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Physics Question on Current Electricity

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

0.6 A

Explanation

Solution

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