A condenser of capacity 16 mF charged to a potential of 20 V is connected to a condenser of capacity C charged to a potential of 10 V as shown in the figure. If the common potential is 14 V, the capacity C is equal to

18 m F

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24 m F

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8 m F

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4 m F

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Solution

The correct option is D $4 m F$
Given, $C_{1} = 16 m F, V_{1} = 20 V C_{2} = C, V_{2} = 10 V$
The common potential, V=14V
$V = \frac{C_{1} V_{1} - C_{2} V_{2}}{C_{1} + C_{2}}$
$V = \frac{(60 \times 20 - C \times 10)}{16 + C}$
$14 (16 + C) = 320 - 10 C => 24 C = 320 - 224 => C = \frac{96}{24}$
C= 4mF

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A condenser of capacity 16 mF charged to a potential of 20 V is connected to a condenser of capacity C charged to a potential of 10 V as shown in the figure. If the common potential is 14 V, the capacity C is equal to

The correct option is D 4mFGiven, C1=16mF,V1=20VC2=C,V2=10VThe common potential, V=14VV=C1V1−C2V2C1+C2V=(60×20−C×10)16+C14(16+C)=320−10C=>24C=320−224=>C=9624C= 4mF