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\[\mu \operatorname{C}\] |
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