Question

If two capacitors C₁ & C₂ are connected in parallel then equivalent capacitance is 10μF. If both capacitance are connected across 1V battery then energy stored by C₂ is 4 times of C₁. Then the equivalent capacitance if they are connected in series is–

(1) 1.6μF 

(2) 16μF 

(3) 4μF 

(4) 1/4μF

Answer 1

Answer is (1)

Given C₁ + C₂ = 10μF    ....(i)

4(1/2C₁V²) = 1/2C₂V²

⇒ 4C₁ = C₂    ....(ii)

from equation (i) & (ii)

C₁ = 2μF

C₂ = 8μF

If they are in series

C_eq = {C₁C₂}/{C₁ + C₂} = 1.6μF