Question

If $100\text{volts}$ of potential difference is applied between ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ in the circuit shown in the figure. The potential difference between '$\text{c}$' and '$\text{d}$' will be (Given $V_a>V_b)$.

• A.
  $\frac{200}{3}\text{V}$
• B.
  $-\frac{100}{3}\text{V}$
• C.
  $\frac{100}{3}\text{V}$
• D.
  $-50\text{V}$

Answer 1

The correct option is C

$\frac{100}{3}\text{V}$
From the circuit given, we can see that the three $6\mu \text{F}$ capacitors connected in upper branch are in series.

Now simplifying the circuit and redrawing it,


Since the combination of three $6\mu \text{F}$ capacitors are connected in parallel with battery. So,

Sum of potential difference across $6\mu \text{F}$ capacitors = potential difference of battery

Since, all the three capacitors are identical, hence

$V_1=V_2=V_3=V^{\prime }$

$\Rightarrow V_1+V_2+V_3=100V$

$\Rightarrow 3V^{\prime }=100\text{V}$

$\therefore V^{\prime }=\frac{100}{3}\text{V}$

From the shown figure, direction of flow of positive charge it is clear that $V_c>V_d$

$\Rightarrow V_c-V_d=\frac{+100}{3}\text{V}$

$\text{Alternative solution:}$

Let the three $6\mu \text{F}$ capacitors be replaced an equivalent capacitor

$\frac{1}{C_{eq}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$

$\Rightarrow C_{eq}=\frac{6}{3}=2\mu F$



Charge flow through the $2\mu \text{F}$, $q_1=100\times 2=200\mu \text{C}$

The potential difference,

$V_c-V_d=\frac{\left(q_1\right)}{6\mu \text{F}}$

$\Rightarrow V_c-V_d=\frac{200\mu \text{C}}{6\mu \text{F}}=\frac{100}{3}\text{V}$

Hence, option (c) is correct answer.