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

To solve the problem, we need to first analyze the given conditions and then apply the relevant formulas to find the current through the coil when connected with a cell. 

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

At this condition:

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

Given:

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

Using the formula:

\( 6 = 0.6 \cdot R \)

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

Now, when the same coil is connected to a cell of 6 V with internal resistance 2 ohms, the total resistance in the circuit is given by:

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

Applying Ohm's Law to find the current \( I \), we have:

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

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

Answer 2

Given AC circuit scenario:

Series RLC circuit with 6V AC source shows maximum current (I_max) = 600 mA at resonance.

Step 1: Determine coil resistance (R)

At resonance (maximum current): \[ I_{max} = \frac{V}{Z} = \frac{V}{R} \] \[ 0.6 = \frac{6}{R} \Rightarrow R = 10\ \Omega \]

Step 2: Analyze DC scenario

When connected to 6V DC source with internal resistance (r) = 2Ω: \[ I = \frac{V}{R + r} = \frac{6}{10 + 2} = 0.5\ \text{A} \]

Final answer: The current through the coil will be 0.5 A.