Question

For a given ac supply voltage $60\sin\omega t$, the reactances offered by $L$ and $C$ are found to be $20,\Omega$ each. If the resistance of the resistor is also $20,\Omega$, then the current in the circuit is

• A. $1,A$
• B. $3,A$
• C. $2.12,A$
• D. $1.73,A$
• E. zero

Hint:

Physics Tip: When $X_L=X_C$, series RLC circuit is at resonance and impedance becomes only $R$.

Answer 1

The Correct Option is C

Concept:
In a series RLC circuit: $$Z=\sqrt{R^2+(X_L-X_C)^2}$$ where $Z$ is impedance.
Step 1: Given values.
Supply voltage: $$v=60\sin\omega t$$ So peak voltage: $$V_0=60V$$ Given: $$X_L=20\Omega,\quad X_C=20\Omega,\quad R=20\Omega$$
Step 2: Find impedance.
Since: $$X_L-X_C=20-20=0$$ Therefore, $$Z=\sqrt{20^2+0}=20\Omega$$ Circuit is at resonance.
Step 3: Calculate current.
Peak current: $$I_0=\frac{V_0}{Z}=\frac{60}{20}=3A$$ RMS current: $$I_{\text{rms}}=\frac{I_0}{\sqrt2}=\frac{3}{\sqrt2}=2.12A$$
Hence correct answer is Option (C). :contentReference[oaicite:0]{index=0}