question_answer

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:               [NCERT Exemp. Q. 2.1, Page 10]

A) 0\[\mu \operatorname{C}\]

B) 4\[\mu \operatorname{C}\]

C) 16\[\mu \operatorname{C}\]

D) 8\[\mu \operatorname{C}\]

Correct Answer: D

Solution :

             

Option [d] is correct.

Explanation: As capacitor offer infinite resistance for DC circuit So current from cell will not flow across branch of 4\[\mu \operatorname{F}\]and 10\[\Omega \]. So current will flow across 2 ohm branch.

So Potential Difference (PD) across 2 \[\Omega \]  resistance V=RI=2\[\times \]1=2 Volt. As battery, capacitor and 2 branches are in parallel.

So PD will remain same across all three branches. As current does not flow through capacitor branch, so no potential drop will be across 10\[\Omega \]. So PD across 4 \[\mu \operatorname{F}\] capacitor = 2 Volt Q = CV=2\[\mu \operatorname{F}\]\[\times \]2V= 8\[\mu \operatorname{C}\]