+0) V⋅Fs=−0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=−0.275×103×10−3 A=−275 mA {∵1 m=103}$

For $8 ms<t<10 ms$ :

Substitute $55 μF$ for $C$ and $−10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−10 V)=(55×10−6)ddt(−10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $10 ms<t<12 ms$ :

Substitute $55 μF$ for $C$ and $(5000 t−60) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t−60) V=(55×10−6)ddt(5000 t−60) V⋅F {∵1 μ=10−6}=(55×10−6)(5000 (1)−0) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

Therefore, the current function of the signal in Figure 1 is,

$i(t)={275 mA, 0 ms<t<2 ms0 A, 2 ms<t<4 ms−275 mA, 4 ms<t<8 ms0 A, 8 ms<t<10 ms275 mA, 10 ms<t<12 ms$

From the current expression $i(t)$ , the current waveform is drawn in Figure 2.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 2

Conclusion:

Thus, the current equation is found and its respective waveform is drawn.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Find the current waveform.

Answer 12

Explanation of Solution

Given data:

The value of the capacitor $(C)$ is $55 μF$ .

Formula used:

Write the expression to calculate the straight line equation for two points $(x1,y1)$ and $(x2,y2)$ .

$(y−y1)=y2−y1x2−x1(x−x1)$ (1)

Refer to Figure 6.46 in the textbook.

From the given graph, substitute $t$ for $x$ and $v(t)$ for $y$ in equation (1).

$(v(t)−y1)=y2−y1x2−x1(t−x1)$ (2)

Write the expression to calculate the current through the inductor.

$i(t)=Cdv(t)dt$ (3)

Here,

$C$ is the value of the capacitor, and

$dv(t)dt$ is the rate of change of voltage with time.

Calculation:

The given voltage waveform is redrawn as Figure 1.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 1

Refer to Figure 1, split up the time period as five divisions such as $0 ms<t<2 ms$ , $2 ms<t<4 ms$ , $4 ms<t<8 ms$ , $8 ms<t<10 ms$ and $10 ms<t<12 ms$ to find the respective voltage value.

Case (i): $0 ms<t<2 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(0 ms,0 V)$ and $(2 ms,10 V)$ .

Substitute $0 ms$ for $x1$ , $0 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=10 V−0 V2 ms−0 ms(t−0 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=10 V2 ms(t)=102×10−3t V {∵1 m=10−3}=(5000 t) V$

Case (ii): $2 ms<t<4 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(4 ms,10 V)$ and $(2 ms,10 V)$ .

Substitute $4 ms$ for $x1$ , $10 V$ for $y1$ , $2 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−10 V=10 V−10 V2 ms−4 ms(t−4 ms)v(t)−10 V=0 V−2 ms(t−4 ms)v(t)−10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=10 V$

Case (iii): $4 ms<t<8 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(4 ms,10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $4 ms$ for $x2$ and $10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−(−10 V))=10 V−(−10 V)4 ms−8 ms(t−8 ms)(v(t)+10 V)=10 V+10 V−4 ms(t−8 ms)(v(t)+10 V)=−20 V4 ms(t−8 ms)$

Simplify the equation to find $v(t)$ .

$v(t)= (−20 V4 ms(t−8 ms)−10 V)=(−(204×10−3)(t−(8×10−3))−10) V {∵1 m=10−3}=(−5000 t+40−10) V=(−5000 t+30) V$

Case (iv): $8 ms<t<10 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(8 ms,−10 V)$ and $(10 ms,−10 V)$ .

Substitute $8 ms$ for $x1$ , $−10 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$v(t)−(−10 V)=(−10 V)−(−10 V)10 ms−8 ms(t−8 ms)v(t)+10 V=−10 V+10 V2 ms(t−8 ms)v(t)+10 V=0 V2 ms(t−8 ms)v(t)+10 V=0$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V$

Case (v): $10 ms<t<12 ms$

The two points $(x1,y1)$ and $(x2,y2)$ are $(12 ms,0 V)$ and $(10 ms,−10 V)$ .

Substitute $12 ms$ for $x1$ , $0 V$ for $y1$ , $10 ms$ for $x2$ and $−10 V$ for $y2$ in equation (2) to find $v(t)$ .

$(v(t)−0 V)=−10 V−0 V10 ms−12 ms(t−12 ms)$

Simplify the equation to find $v(t)$ .

$v(t)=−10 V−2 ms(t−12 ms)=(102×10−3(t−(12×10−3))) V {∵1 m=10−3}=(5000 t−60) V$

Therefore, the voltage function of the signal in Figure 1 is,

$v(t)={(5000 t) V, 0 ms<t<2 ms10 V, 2 ms<t<4 ms(−5000 t+30) V, 4 ms<t<8 ms−10 V, 8 ms<t<10 ms(5000 t−60) V, 10 ms<t<12 ms$

For $0 ms<t<2 ms$ :

Substitute $55 μF$ for $C$ and $5000 t V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t V)=(55×10−6)(5000)ddt(t) V⋅F {∵1 μ=10−6}=(55×10−6)(5000)(1) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

For $2 ms<t<4 ms$ :

Substitute $55 μF$ for $C$ and $10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(10 V)=(55×10−6)ddt(10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $4 ms<t<8 ms$ :

Substitute $55 μF$ for $C$ and $(−5000 t+30) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−5000 t+30) V=(55×10−6)ddt(−5000 t+30) V⋅F {∵1 μ=10−6}=(55×10−6)(−5000 (1)+0) V⋅Fs=−0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=−0.275×103×10−3 A=−275 mA {∵1 m=103}$

For $8 ms<t<10 ms$ :

Substitute $55 μF$ for $C$ and $−10 V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(−10 V)=(55×10−6)ddt(−10) V⋅F {∵1 μ=10−6}=(55×10−6)(0) V⋅Fs=0 A {∵1 A=1V⋅1F1s}$

For $10 ms<t<12 ms$ :

Substitute $55 μF$ for $C$ and $(5000 t−60) V$ for $v(t)$ in equation (3) to find $i(t)$ .

$i(t)=(55 μF)ddt(5000 t−60) V=(55×10−6)ddt(5000 t−60) V⋅F {∵1 μ=10−6}=(55×10−6)(5000 (1)−0) V⋅Fs=0.275 A {∵1 A=1V⋅1F1s}$

Simplify the equation to find $i(t)$ .

$i(t)=0.275×103×10−3 A=275 mA {∵1 m=103}$

Therefore, the current function of the signal in Figure 1 is,

$i(t)={275 mA, 0 ms<t<2 ms0 A, 2 ms<t<4 ms−275 mA, 4 ms<t<8 ms0 A, 8 ms<t<10 ms275 mA, 10 ms<t<12 ms$

From the current expression $i(t)$ , the current waveform is drawn in Figure 2.

Fundamentals of Electric Circuits, Chapter 6, Problem 6P , additional homework tip 2

Conclusion:

Thus, the current equation is found and its respective waveform is drawn.

Answer 13

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The voltage waveform in Fig. 6.46 is applied across a 55- μ F capacitor. Draw the current waveform through it. Figure 6.46 For Prob. 6.6.

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