Two condensers of capacity 0.3 μ F and 0.6 μ F, respectively, are connected in series. The combination is connected across a potential of 6 volts. The ratio of energies stored by the condensers will be

Question

Two condensers of capacity 0.3 μ F and 0.6 μ F, respectively, are connected in series. The combination is connected across a potential of 6 volts. The ratio of energies stored by the condensers will be (A) 1/2 (B) 2/1 (C) 1/4 (D) 4/1

Tardigrade Admin · Accepted Answer

In series combination Q is constant, hence according to U=(Q2/2 C) ⇒ U propto (1/C) ⇒ (U1/U2)=(C2/C1)=(0.6/0.3)=(2/1)

Q. Two condensers of capacity $0.3 μ F$ and $0.6 μ F$ , respectively, are connected in series. The combination is connected across a potential of 6 volts. The ratio of energies stored by the condensers will be

1/2

2/1

1/4

4/1

In series combination $Q$ is constant, hence according to
$U = \frac{Q ^{2}}{2 C}$
$\Rightarrow U \propto \frac{1}{C}$
$\Rightarrow \frac{U _{1}}{U _{2}} = \frac{C _{2}}{C _{1}} = \frac{0.6}{0.3} = \frac{2}{1}$

Q. Two condensers of capacity 0.3μF and 0.6μF, respectively, are connected in series. The combination is connected across a potential of 6 volts. The ratio of energies stored by the condensers will be

1/2

2/1

1/4

4/1

In series combination Q is constant, hence according to U=2CQ2​ ⇒U∝C1​ ⇒U2​U1​​=C1​C2​​=0.30.6​=12​

Q. Two condensers of capacity $0.3 μ F$ and $0.6 μ F$ , respectively, are connected in series. The combination is connected across a potential of 6 volts. The ratio of energies stored by the condensers will be

In series combination $Q$ is constant, hence according to
$U = \frac{Q ^{2}}{2 C}$
$\Rightarrow U \propto \frac{1}{C}$
$\Rightarrow \frac{U _{1}}{U _{2}} = \frac{C _{2}}{C _{1}} = \frac{0.6}{0.3} = \frac{2}{1}$