0.2 F capacitor is charge to 600 V by a battery, on removing the battery, it is connected with another parallel plate condenser of I F . The potential decreases to

Question

0.2 F capacitor is charge to 600 V by a battery, on removing the battery, it is connected with another parallel plate condenser of I F . The potential decreases to (A) 100 V (B) 120 V (C) 300 V (D) 600 V

Tardigrade Admin · Accepted Answer

By using charge conservation 0.2 × 600 = (0.2 + 1)V V = (0.2 × 600/1.2) = 100V

Q. 0.2 F capacitor is charge to 600 V by a battery, on removing the battery, it is connected with another parallel plate condenser of I F . The potential decreases to

100 V

120 V

300 V

600 V

By using charge conservation
$0.2 \times 600 = (0.2 + 1) V$
$V = \frac{0.2 \times 600}{1.2} = 100 V$

Q. 0.2 F capacitor is charge to 600 V by a battery, on removing the battery, it is connected with another parallel plate condenser of I F . The potential decreases to

100 V

120 V

300 V

600 V

By using charge conservation 0.2×600=(0.2+1)V V=1.20.2×600​=100V

By using charge conservation
$0.2 \times 600 = (0.2 + 1) V$
$V = \frac{0.2 \times 600}{1.2} = 100 V$