Chapter 10, Problem 30E

(a)

To determine

Find the instantaneous voltage of $9\angle 65°\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$.

Answer to Problem 30E

The instantaneous voltage of $9\angle 65°\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$ are $\underset{\_}{-3.816\text{V}\text{and}-8.136\text{V}}$, respectively.

Explanation of Solution

Given data:

$9\angle 65°\text{V}$        (1)

$f=50\text{Hz}$

Formula used:

Consider the Euler’s identity,

$e^{j\theta }=\cos \theta +j\sin \theta$

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:

Consider the general expression for frequency in terms of $\text{rad}/\text{s}$.

$\omega =2\pi f$

Substitute $50\text{Hz}$ for $f$ as follows.

$\begin{array}{c}\omega =2\pi \left(50\right)\\ =314.15\\ \cong 314\text{rad}/\text{s}\end{array}$

The cosine function of phasor expression $9\angle 65°\text{V}$ is,

$v\left(t\right)=9\cos \left(314t+65°\right)\text{V}$        (2)

Substitute $10\text{ms}$ for $t$ in equation (2) as follows.

$\begin{array}{c}v\left(10\text{ms}\right)=9\cos \left[314\left(10\text{ms}\right)+65°\right]\\ =9\cos \left[314\left(10\times {10}^{-3}\right)+65°\right]\left\{\because 1\text{m}=1\times {10}^{-3}\right\}\\ =9\cos \left[3.14+65°\right]\\ =9\cos \left[3.14\left(\frac{180}{\pi }\right)+65°\right]\end{array}$

Simplify the equation as follows.

$\begin{array}{c}v\left(10\text{ms}\right)=9\cos \left[179.9°+65°\right]\\ =9\cos \left[244.9°\right]\\ =9\left(-0.424\right)\\ =-3.816\text{V}\end{array}$

Substitute $25\text{ms}$ for $t$ in equation (2) as follows.

$\begin{array}{c}v\left(25\text{ms}\right)=9\cos \left[314\left(25\text{ms}\right)+65°\right]\\ =9\cos \left[314\left(25\times {10}^{-3}\right)+65°\right]\left\{\because 1\text{m}=1\times {10}^{-3}\right\}\\ =9\cos \left[7.85+65°\right]\\ =9\cos \left[7.85\left(\frac{180}{\pi }\right)+65°\right]\end{array}$

Simplify the equation as follows.

$\begin{array}{c}v\left(25\text{ms}\right)=9\cos \left[449.7°+65°\right]\\ =9\cos \left[514.7°\right]\\ =9\left(-0.904\right)\\ =-8.136\text{V}\end{array}$

Conclusion:

Thus, the instantaneous voltage of $9\angle 65°\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$ are $\underset{\_}{-3.816\text{V}\text{and}-8.136\text{V}}$, respectively.

(b)

To determine

Find the instantaneous voltage of $\frac{2\angle 31°}{4\angle 25°}\text{A}$ at $t=10\text{ms}$ and $t=25\text{ms}$.

Answer to Problem 30E

The instantaneous voltage of $\frac{2\angle 31°}{4\angle 25°}\text{A}$ at $t=10\text{ms}$ and $t=25\text{ms}$ are $\underset{\_}{-0.497\text{A}\text{and}-0.0495\text{A}}$, respectively.

Explanation of Solution

Given data:

$\frac{2\angle 31°}{4\angle 25°}\text{A}$        (3)

Calculation:

Simplify equation (2) in single complex form.

$\begin{array}{c}i=\frac{2\angle 31°}{4\angle 25°}\\ =0.5\angle 6°\end{array}$

The cosine function of phasor expression $0.5\angle 6°\text{A}$ is,

$i\left(t\right)=0.5\cos \left(314t+6°\right)\text{A}$        (4)

Substitute $10\text{ms}$ for $t$ in equation (4) as follows.

$\begin{array}{c}v\left(10\text{ms}\right)=0.5\cos \left[314\left(10\text{ms}\right)+6°\right]\\ =0.5\cos \left[314\left(10\times {10}^{-3}\right)+6°\right]\left\{\because 1\text{m}=1\times {10}^{-3}\right\}\\ =0.5\cos \left[3.14+6°\right]\\ =0.5\cos \left[3.14\left(\frac{180}{\pi }\right)+6°\right]\end{array}$

Simplify the equation as follows.

$\begin{array}{c}v\left(10\text{ms}\right)=0.5\cos \left[179.9°+6°\right]\\ =0.5\cos \left[185.9°\right]\\ =0.5\left(-0.994\right)\\ =-0.497\text{A}\end{array}$

Substitute $25\text{ms}$ for $t$ in equation (4) as follows.

$\begin{array}{c}v\left(25\text{ms}\right)=0.5\cos \left[314\left(25\text{ms}\right)+6°\right]\\ =0.5\cos \left[314\left(25\times {10}^{-3}\right)+6°\right]\left\{\because 1\text{m}=1\times {10}^{-3}\right\}\\ =0.5\cos \left[7.85+6°\right]\\ =0.5\cos \left[7.85\left(\frac{180}{\pi }\right)+6°\right]\end{array}$

Simplify the equation as follows.

$\begin{array}{c}v\left(25\text{ms}\right)=0.5\cos \left[449.7°+6°\right]\\ =0.5\cos \left[455.7°\right]\\ =0.5\left(-0.099\right)\\ =-0.0495\text{A}\end{array}$

Conclusion:

Thus, the instantaneous voltage of $\frac{2\angle 31°}{4\angle 25°}\text{A}$ at $t=10\text{ms}$ and $t=25\text{ms}$ are $\underset{\_}{-0.497\text{A}\text{and}-0.0495\text{A}}$, respectively.

(c)

To determine

Find the instantaneous voltage of $22\angle 14°-8\angle 33°\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$.

Answer to Problem 30E

The instantaneous voltage of $22\angle 14°-8\angle 33°\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$ are $\underset{\_}{-14.61\text{V}\text{and}-0.879\text{V}}$, respectively.

Explanation of Solution

Given data:

$22\angle 14°-8\angle 33°\text{V}$        (5)

Calculation:

Simplify equation (5) in single complex form.

$\begin{array}{c}v=22\angle 14°-8\angle 33°\text{V}\\ =14.66\angle 3.77°\text{V}\end{array}$

The cosine function of phasor expression $14.66\angle 3.77°\text{V}$ is,

$v\left(t\right)=14.66\cos \left(314t+3.77°\right)\text{V}$        (6)

Substitute $10\text{ms}$ for $t$ in equation (6) as follows.

$\begin{array}{c}v\left(10\text{ms}\right)=14.66\cos \left[314\left(10\text{ms}\right)+3.77°\right]\\ =14.66\cos \left[314\left(10\times {10}^{-3}\right)+3.77°\right]\left\{\because 1\text{m}=1\times {10}^{-3}\right\}\\ =14.66\cos \left[3.14+3.77°\right]\\ =14.66\cos \left[3.14\left(\frac{180}{\pi }\right)+3.77°\right]\end{array}$

Simplify the equation as follows.

$\begin{array}{c}v\left(10\text{ms}\right)=14.66\cos \left[179.9°+3.77°\right]\\ =14.66\cos \left(183.67°\right)\\ =14.66\left(-0.997\right)\\ =-14.61\text{V}\end{array}$

Substitute $25\text{ms}$ for $t$ in equation (6) as follows.

$\begin{array}{c}v\left(25\text{ms}\right)=14.66\cos \left[314\left(25\text{ms}\right)+3.77°\right]\\ =14.66\cos \left[314\left(25\times {10}^{-3}\right)+3.77°\right]\left\{\because 1\text{m}=1\times {10}^{-3}\right\}\\ =14.66\cos \left[7.85+3.77°\right]\\ =14.66\cos \left[7.85\left(\frac{180}{\pi }\right)+3.77°\right]\end{array}$

Simplify the equation as follows.

$\begin{array}{c}v\left(25\text{ms}\right)=14.66\cos \left[449.7°+3.77°\right]\\ =14.66\cos \left[453.47°\right]\\ =14.66\left(-0.06\right)\\ =-0.879\text{V}\end{array}$

Conclusion:

Thus, the instantaneous voltage of $22\angle 14°-8\angle 33°\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$ are $\underset{\_}{-14.61\text{V}\text{and}-0.879\text{V}}$, respectively.