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The effective capacitance of parallel combination of two capacitors C1 and C2 is 10\mu F. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor C2 is 4 times that of C1. If these capacitors are connected in series, their effective capacitance will be (in \ \mu F): 
Option: 1 1.6
Option: 2 8.4
Option: 3 3.2
Option: 4 4.2
 

Answers (1)

best_answer

When capacitors are in parallel, 

C_{eq}=C_1+C_2

So, C_1+C_2=10 \mu F

Now when they are individually connected to 1V battery, then 

The energy of C2=4 x energy of C1

\Rightarrow \frac{1}{2}C_2V^2=4\times \frac{1}{2}C_1V^2

\Rightarrow C_2=4C_1

So, C2=8 and C1=2

So when both are in series, then C_{eq}=\frac{C_1C_2}{C_1+C_2}=\frac{8}{5}=1.6 \ \mu F

So the correct option is 1.

Posted by

vishal kumar

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