Question

A series LCR circuit containing a resistance of $120\Omega$ has resonance frequency $4\times {10}^5\text{rad/s}$. At resonance, the voltage across resistor and inductor are $60\text{V}$ and $40\text{V}$, respectively. The value of capacitance is -

• A. $1.125\times {10}^{-8}\text{F}$
• B. $2.125\times {10}^{-8}\text{F}$
• C. $3.125\times {10}^{-8}\text{F}$
• D. $4.125\times {10}^{-8}\text{F}$

Answer 1

The correct option is C $3.125\times {10}^{-8}\text{F}$

At resonance, $X_L=X_C$ and $Z=R=120\Omega$

So, $i_{\text{rms}}=\frac{(V_R{)}_{\text{rms}}}{R}=\frac{60}{120}=0.5\text{A}$

Also, $i_{\text{rms}}=\frac{(V_L{)}_{\text{rms}}}{\omega L}$

$\Rightarrow L=\frac{(V_L{)}_{\text{rms}}}{\omega i_{\text{rms}}}=\frac{40}{4\times {10}^5\times 0.5}=2\times {10}^{-4}\text{H}$

Further, the resonance frequency is,

$\omega =\frac{1}{\sqrt{LC}}$

$\Rightarrow C=\frac{1}{{\omega }^{2}L}=\frac{1}{(4\times {10}^5{)}^2\times 2\times {10}^{-4}}=3.125\times {10}^{-8}\text{F}$

Hence, option $\left(C\right)$ is the correct answer.