Chapter 10, Problem 29E

Assuming an operating frequency of 50 Hz, compute the instantaneous voltage at t = 10 ms and t = 25 ms for each of the phasor quantities represented in Exercise 28.

To determine

Find the instantaneous voltage of $\frac{2-j}{5\angle 45°}\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$.

Answer to Problem 29E

The instantaneous voltage of $\frac{2-j}{5\angle 45°}\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$ are $\underset{\_}{-0.14\text{V}\text{and}0.423\text{V}}$, respectively.

Explanation of Solution

Given data:

$\frac{2-j}{5\angle 45°}\text{V}$        (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:

Simplify equation (1) as follows.

$\begin{array}{c}\frac{2-j}{5\angle 45°}=\frac{2.23\angle -26.56°}{5\angle 45°}\\ =0.141-j0.423\text{V}\\ =0.446\angle -71.56°\text{V}\end{array}$

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 $0.446\angle -71.56°\text{V}$ is,

$v\left(t\right)=0.446\cos \left(314t-71.56°\right)\text{V}$        (2)

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

$\begin{array}{c}v\left(10\text{ms}\right)=0.446\cos \left[314\left(10\text{ms}\right)-71.56°\right]\\ =0.446\cos \left[314\left(10\times {10}^{-3}\right)-71.56°\right]\left\{\because 1\text{m}=1\times {10}^{-3}\right\}\\ =0.446\cos \left[3.14-71.56°\right]\\ =0.446\cos \left[3.14\left(\frac{180}{\pi }\right)-71.56°\right]\end{array}$

Simplify the equation as follows.

$\begin{array}{c}v\left(10\text{ms}\right)=0.446\cos \left[179.9°-71.56°\right]\\ =0.446\cos \left[108.34°\right]\\ =0.446\left(-0.314\right)\\ =-0.14\text{V}\end{array}$

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

$\begin{array}{c}v\left(25\text{ms}\right)=0.446\cos \left[314\left(25\text{ms}\right)-71.56°\right]\\ =0.446\cos \left[314\left(25\times {10}^{-3}\right)-71.56°\right]\left\{\because 1\text{m}=1\times {10}^{-3}\right\}\\ =0.446\cos \left[7.85-71.56°\right]\\ =0.446\cos \left[7.85\left(\frac{180}{\pi }\right)-71.56°\right]\end{array}$

Simplify the equation as follows.

$\begin{array}{c}v\left(25\text{ms}\right)=0.446\cos \left[449.7°-71.56°\right]\\ =0.446\cos \left[378.2°\right]\\ =0.446\left(0.949\right)\\ =0.423\text{V}\end{array}$

Conclusion:

Thus, the instantaneous voltage of $\frac{2-j}{5\angle 45°}\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$ are $\underset{\_}{-0.14\text{V}\text{and}0.423\text{V}}$, respectively.

To determine

Find the instantaneous voltage of $\frac{6\angle 20°}{1000}-j\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$.

Answer to Problem 29E

The instantaneous voltage of $\frac{6\angle 20°}{1000}-j\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$ are $\underset{\_}{-0.004\text{V}\text{and}0.987\text{V}}$, respectively.

Explanation of Solution

Given data:

$\frac{6\angle 20°}{1000}-j\text{V}$        (3)

Calculation:

Simplify equation (3) as follows.

$\begin{array}{c}\frac{6\angle 20°}{1000}-j\text{V}=\left(0.006\angle 20°\right)-j\\ =5.638\times {10}^{-3}-j0.997\\ =0.005638-j0.997\text{V}\\ =0.997\angle -89.67°\text{V}\end{array}$

The cosine function of phasor expression $0.997\angle -89.67°\text{V}$ is,

$v\left(t\right)=0.997\cos \left(314t-89.67°\right)\text{V}$        (4)

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

$\begin{array}{c}v\left(10\text{ms}\right)=0.997\cos \left[314\left(10\text{ms}\right)-89.67°\right]\\ =0.997\cos \left[314\left(10\times {10}^{-3}\right)-89.67°\right]\left\{\because 1\text{m}=1\times {10}^{-3}\right\}\\ =0.997\cos \left[3.14-89.67°\right]\\ =0.997\cos \left[3.14\left(\frac{180}{\pi }\right)-89.67°\right]\end{array}$

Simplify the equation as follows.

$\begin{array}{c}v\left(10\text{ms}\right)=0.997\cos \left[179.9°-89.67°\right]\\ =0.997\cos \left[90.23°\right]\\ =0.997\left(-4.014\times {10}^{-3}\right)\\ =-0.004\text{V}\end{array}$

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

$\begin{array}{c}v\left(25\text{ms}\right)=0.997\cos \left[314\left(25\text{ms}\right)-89.67°\right]\\ =0.997\cos \left[314\left(25\times {10}^{-3}\right)-89.67°\right]\left\{\because 1\text{m}=1\times {10}^{-3}\right\}\\ =0.997\cos \left[7.85-89.67°\right]\\ =0.997\cos \left[7.85\left(\frac{180}{\pi }\right)-89.67°\right]\end{array}$

Simplify the equation as follows.

$\begin{array}{c}v\left(25\text{ms}\right)=0.997\cos \left[449.7°-89.67°\right]\\ =0.997\cos \left[360.03°\right]\\ =0.997\left(0.99\right)\\ =0.987\text{V}\end{array}$

Conclusion:

Thus, the instantaneous voltage of $\frac{6\angle 20°}{1000}-j\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$ are $\underset{\_}{-0.004\text{V}\text{and}0.987\text{V}}$, respectively.

To determine

Find the instantaneous voltage of $\left(j\right)\left(52.5\angle -90°\right)\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$.

Answer to Problem 29E

The instantaneous voltage of $\left(j\right)\left(52.5\angle -90°\right)\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$ are $\underset{\_}{-52.4\text{V}\text{and}0.2793\text{V}}$, respectively.

Explanation of Solution

Given data:

$\left(j\right)\left(52.5\angle -90°\right)\text{V}$        (5)

Calculation:

Simplify equation (5) as follows.

$\begin{array}{c}\left(j\right)\left(52.5\angle -90°\right)=52.5+j0\text{V}\\ =52.5\angle 0°\text{V}\end{array}$

The cosine function of phasor expression $52.5\angle 0°\text{V}$ is,

$v\left(t\right)=52.5\cos \left(314t+0°\right)\text{V}$        (6)

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

$\begin{array}{c}v\left(10\text{ms}\right)=52.5\cos \left[314\left(10\text{ms}\right)\right]\\ =52.5\cos \left[314\left(10\times {10}^{-3}\right)\right]\left\{\because 1\text{m}=1\times {10}^{-3}\right\}\\ =52.5\cos \left[3.14\right]\\ =52.5\cos \left[3.14\left(\frac{180}{\pi }\right)\right]\end{array}$

Simplify the equation as follows.

$\begin{array}{c}v\left(10\text{ms}\right)=52.5\cos \left[179.9°\right]\\ =52.5\left(-0.999\right)\\ =-52.4\text{V}\end{array}$

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

$\begin{array}{c}v\left(25\text{ms}\right)=52.5\cos \left[314\left(25\text{ms}\right)\right]\\ =52.5\cos \left[314\left(25\times {10}^{-3}\right)\right]\left\{\because 1\text{m}=1\times {10}^{-3}\right\}\\ =52.5\cos \left[7.85\right]\\ =52.5\cos \left[7.85\left(\frac{180}{\pi }\right)\right]\end{array}$

Simplify the equation as follows.

$\begin{array}{c}v\left(25\text{ms}\right)=52.5\cos \left[449.7°\right]\\ =52.5\left(5.23\times {10}^{-3}\right)\\ =0.2793\text{V}\end{array}$

Conclusion:

Thus, the instantaneous voltage of $\left(j\right)\left(52.5\angle -90°\right)\text{V}$ at $t=10\text{ms}$ and $t=25\text{ms}$ are $\underset{\_}{-52.4\text{V}\text{and}0.2793\text{V}}$, respectively.