Question

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

• A. $40\mu \text{F}$ and $10\mu \text{F}$
• B. $60\mu \text{F}$ and $40\mu \text{F}$
• C. $50\mu \text{F}$ and $30\mu \text{F}$
• D. $20\mu \text{F}$ and $30\mu \text{F}$

Answer 1

The correct option is A $40\mu \text{F}$ and $10\mu \text{F}$

Given:
Graph between the series and parallel combination of the two capacitors. As we know that slope of the parallel combination will be more than the series combination.

Hence, from the given graph we can conclude the following,
Voltage $V=10\text{V}$
Parallel equivalent charge $q_1=500\mu \text{C}$
Series equivalent charge $q_2=80\mu \text{C}$

Also we know that,
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

$C^{\prime \prime }=C_1+C_2$

Now, using $q=CV$ for both the conditions

$\left(1\right)$ For parallel combination

$q_1=10(C_1+C_2)$

$q_1=500\mu \text{C}$

$500=10(C_1+C_2)$

$C_1+C_2=50$ ... $\left(i\right)$

For series combination

$q_2=10\frac{C_1{C}_2}{(C_1+C_2)}$

$80=10\frac{C_1{C}_2}{50}$ $($From equation ... $\left(i\right))$

$C_1{C}_2=400$ ... $\left(ii\right)$

From equation $\left(i\right)$ and $\left(ii\right)$

$C_1=10\mu \text{F}$

$C_2=40\mu \text{F}$

Hence option $\left(A\right)$ is correct.