Question

A circuit of resistance $5\Omega$ and inductance $0.6H$ is in series with a capacitance of $10\mu F$. If a voltage of $200\text{V}$ is applied and the frequency is adjusted to resonance. The current in the circuit is $I_0$ and voltage across the inductor and capacitor are $V_0$, and $V_1$ respectively. We have

• A. $I_0=40A$
• B. $V_0=9.8kV$
• C. $V_1=9.8kV$
• D. $V_1=19.6kV$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $I_0=40A$
B $V_0=9.8kV$
C $V_1=9.8kV$


We know that
$I_0=\frac{V}{\sqrt{R^2+(X_L-X_C{)}^2}}$
Since it is given that circuit is at resosnance
$\therefore X_L=X_C\Rightarrow \omega L=\frac{1}{\omega C}\Rightarrow \omega =\frac{1}{\sqrt{LC}}\Rightarrow \omega =\frac{{10}^3}{\sqrt{6}}$

Since $X_L=X_C$
We get
$I_0=\frac{V}{R}=\frac{200}{5}=40A$

Now for voltage across inductor($V_0$)

$V_0=i_0{X}_L$
And we have $X_L=\omega L=\frac{{10}^3}{\sqrt{6}}\times 0.6$
$\Rightarrow V_0=40\times 0.6\times \frac{{10}^3}{\sqrt{6}}\Rightarrow V_0=9.8kV$

For voltage across capacitor ($V_1$)
$V_1=i_0{X}_C$
We know that $X_L=X_C$
$\therefore V_1=9.8kV$