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Electrostatics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsElectrostatics4 Marks
Question
Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitances in the ratio 1 : 2, so that the energy stored in these two cases becomes the same.
✓

Answer

Data $: \frac{C_1}{C_2}=\frac{1}{2}, U_1$ (for parallel) $=U_2$ (for series)
$
\frac{C_1}{C_2}=\frac{1}{2} \quad \therefore C_2=2 C_1
$
For the parallel combination of $C_1$ and $C_2$,
$
C_p=C_1+C_2=3 C_1
$
and charged to a potential $V_1$, the energy stored is
$U_1=\frac{1}{2} C_{ p } V_1^2=\frac{3}{2} C_1 V_1^2$
For the series combination of $C_1$ and $C_2$,
$
C_5=\frac{C_1 C_2}{C_1+C_2}=\frac{2 C_1^2}{3 C_1}=\frac{2}{3} C_1
$
and charged to a potential $V_2$, the energy stored is
$
\begin{array}{c}
U_2=\frac{1}{2} C_s V_2^2=\frac{1}{3} C_1 V_2^2 \\
\therefore \text { For } U_1=U_2, \frac{3}{2} C_1 V_1^2=\frac{1}{3} C_1 V_2^2 \\
\therefore\left(\frac{V_1}{V_2}\right)^2=\frac{2}{9} \quad \therefore \frac{V_1}{V_2}=\frac{\sqrt{2}}{3}=\frac{1.414}{3}=0.471
\end{array}
$
This gives the required ratio.

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