Question

A series AC circuit contains L = 0.11mH, $R = 15\Omega $ and $C = 60\mu F$. Find the resonance frequency of the circuit and the current at the resonance. Peak E.M.F of AC source $ = 240V$.

Answer 1

Hint: We can use the values of the inductance and the capacitance and substitute it in the formula for the resonant frequency to find the answer. And at the resonance, the capacitive impedance and the inductive impedance cancel each other out and the only impedance will be the resistor. So the current will be the peak voltage divided by the resistance.
Formula used: In this solution we will be using the following formula,
$\Rightarrow f = \dfrac{1}{{2\pi \sqrt {LC} }}$
where $f$ is the frequency, $L$ is the inductance and $C$ is the capacitance.
$\Rightarrow {I_{\max }} = \dfrac{{{V_{\max }}}}{Z}$
where ${I_{\max }}$ is the maximum current, ${V_{\max }}$ is the peak voltage and $Z$ is the impedance.

Complete step by step solution:
In the question we are given the values of the inductance as $L = 0.11mH$ and the capacitance as $C = 60\mu F$. We can write the inductance as, $L = 0.11 \times {10^{ - 3}}H$ and the capacitance as $C = 60 \times {10^{ - 6}}F$
Now using these 2 values we can find the value of the frequency from the formula,
$\Rightarrow f = \dfrac{1}{{2\pi \sqrt {LC} }}$
Substituting the values we get,
$\Rightarrow f = \dfrac{1}{{2\pi \sqrt {0.11 \times {{10}^{ - 3}} \times 60 \times {{10}^{ - 6}}} }}$
On calculating we get,
$\Rightarrow f = \dfrac{1}{{2\pi \sqrt {6.6 \times {{10}^{ - 9}}} }}$
On doing the square root and multiplying,
$\Rightarrow f = \dfrac{1}{{5.1 \times {{10}^{ - 4}}}}$
On taking the inverse we get the resonance frequency as,
$\Rightarrow f = 1.9kHz$
The impedance of a series LCR circuit is given by the formula,
$\Rightarrow Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} $
At the resonance condition ${X_L} = {X_C}$
So the second term vanishes and the impedance becomes,
$\Rightarrow Z = R$
In the question we are given $R = 15\Omega $. So $Z = R = 15\Omega $
Now the maximum current is at the resonance condition given by,
$\Rightarrow {I_{\max }} = \dfrac{{{V_{\max }}}}{Z}$
We are given ${V_{\max }} = 240V$. Hence on substituting,
$\Rightarrow {I_{\max }} = \dfrac{{240}}{{15}}A$
Hence on calculating we get, ${I_{\max }} = 16A$. This is the current at the resonance.

Note:
In a series LCR circuit, the resonance occurs when the supply frequency causes the voltage across the inductor and the capacitor to be equal and opposite in phase. These series resonance circuits can be found in various forms such as in AC mains filter, noise filter, radio etc.