Question

A capacitor of $4\mu F$ is connected as shown in the circuit (figure). The internal resistance of the battery is $0.5\Omega$. The amount of charge on the capacitor plates will be

• A. $8\mu C$
• B. $0$
• C. $16\mu C$
• D. $4\mu C$

Answer 1

The correct option is A $8\mu C$

Step 1: Find the current flowing through $2\Omega$ resistance.



Given, capacitance of capacitor $=4\mu F$
Internal resistance of the battery is $0.5\Omega$.

Initially the capacitor gets charged. When its charged, no current will flow through the branch with the capacitor and $10\Omega$ resistor.

As internal resistor of battery is $0.5\Omega$,

so, the current through the $2\Omega$ resistor will be

$i=\frac{V}{R}=\frac{2.5}{2+0.5}=1A$

Or $i_{cd}=1A$

Step 2: Find the voltage across $CD$.

As we can see that the branches $AB$ and $CD$ are connected in parallel, the voltage across $AB$ and $CD$ is same.

Voltage across $CD$

$V_{CD}=i_{CD}\times R_{CD}=1\times 2=2V$

Here the voltage across $AB$ will be only because of the capacitor because no current will flow through $10\Omega$ resistor.

Step 3: Find the charge on capacitor

The charge on capacitor is

$q_C=V_{AB}\times C$

$=(2\times 4)\mu C$

$q_C=8\mu C$

Final Answer: (d) 8 μC