Problem

The AC voltage across a $0.5 \, \mu \mathrm{F}$ capacitor is given by $V_c = 16 \sin(2 \times 10^3 t) \, \mathrm{V}$.
Find:
(a) The capacitive reactance, $X_C$.
(b) The peak value of the current through the capacitor, $I_{\max}$.

Data

• Capacitance: $C = 0.5 \, \mu \mathrm{F} = 0.5 \times 10^{-6} \, \mathrm{F}$
• Instantaneous voltage: $V_c = 16 \sin(2 \times 10^3 t) \, \mathrm{V}$
• Peak voltage: $V_{\max} = 16 \, \mathrm{V}$
• Angular frequency: $\omega = 2 \times 10^3 \, \mathrm{rad/s}$

Prerequisite Concepts

1. Capacitive reactance, $X_C$, is given by:

$$X_C = \frac{1}{\omega C}$$

2. Peak current, $I_{\max}$, is related to peak voltage and reactance by Ohm’s Law:

$$I_{\max} = \frac{V_{\max}}{X_C}$$

Solution

Part (a): Calculate Capacitive Reactance

Using the formula:

$$X_C = \frac{1}{\omega C}$$

Substitute the values:

$$X_C = \frac{1}{(2 \times 10^3) \times (0.5 \times 10^{-6})}$$

Simplify:

$$X_C = \frac{1}{1 \times 10^{-3}} = 1 \times 10^3 \, \Omega$$ $$X_C = 1 \, \mathrm{k\Omega}$$

Part (b): Calculate Peak Current

Using the formula:

$$I_{\max} = \frac{V_{\max}}{X_C}$$

Substitute the values:

$$I_{\max} = \frac{16}{10^3} = 16 \times 10^{-3} \, \mathrm{A}$$ $$I_{\max} = 16 \, \mathrm{mA}$$

Answer

(a) The capacitive reactance is $X_C = 1 \, \mathrm{k\Omega}$.
(b) The peak current is $I_{\max} = 16 \, \mathrm{mA}$.

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