Answer to Question #134351 in Electric Circuits for sai sudeep

Question

A series circuit with a resistor of 100 Ω
capacitor of 25 μF and inductance of
0.15 H is connected across 220-V, 60-Hz
supply. Calculate (i) current (ii) power
and (iii) power factor in the circuit.

Answer 1

Calculating the impedance of the circuit-

X_L = 2pifL = (2)(3.14)(60)(0.15)

= 56.55 ohm

X_C = $\frac{1}{2fC\pi}$ = $\frac{10^6}{(2)(3.14)(25)(60)}$

= 106.10 ohm

R = 100 ohm

Total Impedance(Z) = (R² + (X_L - X_C)²)^0.5

Z = (10000 + 2455.2025)^0.5

= 111.6 ohm

(i). Current in the circuit(I)-

I = V_S/Z

= $\frac{220}{111.6}$

= 1.97 A, 60 Hz

(ii). Power dissipated(P) in series LCR circuit is-

P = $\frac{V^2R}{|Z|^2}$

= $\frac{(48400)(100)}{12455.2025}$

= 388.59 W

(iii). Power factor = R/|Z|

cos$\phi$ = $\frac{100}{111.6}$

= 0.896