Ncert Exercises & Solutions

A capacitor of 4 $μF$ $μF$ is connected as shown in the circuit. The internal resistance of the battery is 0.5 $Ω$ $Ω$ . The amount of charge on the capacitor plates will be:

1. 0 $μC$ $μC$

2. 4 $μC$ $μC$

3. 16 $μC$ $μC$

4. 8 $μC$ $μC$

NEETprep Answer:

(D) Hint: Considering the steady-state condition, there will be no current through 10

Ω

$Ω$ resistance.

Step 1: Find the potential difference across the capacitor.

Current flowing through 2

Ω

$Ω$ resistance from left to right, is given by:

I = \frac{V}{R + r} = \frac{2.5}{2 + 0.5} = 1 A

$I = \frac{V}{R + r} = \frac{2.5}{2 + 0.5} = 1 A$
The potential difference across 2

Ω

$Ω$ resistance, V= IR = 1x2 = 2 V
Since the capacitor is in parallel with 2

Ω

$Ω$ resistance, so it also has a 2 V potential difference across it.

Step 2: Find the charge on the capacitor.
The charge on the capacitor,
Q=CV=(2

μF

$μF$ )x2 V = 8

μC

$μC$

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