Question

A sinewave voltage with amplitude 2 V and frequency 1 kHz is applied across a capacitor. The capacitance is 0.1 uF. What is the amplitude of the current?

Answer 1

The amplitude of the current across the capacitor is 0.4 mA.

Step-by-step explanation:

When a sinusoidal voltage is applied across a capacitor, the current across the capacitor is given by the formula:

I = C * dV/dt

Where I is the current, C is the capacitance, and dV/dt is the rate of change of voltage. In this case, the amplitude of the voltage is 2 V and the frequency is 1 kHz. To find the amplitude of the current, we need to find the rate of change of voltage.

The rate of change of voltage can be calculated by taking the derivative of the sinusoidal voltage equation. For a sinusoidal voltage of the form V = A * sin(2πft), where A is the amplitude, f is the frequency, and t is the time, the rate of change of voltage is given by:

dV/dt = 2πfA * cos(2πft)

Plugging in the values, we have:

dV/dt = 2π * 1 kHz * 2 V * cos(2π * 1 kHz * t)

Now we can find the amplitude of the current:

I = C * 2πfA

Plugging in the values, we get:

I = 0.1 μF * 2π * 1 kHz * 2 V

I = 0.4 mA