Question

A capacitor and a coil with resistance R are in series and connected to a 6 V AC source. By varying the frequency of the source, a maximum current of 600 mA is observed. If the same coil is now connected to a cell of emf 6 V and internal resistance of 2 ohms, the current through it will be:

• A. 0.6 A
• B. 0.5 A
• C. 1.0 A
• D. 2.0 A

Answer 1

The Correct Option is B

The problem is solved by first analyzing the given conditions and then applying relevant formulas to determine the current through the coil when connected to a cell.

Initially, when the capacitor and coil are in resonance with the AC source, maximum current occurs at the resonant frequency. At this frequency, the reactance of the coil and capacitor cancel each other out, making the impedance Z purely resistive and equal to R.

Under this condition:

\( V = I_{max} \cdot R \)

Given values:

• \( V = 6 \, \text{V} \)
• \( I_{max} = 600 \, \text{mA} = 0.6 \, \text{A} \)

Using the formula:

\( 6 = 0.6 \cdot R \)

Solving for R:

\( R = \frac{6}{0.6} = 10 \, \Omega \)

Subsequently, when the same coil is connected to a 6 V cell with an internal resistance of 2 ohms, the total resistance in the circuit is calculated as:

\( R_{total} = 10 \, \Omega + 2 \, \Omega = 12 \, \Omega \)

Ohm's Law is applied to find the current \( I \):

\( I = \frac{V}{R_{total}} = \frac{6 \, \text{V}}{12 \, \Omega} = 0.5 \, \text{A} \)

Therefore, the current through the coil when connected with the cell is 0.5 A.