Question

If at $t=0$, switch $\text{S}$ is closed, at steady state find heat generated $\left(H\right)$ in the given circuit having a battery of $16\text{volts}$.

• A. $640\mu \text{J}$
• B. $320\mu \text{J}$
• C. $1280\mu \text{J}$
• D. $192\mu \text{J}$

Answer 1

The correct option is A $640\mu \text{J}$

Let us solve this problem using potential method, assigning potential to the known points, as shown in figure.


Using Kirchoff`s law at junction $\text{P}$,

$(V_0-16)\times 10+(V_0-0)\times 4+(V_0-0)\times 6=0$

$\Rightarrow 10V_0-160+4V_0+6V_0=0$

$\Rightarrow 20V_0=160$

$\therefore V_0=8\text{V}$

Using $V_0$ we can get potential across $C_1,$ $C_2$ and $C_3$ as follows,

Potential difference across $C_1$ is,

$\Rightarrow V_1=(16-8)=80\text{V}$

Similarly $V_2$ and $V_3$ as,

$V_2=(8-0)=8\text{V}$

$V_3=(8-0)=8\text{V}$

Now using $V_1,$ $V_2$ and $V_3$ we can calculate energy stored in each capacitor as follows,

$U_1=\frac{1}{2}{C}_{1}V{{}_1}^2=\frac{1}{2}(10\times {10}^{-6})\left(64\right)=320\mu \text{J}$

$U_2=\frac{1}{2}{C}_{2}V{{}_2}^2=\frac{1}{2}(4\times {10}^{-6})\left(64\right)=128\mu \text{J}$

$U_3=\frac{1}{2}{C}_{3}V{{}_3}^2=\frac{1}{2}(6\times {10}^{-6})\left(64\right)=192\mu \text{J}$

Work done by battery:

$W_B=\left(Q_{throughcircuit}\right)\left(V_{Battery}\right)$

$Q_{throughcircuit}=Q_{across{C}_1}=(10\times {10}^{-6})\left(8\right)=80\mu \text{C}$

$\Rightarrow W_B=\left(80\mu \text{C}\right)\left(16V\right)=1280\mu \text{J}$

As we know that,
$W_B=U_{system}+H$

$\Rightarrow H=W_B-U_{system}$

$\Rightarrow H=W_B-(U_1+U_2+U_3)$

$\Rightarrow H=1280-[320+128+192]$

$\therefore H=640\mu \text{J}$

Hence, option (b) is the correct answer.