A circuit consist of a capacitor of 10^-4F and a resistor of 100 ohms an alternating emf of 20Vrms and frequency of 50/π H is applied determine

Question

The peak current flowing
The voltage across the capacitor
The average power supplied

Answer 1

To determine the peak current flowing in the circuit, we can use Ohm's Law. The formula to calculate peak current is:

I_peak = V_peak / R

Where V_peak is the peak voltage and R is the resistance.

Since the given voltage is in RMS (Root Mean Square), we need to convert it to peak voltage by multiplying it by the square root of 2:

V_peak = V_rms * √2

V_rms = 20V (given)

V_peak = 20V * √2
= 20V * 1.414
≈ 28.28V

Now, we can calculate the peak current as follows:

I_peak = V_peak / R
= 28.28V / 100Ω
= 0.2828A

The peak current flowing through the circuit is approximately 0.2828A.

To determine the voltage across the capacitor, we can use the formula for the impedance of a capacitor in an AC circuit:

Z_c = 1 / (2πfC)

Where Z_c is the impedance of the capacitor, f is the frequency, and C is the capacitance.

Given:
C = 10^-4F
f = 50/π

Substituting the values:

Z_c = 1 / (2π * (50/π) * 10^-4)
= 1 / (2 * 50 * 10^-4)
= 1 / (100 * 10^-4)
= 1 / 10^-2
= 100Ω

Since the impedance of the capacitor is equal to the resistance, the voltage across the capacitor will be equal to the RMS voltage, which is 20V.

The voltage across the capacitor is 20V.

To determine the average power supplied to the circuit, we can use the formula:

P_avg = (V_rms^2) / R

Substituting the values:

P_avg = (20V)^2 / 100Ω
= 400V^2 / 100Ω
= 400W / 100Ω
= 4W

The average power supplied to the circuit is 4W.