Two capacitors of capacitances C1 and C2 are connected across 200 V power supply. The potential drop across C1 is 120 V. A capacitor of capacitance 2 μ F is connected in parallel with C1 and the potential drop across C2 becomes 160 V. What are the values of C1 and C2 in μ F ?

Question

Two capacitors of capacitances C1 and C2 are connected across 200 V power supply. The potential drop across C1 is 120 V. A capacitor of capacitance 2 μ F is connected in parallel with C1 and the potential drop across C2 becomes 160 V. What are the values of C1 and C2 in μ F ? (A) 0.4 and 0.6 (B) 0.6 and 0.4 (C) 0.8 and 1.2 (D) 1.2 and 0.8

Tardigrade Admin · Accepted Answer

Initially the charge is same on both capacitors. Therefore, 120 C1 = 80 C2. Afterwards (C1+2) 40 = 160 C2. This gives C1 = 0.4 μ F and C2 = 0.6 μ F.

Q. Two capacitors of capacitances C1​ and C2​ are connected across 200V power supply. The potential drop across C1​ is 120V.A capacitor of capacitance 2μF is connected in parallel with C1​ and the potential drop across C2​ becomes 160V. What are the values of C1​ and C2​ in μF ?

0.4 and 0.6

0.6 and 0.4

0.8 and 1.2

1.2 and 0.8

Initially the charge is same on both capacitors. Therefore, 120 C1​ = 80 C2​. Afterwards (C1​+2) 40 = 160 C2​. This gives C1​=0.4μF and C2​=0.6μF.

Initially the charge is same on both capacitors.
Therefore, 120 $C_{1}$ = 80 $C_{2}$ . Afterwards $(C_{1} + 2)$ 40 = 160 $C_{2}$ .
This gives $C_{1} = 0.4 μ$ F and $C_{2} = 0.6 μ$ F.