Problem 9: A resistor of 30 ohm is connected in series with a capacitor of 79.5 μF across a power supply of 50 Hz and 100 V. Find (a) impedance (b) current (c) phase angle and (d) equation of instantaneous current.Add Your Heading Text Here

Solution

Given

Capacitance, C = 79.5 μF = 79.5 × 10-6 F
Resistance, R = 30 Ω
Supply frequency, f = 50 Hz
Supply voltage, V = 100 V

Find: Z, I, ϕ, Iinst

(a) Equation for the impedance of a capacitive circuit is, Now, Put the values, Therefore, (b) Current in the circuit is determined by Ohm’s law, (c) Phase angle ϕ is given by, Put the values, (d) Instantaneous current in a capacitive circuit is given by, Which means the current leads the voltage by phase angle ϕ. Im is not given, therefore, we calculate it from the formula, Im = Irms/0.707. Put the value of effective current calculated in (b), we have, Im = 2/0.707 = 2.828 A. Similarly, ω = 2πf. Put the value of frequency, ω = 2× 3.14× 50 = 314 rad/s. Put all these values in the above equation, we have This is the required equation.