Question

An inductor coil when connected to a $12 \, V$ batt.ery draws a steady current $6 \, A$. This coil when connected in series with a capacitor and an $AC$ source of $10 \, V$, 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)

5 A

Answer 2

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, $Z = R = 2 \, \Omega \, V = 10 \, V$
Current at resonance, $I = \frac{V}{R} = \frac{10}{2} = 5 \, A$