Capacitors

JEE Main level, single correct, medium

49 of 180

Question 49 of 180

A 10μF10\mu F capacitor is fully charged to a potential difference of 50V50V After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20V20V. The capacitance of the second capacitor is :

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Answer: (A) 15μF15\mu F

V=C1V1+C2V2C1+C2V=\dfrac{{C}_{1}{V}_{1}+{C}_{2}{V}_{2}}{{C}_{1}+{C}_{2}}
20=10×50+020+C20=\dfrac{10\times 50+0}{20+C}
C=15μFC=15\mu F

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