Figure shows charge ( q ) versus voltage ( V ) graph for series a…

MCQ

Figure shows charge $( q )$ versus voltage $( V )$ graph for series and parallel combination of two given capacitors. The capacitances are:

A
$60 \mu F$ and $40 \mu F$
B
$40 \mu F$ and $10 \mu F$
C
$50 \mu F$ and $30 \mu F$
D
$20 \mu F$ and $30 \mu F$

✓

Answer

B. $40 \mu F$ and $10 \mu F$
Explanation:
Equivalent capacitance in series combination $\left( C ^{\prime}\right)$ is given by
$
\frac{1}{C^{\prime}}=\frac{1}{C_1}+\frac{1}{C_2} \Rightarrow C^{\prime}=\frac{C_1 C_2}{C_1+C_2}
$
For parallel combination equivalent capacitance
$
C^{\prime}=C_1+C_2
$
For parallel combination
$
\begin{array}{l}
q=10\left(C_1+C_2\right) \\
q_1=500 \mu C \\
500=10\left(C_1+C_2\right) \\
C_1+C_2=50 \mu F \ldots \text { (i) }
\end{array}
$
For Series Combination-
$
q_2=10 \frac{C_1 C_2}{\left(C_1+C_2\right)}
$
$
\begin{array}{l}
80=10 \frac{C_1 C_2}{50} \text { From equation }\ldots\text{(ii)} \\
C_1 C_2=400 \ldots \text { (iii) }
\end{array}
$
From equation (i) and (ii)
$
\begin{array}{l}
C_1=10 \mu F, C_2=40 \mu F \\
=40 \mu F \text { and } 10 \mu F
\end{array}
$

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