Resistor of $2 \Omega$, inductor of $100 \mu \mathrm{H}$ and capacitor of $400 \mathrm{pF}$ are connected in…
Resistor of $2 \Omega$, inductor of $100 \mu \mathrm{H}$ and capacitor of $400 \mathrm{pF}$ are connected in series across an a.c. source of $e_{r m s}=0.1$ volt.
At resonance, voltage drop across inductor is
20V
25 V
2.5 V
250 V
Solution
We know, at resonance $X_C=X_L$
$\Rightarrow \omega L=\frac{1}{\omega C} \Rightarrow \omega=\left(\frac{1}{\sqrt{L C}}\right)$
The potential drop across inductor is
$V_L=i X_L=i \sqrt{\frac{L}{C}}$
$\because$ LCR circuit in series and at resonance
$i=\frac{V}{R}=\frac{0.1}{2} \mathrm{~A}=0.05 \mathrm{~A}$
Now, $V_L=0.05 \sqrt{\left(\frac{100 \times 10^{-6}}{400 \times 10^{-12}}\right)} \mathrm{V}=2.5 \times 10^{-2} \times 10^3 \mathrm{~V}=25 \mathrm{~V}$