Question

A series LCR circuit containing a resistor of $120\Omega$ has an angular resonant frequency $4\times {10}^{5}rad{s}^{-1}$. At resonance the voltages across resistor and inductor are 60 V and $40\text{V}$ respectively.
The value of capacitance C is

• A. $\frac{1}{32}\mu F$
• B. $\frac{1}{16}\mu F$
• C. $32\mu F$
• D. $16\mu F$

Answer 1

The correct option is A $\frac{1}{32}\mu F$

Voltage across a capacitor is given by
$V_C=I_{rms}{X}_C$

Now the circuit is given at resonance
So $X_L=X_C$ and also $V_L=V_C$
and the circuit will behave as purely resistive.
$\Rightarrow Z=R$
$\therefore I_{rms}=\frac{V_{rms}}{Z}=\frac{V_{rms}}{R}\Rightarrow I_{rms}=\frac{60}{120}=\frac{1}{2}\text{A}$

Now putting the values
$V_C=i_{rms}\times \frac{1}{\omega C}\Rightarrow 40=\frac{1}{2}\times \frac{1}{\omega C}\Rightarrow C=\frac{1}{32}\mu F$