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JEE Main 20202 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
2 September
Shift
Evening Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2020 · 2 September · Evening Shift · Physics · Capacitance

A 10μ F capacitor is fully charged to a potential difference of 50V...

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 :

Options

  1. A15μF15\mu F
  2. B30μF30\mu F
  3. C20μF20\mu F
  4. D10μF10\mu F

Correct answer

A. 15μF15\mu F

Step-by-step solution

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=15\muFC = 15\muF
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