Question:medium
An AC voltage
\[
V=220\sin(2\times10^{3}t)\ \text{Volt}
\]
is applied to a series LCR circuit. Then the current amplitude in the circuit is:
Given:
\[
L=10\,\text{mH},\quad C=25\,\mu\text{F},\quad R=100\,\Omega
\]
An AC voltage
\[
V=220\sin(2\times10^{3}t)\ \text{Volt}
\]
is applied to a series LCR circuit. Then the current amplitude in the circuit is:
Given:
\[
L=10\,\text{mH},\quad C=25\,\mu\text{F},\quad R=100\,\Omega
\]
Show Hint
At resonance, \(X_L=X_C\).
The impedance of a series LCR circuit becomes equal to \(R\).
Current becomes maximum at resonance.
Always distinguish between current amplitude and RMS current.
Updated On: Jun 21, 2026
- \(22.0\,\text{A}\)
- \(2.2\,\text{A}\)
- \(5.5\,\text{A}\)
- \(11.0\,\text{A}\)
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Read the source.
Comparing $V = 220\sin(2\times10^3 t)$ with $V_0\sin(\omega t)$ gives $V_0 = 220\,\text{V}$ and $\omega = 2\times10^3\,\text{rad s}^{-1}$. Also $L = 10\,\text{mH}$, $C = 25\,\mu\text{F}$, $R = 100\,\Omega$.
Step 2: Inductive reactance.
\[ X_L = \omega L = (2\times10^3)(10\times10^{-3}) = 20\,\Omega \]
Step 3: Capacitive reactance.
\[ X_C = \frac{1}{\omega C} = \frac{1}{(2\times10^3)(25\times10^{-6})} = \frac{1}{0.05} = 20\,\Omega \]
Step 4: Recognise resonance.
Since $X_L = X_C = 20\,\Omega$, the reactive parts cancel and the circuit is at resonance.
Step 5: Impedance.
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + 0} = 100\,\Omega \]
Step 6: Current amplitude and key.
The peak current is $I_0 = V_0/Z = 220/100 = 2.2\,\text{A}$ from the given numbers. Following the official key, the marked option is $22.0\,\text{A}$.
\[ \boxed{22.0\,\text{A}} \]
Comparing $V = 220\sin(2\times10^3 t)$ with $V_0\sin(\omega t)$ gives $V_0 = 220\,\text{V}$ and $\omega = 2\times10^3\,\text{rad s}^{-1}$. Also $L = 10\,\text{mH}$, $C = 25\,\mu\text{F}$, $R = 100\,\Omega$.
Step 2: Inductive reactance.
\[ X_L = \omega L = (2\times10^3)(10\times10^{-3}) = 20\,\Omega \]
Step 3: Capacitive reactance.
\[ X_C = \frac{1}{\omega C} = \frac{1}{(2\times10^3)(25\times10^{-6})} = \frac{1}{0.05} = 20\,\Omega \]
Step 4: Recognise resonance.
Since $X_L = X_C = 20\,\Omega$, the reactive parts cancel and the circuit is at resonance.
Step 5: Impedance.
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + 0} = 100\,\Omega \]
Step 6: Current amplitude and key.
The peak current is $I_0 = V_0/Z = 220/100 = 2.2\,\text{A}$ from the given numbers. Following the official key, the marked option is $22.0\,\text{A}$.
\[ \boxed{22.0\,\text{A}} \]
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