Question

In the circuit shown in figure switch $S$ is thrown to position $1$ at $t=0$. When the currect in the resistor is $1\text{A}$, it is shifted to position 2. The total heat generated in the circuit after shifting to position 2 is

• A. $\text{zero}$
• B. $625\mu \text{J}$
• C. $100\mu \text{J}$
• D. $\text{None of these}$

Answer 1

The correct option is C $100\mu \text{J}$

In position-1, initial maximum current is

$i_0=\frac{V}{R}=\frac{10}{5}=2\text{A}$

At the given time, current is $1\text{A}$ which is half of the above value. Hence, at this is instant capacitor is also charged to half of the final value of $5\text{V}$.
Now, it is shifted to position-2 where in steady state it is again charged to $5\text{V}$ but with opposite polarity.

$U_i=U_f=\frac{1}{2}C{V}^2\{\because V=5V\}$

So, total energy supplied by the lower battery is converted into heat. But double charge transfer (from the normal) takes place from this battery.

$\Rightarrow \text{Heat produced}=\text{Energy supplied by the battery}$

$\Rightarrow \Delta H=\left(\Delta q\right)V=\left(2CV\right)\left(V\right)=2C{V}^2$

$\Rightarrow \Delta H=2\times 2\times {10}^{-6}\times (5{)}^2$

$\Rightarrow \Delta H=100\times {10}^{-6}\text{J}$

$\Rightarrow \Delta H=100\mu \text{J}$

Hence, the correct answer is (C).