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

Answer 1

The Correct Option is B

In this scenario, the coil is connected to a DC source with internal resistance. The current can be found using Ohm’s law:
\(I = \frac{V}{R + r}\)
\(\text{where: - } V = 6 \, \text{V}, \, R \text{ is the coil's resistance, - } r = 2 \, \Omega \text{ is the internal resistance of the cell.}\\\)
\(\text{The current is calculated as } I = \frac{6}{R + 2}, \text{ which gives } I = 0.6 \, \text{A}.\)