Question

Given two condensers, each of capacitance $5\mu F$ and battery of emf $240V$ What arrangement (series or parallel) using the condensers would give minimum energy? What is the minimum value?
A. Parallel Combination, ${E_{\min }} = 7.2 \times {10^{ - 2}}J$
B. Parallel Combination, ${E_{\min }} = 6.82 \times {10^{ - 2}}J$
C. Series Combination, ${E_{\min }} = 7.2 \times {10^{ - 2}}J$
D. Series Combination, ${E_{\min }} = 6.82 \times {10^{ - 2}}J$

Answer 1

Hint: In order to solve this question we need to understand the definition of capacitors. Capacitors are the device which is used to store energy in the form of charges on its plates. We will examine two cases of capacitor combination both parallel and series and then we will find energy in each case of combination of capacitors and will check which option is most correct.

Complete step by step answer:
Let us take the case of two capacitors connected in series.Let the voltage around first capacitor is given by formula: ${V_1} = \dfrac{{V{C_2}}}{{{C_1} + {C_2}}}$ where $V$ is emf of battery and ${C_1} = {C_2} = 5\mu F$ (Given)
$V = 240V$
Putting values on formula we get
${V_1} = \dfrac{{(240 \times 5)}}{{5 + 5}} = 120V$
Similarly around second capacitor voltage developed is: ${V_2} = \dfrac{{V{C_1}}}{{{C_1} + {C_2}}}$
Putting values we get
${V_2} = \dfrac{{(240 \times 5)}}{{5 + 5}} = 120V$
So Energy stored is given by formula $E = \dfrac{1}{2}C{V^2}$.
So for first capacitor it is ${E_1} = \dfrac{1}{2}{C_1}{V_1}^2$.

Putting values we get:
${E_1} = \dfrac{1}{2} \times (5 \times {10^{ - 6}}) \times {(120)^2}$
$\Rightarrow {E_1} = 3.6 \times {10^{ - 2}}J$
So for second capacitor it is ${E_2} = \dfrac{1}{2}{C_2}{V_2}^2$
Putting values we get:
${E_2} = \dfrac{1}{2} \times (5 \times {10^{ - 6}}) \times {(120)^2}$
$\Rightarrow {E_2} = 3.6 \times {10^{ - 2}}J$
Total energy stored is $E = {E_1} + {E_2}$
$E = (3.6 + 3.6) \times {10^{ - 2}} \\
\Rightarrow E= 7.2 \times {10^{ - 2}}J$
Now consider the case of two capacitors in parallel with each other.
Since they are in parallel combination so voltage across each is same ${V_1} = {V_2} = V = 240J$

So Energy stored is given by formula $E = \dfrac{1}{2}C{V^2}$.
So for the first capacitor it is ${E_1} = \dfrac{1}{2}{C_1}{V_1}^2$.
Putting values we get:
${E_1} = \dfrac{1}{2} \times (5 \times {10^{ - 6}}) \times {(240)^2}$
$\Rightarrow {E_1} = 14.4 \times {10^{ - 2}}J$
So for second capacitor it is ${E_2} = \dfrac{1}{2}{C_2}{V_2}^2$
Putting values we get:
${E_2} = \dfrac{1}{2} \times (5 \times {10^{ - 6}}) \times {(240)^2}$
$\Rightarrow {E_2} = 14.4 \times {10^{ - 2}}J$
Total energy stored is $E = {E_1} + {E_2}$
$E = (14.4 + 14.4) \times {10^{ - 2}} \\
\therefore E= 28.8 \times {10^{ - 2}}J$
So in both combination we see total energy in series combination is less than parallel combination

So the correct option is C.

Note:It should be remembered that net capacitance in parallel combination is greater than series combination and because of this parallel combination stores more energy than series so in general in household capacitive devices are connected in parallel in order to increase its efficiency. In series combination charge stored on each capacitor is same whereas in parallel combination capacitors voltage of each capacitor is same.