Question

Find potential drop across capacitors $C_2$, $C_4$ and $C_5$ in the given circuit?

• A. $0.6\text{V},3.5\text{V},1.5\text{V}$
• B. $3\text{V},0.6\text{V},15\text{V}$
• C. $15\text{V},3.5\text{V},9\text{V}$
• D. $1.5\text{V},12\text{V},9\text{V}$

Answer 1

The correct option is A $0.6\text{V},3.5\text{V},1.5\text{V}$

We can solve this problem by potential method.

Assigning potential to known point and identifying isolated parts in the given circuit and applying Kirchoff`s junction law to it as shown in figure.


For isolated system $\text{I}$:

$(x-10)\times 6+(x-y)\times 10+(x-5)\times 8=0$

$\Rightarrow 6x-60+10x-10y+8x-40=0$

$\Rightarrow 24x-10y-100=0......\left(1\right)$

For isolated system $\text{II}:$

$(y-5)\times 14+(y-x)\times 10+(y-0)\times 12=0$

$\Rightarrow 14y-70+10y-10x+12y=0$

$\Rightarrow 36y-10x-70=0......\left(2\right)$

Solving equations (1) and (2): Multiplying equation $\left(2\right)$ by $2.4$ and adding to equation $\left(1\right)$, we get

$y=3.5\text{V}$

Substituing $y$ value in equation (1)

$24x=100+10\times \left(3.5\right)$

$x=\frac{135}{24}=5.6\text{V}$

Using the value of $x$ and $y$ we can get the potential drop across capacitors $C_2,C_4$ and $C_5$ as follows.

$V_2=(5.6-5)=0.6\text{V}$

$V_4=(3.5-0)=3.5\text{V}$

$V_5=(5-3.5)=1.5\text{V}$

Hence, option (c) is the correct answer.