Question

A sinusoidal voltage is applied to a series circuit containing a resistor of 25 Ω, a capacitor of 0.90 μF and an inductor of 100 mH. The frequency of the source can be varied. The frequency of the source that will create a resonance condition in the circuit is: 

1. 14.7 Hz
2. 25.9 Hz
3. 46.3 Hz
4. 530.1 Hz

Answer 1

Correct Answer - Option 4 : 530.1 Hz

Concept:

Resonance frequency: The frequency of an AC circuit at which the impedance of the circuit becomes minimum or current in the circuit becomes maximum is called resonance frequency.

The resonance frequency is given by:

\(f = \frac{1}{{2\pi \;\sqrt {LC} }}\)

Where L is the inductance and C is the capacitance of the circuit.

Calculation:

Given that:

Capacitance (C) = 0.90 μF

Inductance (L) = 100 mH

\(f = \frac{1}{{2\pi \sqrt {LC} }} = \frac{1}{{2\pi \;\sqrt {100\times 10^{-3}\; \times \;0.90\; \times \;{{10}^{ - 6}}} }} \)

f = 530.1 Hz

Notes:

In a series RLC circuit, the impedance is given by:

\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}\)

R is resistance

X­L is inductive reactance = ωL

XC is capacitive reactance = -1/ωC

(XL – XC) is net reactance

At resonant frequency, inductive reactance is equal to capacitive reactance i.e. XL = XC

So, at this condition the impedance is minimum, and it is equivalent to R.