Question

A 400 Ω resistor, a 3 H inductor and a 5 µF capacitor are connected in series to a 220 V, 50 Hz ac source. Calculate the:

1. Impedance of the circuit.
2. Current flowing through the circuit.

Answer 1

Given: Resistor (R) = 400 Ω

Inductor (L) = 3 H

Capacitor (C) = 5 µF = 5 × 10⁻⁶ F

V = 200 Volt

f = 50 Hz

1. Impedance (Z) = `sqrt(R^2 + (X_L - X_C)^2)`

X_L = 2 π f L

= 2 π × 50 × 3

= 300 π Ω

X_C = `1/(2 pi f C)`

= `1/(2 pi xx 50 xx 5 xx 10^-6)`

= `2/pi xx 10^3 Omega`

Z = `sqrt(R^2 + (X_L - X_C)^2)`

= `sqrt(400^2 + (300 pi - 2/pi xx 10^3)^2)`

= 503.54 Ω

2. Current (I) = `V/Z`

= `220/503.54`

= 0.44 A