Problem

The voltage across a $0.01 \, \mu \mathrm{F}$ capacitor is given by $V_C = 240 \sin(1.25 \times 10^4 t - 30^\circ)$.
Find the mathematical expression for the instantaneous current through the capacitor.

Data

• Capacitance: $C = 0.01 \, \mu \mathrm{F} = 0.01 \times 10^{-6} \, \mathrm{F}$
• Instantaneous voltage: $V_C = 240 \sin(1.25 \times 10^4 t - 30^\circ)$
• Peak voltage: $V_{\max} = 240 \, \mathrm{V}$
• Angular frequency: $\omega = 1.25 \times 10^4 \, \mathrm{rad/s}$

Prerequisite Concepts

1. For a capacitor, the instantaneous current is given by:

$$I_C = I_{\max} \sin(\omega t - \varphi + 90^\circ)$$

2. The peak current is related to peak voltage and capacitance by:

$$I_{\max} = V_{\max} \omega C$$

Solution

Step 1: Identify the peak voltage and angular frequency

From the given equation:

$$V_C = 240 \sin(1.25 \times 10^4 t - 30^\circ)$$

• Peak voltage: $V_{\max} = 240 \, \mathrm{V}$
• Angular frequency: $\omega = 1.25 \times 10^4 \, \mathrm{rad/s}$

Step 2: Calculate the peak current

Using the formula:

$$I_{\max} = V_{\max} \omega C$$

Substitute the values:

$$I_{\max} = 240 \times 1.25 \times 10^4 \times 0.01 \times 10^{-6}$$

Simplify:

$$I_{\max} = 3 \times 10^{-2} \, \mathrm{A} = 30 \, \mathrm{mA}$$

Step 3: Write the expression for instantaneous current

The instantaneous current for a capacitor is given by:

$$I_C = I_{\max} \sin(\omega t - \varphi + 90^\circ)$$

Substitute the values:

$$I_C = 30 \times 10^{-3} \sin\left(1.25 \times 10^4 t - 30^\circ + 90^\circ\right)$$

Simplify the phase angle:

$$I_C = 30 \times 10^{-3} \sin\left(1.25 \times 10^4 t + 60^\circ\right)$$

Answer

The mathematical expression for the instantaneous current is:

$$I_C = 30 \, \mathrm{mA} \, \sin\left(1.25 \times 10^4 t + 60^\circ\right)$$

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