Question

An inductor coil when connected to a $12V$ batt.ery draws a steady current $6A$. This coil when connected in series with a capacitor and an $AC$ source of $10V$, current is in phase with applied voltage. $AC$ meter included in the circuit reads

• A. 5 A
• B. 10 A
• C. 2 A
• D. 1 A

Answer 1

Answer: (A)

Resistance of the coil $=\frac{\text{Battery voltage}}{\text{Steady current}}=\frac{12}{6}=2\Omega$
After connecting capacitor, circuit becomes as shown in the figure, and this figure is in resonance.
Impedance in the circuit, $Z=R=2\Omega V=10V$
Current at resonance, $I=\frac{V}{R}=\frac{10}{2}=5A$