Chapter 10, Problem 1RQ

The voltage V_o across the capacitor in Fig. 10.43 is:

Figure 10.43

For Review Question 10.1.

To determine

Choose the correct option to find the voltage $V_o$ across the capacitor in Figure 10.43.

Answer to Problem 1RQ

The correct option from the given choices is $\text{(c)}7.071\angle -45°\text{V}$.

Explanation of Solution

Given data:

Refer to Figure 10.43 in the textbook for circuit analysis.

Calculation:

From Figure 10.43, write the expression for voltage $V_o$ using voltage division rule.

$\begin{array}{c}{V}_o=\left(10\angle 0°\text{V}\right)\frac{\left(-j1\right)\Omega }{\left(1-j1\right)\Omega }\\ =\left(10\angle 0°\text{V}\right)\frac{1\angle -90°\Omega }{1.41421\angle -45°\Omega }\\ =\left(10\angle 0°\text{V}\right)\left[\frac{1}{1.41421}\angle \left(-90°-\left(-45°\right)\right)\right]\\ =\left(10\right)\frac{1}{1.41421}\angle -45°\text{ V}\end{array}$

Reduce the equation as follows.

$\begin{array}{c}{V}_o=\text{7}\text{.07108}\angle -45°\text{V}\\ \cong \text{7}\text{.071}\angle -45°\text{V}\end{array}$

Hence, the voltage across the capacitor is $\text{7}\text{.071}\angle -45°\text{V}$. Therefore, option (c) is correct and options (a), (b), and (d) are incorrect.

Conclusion:

Thus, the correct option from the given choices is $\text{(c)}7.071\angle -45°\text{V}$.