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

Question

A capacitor of capacitance 4 μ F is charged to 80 V and another capacitor of capacitance 6 μ F is charged to 30 V. When they are connected together, the energy lost by the 4 μ F capacitor is (A) 7.8 mJ (B) 4.6 mJ (C) 3.2 mJ (D) 2.5 mJ

Tardigrade Admin · Accepted Answer

Common potential, V=(C1 V1+C2 V2/C1+C2) =((4 × 10-6) × 80+(6 × 10-6) × 30/4 × 10-6+6 × 10-6)=50 V ∴ Energy lost by 4 μ F capacitor is =(1/2) C1 V12-(1/2) C1 V2= (1/2) C1(V12-V2) =(1/2) ×(4 × 10-6) × (80)2-(50)2 =7.8 × 10-3 J =7.8 mJ

Q. A capacitor of capacitance 4μF is charged to 80V and another capacitor of capacitance 6μF is charged to 30V. When they are connected together, the energy lost by the 4μF capacitor is

7.8 mJ

4.6 mJ

3.2 mJ

2.5 mJ

Common potential, V=C1​+C2​C1​V1​+C2​V2​​ =4×10−6+6×10−6(4×10−6)×80+(6×10−6)×30​=50V ∴ Energy lost by 4μF capacitor is =21​C1​V12​−21​C1​V2= 21​C1​(V12​−V2) =21​×(4×10−6)×{(80)2−(50)2} =7.8×10−3J=7.8mJ

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