Question

Several capacitors are connected as shown in the given figure. If the charge on the $5\mu \text{F}$ capacitor is $120\mu \text{C}$ , the potential between points $A$ and $D$ is

• A. $24\text{V}$
• B. $96\text{V}$
• C. $16\text{V}$
• D. $0\text{V}$

Answer 1

The correct option is C $16\text{V}$

$\text{STEP-1: Analysis of given data/diagram in the question.}$

Let, $C_1=3\mu \text{F},C_2=5\mu \text{F},C_3=4\mu \text{F},C_4=4\mu \text{F},C_5=30\mu \text{F},C_6=4\mu \text{F},C_7=2\mu \text{F}$

From the data given in the question,

Potential drop across the $5\mu \text{F}$ capacitor $V_5=\frac{120}{5}=24\text{V}$


Thus, from the figure we can conclude that, potential difference across all three capacitors connected in parallel is $V^{\prime }=24\text{V}$

$\text{STEP-2: Finding the relation between}{V}^{\prime }\text{and potential difference between A and B}$

Let $V_{AB}$ be the potential difference across points $A$ and $B$ given in the figure.

Since, the charge passing through the equivalent capacitance $C^{\prime }$ and $C_4$ is same. So, $$V^{\prime }=V_{AB}\left(\frac{C^{\prime }+C_4}{C_4}\right)$$$\Rightarrow V_{AB}=24[\frac{12+4}{4}]=96\text{V}$

$\text{STEP-3: Using}{V}_{AB}\text{to find}{V}_{AD}$


Since, the charge passing through $C_5$ and equivalent capacitance $C^{\prime \prime }$ of $C_6$ and $C_7$ is same.

we can say that, $V_{AD}=V_{AB}\left(\frac{C^{\prime \prime }}{C_5+C^{\prime \prime }}\right)$

Substituting the given data we get

$V_{AD}=96[\frac{4+2}{30+4+2}]=16\text{V}$

Hence, option (c) is the correct answer.