Chapter 10, Problem 33E

Assuming the passive sign convention and an operating frequency of 314 rad/s, calculate the phasor voltage V which appears across each of the following when driven by the phasor current : (a) a 2 Ω resistor; (b) a 1 F capacitor; (c) a 1 H inductor; (d) a 2 Ω resistor in series with a 1 F capacitor; (e) a 2 Ω resistor in series with a 1 H inductor. (f) Calculate the instantaneous value of each voltage determined in parts (a) to (e) at t = 0.

To determine

Find the phasor voltage across $2\Omega$ resistor.

Answer to Problem 33E

The phasor voltage across $2\Omega$ resistor is $\underset{\_}{0.02\angle 0°\text{V}}$.

Explanation of Solution

Given data:

$I=10\angle 0°\text{mA}$

$\omega =314\text{rad}/\text{s}$

Formula used:

Consider the expression of phasor voltage across resistor.

$V_R=RI$        (1)

Here,

$R$ is resistance.

$I$ is the phasor current.

Calculation:

Substitute $2\Omega$ for $R$ and $10\angle 0°\text{mA}$ for $I$ in equation (1) as follows.

$\begin{array}{c}{V}_R=\left(2\Omega \right)\left(10\angle 0°\text{mA}\right)\\ =0.02\angle 0°\text{V}\end{array}$

Conclusion:

Thus, the phasor voltage across $2\Omega$ resistor is $\underset{\_}{0.02\angle 0°\text{V}}$.

To determine

Find the phasor voltage across $1\text{F}$ capacitor.

Answer to Problem 33E

The phasor voltage across $1\text{F}$ capacitor is $\underset{\_}{3.18\times {10}^{-5}\angle -90°\text{V}}$.

Explanation of Solution

Formula used:

Consider the expression of phasor voltage across capacitor.

$V_C=\frac{I}{j\omega C}$        (2)

Here,

$C$ is Capacitor.

$I$ is the phasor current.

Calculation:

Substitute $1\text{F}$ for $C$, $314\text{rad}/\text{s}$ for $\omega$, and $10\angle 0°\text{mA}$ for $I$ in equation (2) as follows.

$\begin{array}{c}{V}_C=\frac{10\angle 0°\text{mA}}{j\left(314\text{rad}/\text{s}\right)\left(1\text{F}\right)}\\ =\frac{10\times {10}^{-3}\angle 0°}{j314}\text{V}\\ =\frac{0.01\angle 0°}{314\angle 90°}\text{V}\\ =3.18\times {10}^{-5}\angle -90°\text{V}\end{array}$

Conclusion:

Thus, the phasor voltage across $1\text{F}$ capacitor is $\underset{\_}{3.18\times {10}^{-5}\angle -90°\text{V}}$.

To determine

Find the phasor voltage that developed across $1\text{H}$ inductor.

Answer to Problem 33E

The phasor voltage that developed across $1\text{H}$ inductor is $\underset{\_}{3.14\angle 90°\text{V}}$.

Explanation of Solution

Formula used:

Consider the expression of phasor voltage across inductor.

$V_L=j\omega LI$        (3)

Here,

$L$ is Inductor.

$I$ is the phasor current.

Calculation:

Substitute $1\text{H}$ for $L$, $314\text{rad}/\text{s}$ for $\omega$, and $10\angle 0°\text{mA}$ for $I$ in equation (3) as follows.

$\begin{array}{c}{V}_L=j\left(314\text{rad}/\text{s}\right)\left(1\text{H}\right)\left(10\angle 0°\text{mA}\right)\\ =\left(j314\right)\left(0.01\angle 0°\right)\text{V}\\ =3.14\angle 90°\text{V}\end{array}$

Conclusion:

Thus, the phasor voltage that developed across $1\text{H}$ inductor is $\underset{\_}{3.14\angle 90°\text{V}}$.

To determine

Find the phasor voltage across series connected $2\Omega$ resistor and $1\text{F}$ capacitor.

Answer to Problem 33E

The phasor voltage across a series connected $2\Omega$ resistor and $1\text{F}$ capacitor is $\underset{\_}{\text{0}\text{.02}\angle -\text{0}\text{.091}°\text{V}}$.

Explanation of Solution

Calculation:

Find total voltage across a series connected $2\Omega$ resistor and $1\text{F}$ capacitor, $\left(V_{RC}\right)$.

$V_{RC}=V_R+V_C$

Substitute $3.18\times {10}^{-5}\angle -90°\text{V}$ for $V_C$ and $0.02\angle 0°\text{V}$ for $V_R$ as follows.

$\begin{array}{c}{V}_{RC}=0.02\angle 0°\text{V}+3.18\times {10}^{-5}\angle -90°\text{V}\\ =\text{0}\text{.02}\angle -\text{0}\text{.091}°\text{V}\end{array}$

Conclusion:

Thus, the phasor voltage across a series connected $2\Omega$ resistor and $1\text{F}$ capacitor is $\underset{\_}{\text{0}\text{.02}\angle -\text{0}\text{.091}°\text{V}}$.

To determine

Find the phasor voltage across series connected $2\Omega$ resistor and $1\text{H}$ inductor.

Answer to Problem 33E

The phasor voltage across a series connected $2\Omega$ resistor and $1\text{H}$ inductor is $\underset{\_}{\text{3}\text{.14}\angle 89.63°\text{V}}$.

Explanation of Solution

Calculation:

Find total voltage across a series connected $2\Omega$ resistor and $1\text{H}$ inductor, $\left(V_{RL}\right)$.

$V_{RL}=V_R+V_L$

Substitute $3.14\angle 90°\text{V}$ for $V_L$ and $0.02\angle 0°\text{V}$ for $V_R$ as follows.

$\begin{array}{c}{V}_{RL}=0.02\angle 0°\text{V}+3.14\angle 90°\text{V}\\ =\text{3}\text{.14}\angle 89.63°\text{V}\end{array}$

Conclusion:

Thus, the phasor voltage across a series connected $2\Omega$ resistor and $1\text{H}$ inductor is $\underset{\_}{\text{3}\text{.14}\angle 89.63°\text{V}}$.

To determine

Find the instantaneous value of each voltage obtained in parts (a) to (e) at $t=0$.

Answer to Problem 33E

The instantaneous value of each voltage obtained in parts (a) to (e) at $t=0$ are $\underset{\_}{0.02\text{V},0\text{V},0\text{V},0.0198\text{V}\text{and}0.02\text{V}}$, respectively.

Explanation of Solution

Formula used:

Consider the general expression for voltage response.

$v\left(t\right)=V_m\cos \left(\omega t+\varphi \right)$

The complex form of voltage response is, $v\left(t\right)=V_m\angle \varphi$.

Calculation:

Part (a):

The phasor voltage across $2\Omega$ resistor is $0.02\angle 0°\text{V}$.

The instantaneous voltage across $2\Omega$ resistor at $t=0$ is calculated as follows,

$\begin{array}{c}{v}_R\left(0\right)=0.02\cos \left(314\left(0\right)+0°\right)\\ =0.02\cos \left(0\right)\\ =0.02\left(1\right)\\ =0.02\text{V}\end{array}$

Part (b):

The phasor voltage across $1\text{F}$ capacitor is $3.18\times {10}^{-5}\angle -90°\text{V}$.

The instantaneous voltage across $1\text{F}$ capacitor at $t=0$ is calculated as follows,

$\begin{array}{c}{v}_C\left(0\right)=3.18\times {10}^{-5}\cos \left(314\left(0\right)-90°\right)\\ =3.18\times {10}^{-5}\cos \left(0-90°\right)\\ =3.18\times {10}^{-5}\cos \left(-90°\right)\\ =3.18\times {10}^{-5}\left(0\right)\end{array}$

$v_C\left(0\right)=0\text{V}$

Part (c):

The phasor voltage across $1\text{H}$ inductor is $3.14\angle 90°\text{V}$.

The instantaneous voltage across $1\text{H}$ inductor at $t=0$ is calculated as follows,

$\begin{array}{c}{v}_L\left(0\right)=3.14\cos \left(314\left(0\right)+90°\right)\\ =3.14\cos \left(0+90°\right)\\ =3.14\cos \left(+90°\right)\\ =3.14\left(0\right)\end{array}$

$v_L\left(0\right)=0\text{V}$

Part (d):

The phasor voltage across a series connected $2\Omega$ resistor and $1\text{F}$ capacitor is $\text{0}\text{.02}\angle -\text{0}\text{.091}°\text{V}$.

The instantaneous voltage across series connected $2\Omega$ resistor and $1\text{F}$ capacitor at $t=0$ is calculated as follows,

$\begin{array}{c}{v}_{RC}\left(0\right)=0.02\cos \left(314\left(0\right)-0.091°\right)\\ =0.02\cos \left(0-0.091°\right)\\ =0.02\cos \left(-0.091°\right)\\ =0.02\left(0.99\right)\end{array}$

$v_{RC}\left(0\right)=0.0198\text{V}$

Part (e):

The phasor voltage across a series connected $2\Omega$ resistor and $1\text{H}$ inductor is $\text{3}\text{.14}\angle 89.63°\text{V}$.

The instantaneous voltage across series connected $2\Omega$ resistor and $1\text{H}$ inductor at $t=0$ is calculated as follows,

$\begin{array}{c}{v}_{RL}\left(0\right)=3.14\cos \left(314\left(0\right)+89.63°\right)\\ =3.14\cos \left(0+89.63°\right)\\ =3.14\cos \left(89.63°\right)\\ =3.14\left(6.45\times {10}^{-3}\right)\end{array}$

$v_{RL}\left(0\right)=0.02\text{V}$

Conclusion:

Thus, the instantaneous value of each voltage obtained in parts (a) to (e) at $t=0$ are $\underset{\_}{0.02\text{V},0\text{V},0\text{V},0.0198\text{V}\text{and}0.02\text{V}}$, respectively.