A condenser of capacitance 12 μ F was initially charged to 20 V. Now potential difference is made 40 V. The increase in potential energy is

Question

A condenser of capacitance 12 μ F was initially charged to 20 V. Now potential difference is made 40 V. The increase in potential energy is (A) 7.2 × 10-3 J (B) 4 × 10-3 J (C) 3 × 10-4 J (D) 5 × 10-6 J

Tardigrade Admin · Accepted Answer

Δ E =(1/2) CV 22-(1/2) CV 12 ∴ Δ E =(1/2) C ( V 22- V 12) ∴ Δ E =(1/2) × 12 × 10-6((40)2-(20)2) Δ E =6 × 10-6(1600-400) Δ E =6 × 10-6(1200) Δ E =72 × 10-4 J ≃ 7.2 × 10-3 J

Q. A condenser of capacitance $12 μ F$ was initially charged to $20 V$ . Now potential difference is made $40 V$ . The increase in potential energy is

$7.2 \times 1 0^{- 3} J$

$4 \times 1 0^{- 3} J$

$3 \times 1 0^{- 4} J$

$5 \times 1 0^{- 6} J$

Q. A condenser of capacitance 12μF was initially charged to 20V. Now potential difference is made 40V. The increase in potential energy is

7.2×10−3J

4×10−3J

3×10−4J

5×10−6J

ΔE=21​CV22​−21​CV12​ ∴ΔE=21​C(V22​−V12​) ∴ΔE=21​×12×10−6((40)2−(20)2) ΔE=6×10−6(1600−400) ΔE=6×10−6(1200) ΔE=72×10−4J ≃7.2×10−3J

Q. A condenser of capacitance $12 μ F$ was initially charged to $20 V$ . Now potential difference is made $40 V$ . The increase in potential energy is

$7.2 \times 1 0^{- 3} J$

$4 \times 1 0^{- 3} J$

$3 \times 1 0^{- 4} J$

$5 \times 1 0^{- 6} J$