CBSE Class 12 · 2017 · CapacitanceShort Answer

A capacitor of unknown capacitance is connected across a battery of

V

volt. A charge of

360 μ C

is stored in it. When the potential across the capacitor is reduced by 120 V, the charge stored in the capacitor becomes

120 μ C

. Calculate

V

and the unknown capacitance. What would have been the charge on the capacitor if the voltage were increased by 120 V?

Answer: $V = 180 V$ , $C = 2 μ F$ ; if the voltage is increased by 120 V (to 300 V), $Q = 600 μ C$ .

Solution

The same capacitor gives two charge–voltage readings:

C = \frac{360 μ C}{V} = \frac{120 μ C}{V - 120}

Cross-multiplying:

360 (V - 120) = 120 V

3 (V - 120) = V \Rightarrow 3 V - 360 = V

V = 180 V

Then the capacitance:

C = \frac{360 μ C}{180 V} = 2 μ F

V + 120 = 300 V

Q = C V = 2 \times 300 = 600 μ C

CBSE Class 12 Board · 2017 · Subjective