Two capacitors of capacitances \(0.3 \mu \mathrm{F}\) and \(0.6 \mu \mathrm{F}\) are connected in series…

Two capacitors of capacitances \(0.3 \mu \mathrm{F}\) and \(0.6 \mu \mathrm{F}\) are connected in series across 6 volts. The ratio of energies stored in them will be
  1. \(2: 1\)
  2. \(1: 2\)
  3. \(2: 3\)
  4. \(4: 1\)

Solution

$\begin{aligned} C_{1} &= 0.3 \mu F \\ C_{2} &= 0.6 \mu F \\ \frac{C_{1}}{C_{2}} &= \frac{0.3 \mu F}{0.6 \mu F} \\ &= \frac{1}{2} \end{aligned}$

Asked in: JEE Mains - Capacitance - Chapter Test

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