Question

A coil having an inductance of 0.5 H and a resistance of 60 Ω is connected in series with a capacitance of 10 μF. The coil is connected to 100 V ac supply. What is the source frequency and current flowing in the circuit under resonance condition?
1. 7.121 Hz and 16.7 A
2. 7.121 Hz and 1.67 A
3. 71.21 Hz and 16.7 A
4. 71.21 Hz and 1.67 A

Answer 1

Correct Answer - Option 4 : 71.21 Hz and 1.67 A

Concept:

The net impedance for a series circuit is given by:

Z = Rnet + j (XL - XC)

Rnet = Net resistance of the series circuit.

XL = Net inductive Reactance defined as:

XL = ωL

XC = Net capacitive Reactance defined as:

\(X_C=\frac{1}{ω C}\)

The magnitude of the impedance will be:

\(|Z|=\sqrt{R_{net}^2+(X_L-X_C)^2}\)

For a series RLC circuit, the resonant frequency is given by:

\(f_{r}=\frac{1}{2\pi\sqrt{LC}}\)

At this condition, the net impedance is equal to its resistance. Hence the current in this condition is given by

I = V/R

Calculation:

Given that, supply voltage = 100 V

Resistance (R) = 60 Ω

Inductance (L) = 0.5 H

Capacitance (C) = 10 μF

Resonant frequency \({f_r} = \frac{1}{{2\pi \sqrt {0.5 \times 10 \times {{10}^{ - 6}}} }} = 71.17\;kHz\)

Current (I) = 100/60 = 1.67 A