First slide

Capacitance

Question

A capacitor of capacitance C₁ = 4 $μ$ F is charged to V₁ = 80V and another capacitor of capacitance C = 6 $μ$ F is charged to V₂ = 30V. When they are connected together, the energy lost by the 4 $μ$ F capacitor is

Easy

A

7.8 mJ
B

4.6 mJ
C

3.2 mJ
D

2.5 mJ

Solution

$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} = 50 V ∴ Energy lost by 4 μF capacitor = \frac{1}{2} C_{1} V_{1}^{2} - \frac{1}{2} C_{1} V^{2} = 7 .8 mJ$

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