Question

A $5.0\mu \text{F}$ capacitor having a charge of $20\mu \text{C}$ is discharged through a wire of resistance of $5.0\Omega$. Find the heat dissipated in the wire between $25\mu \text{s}$ to $50\mu \text{s}$ after the connections are made. (Take, $1/e^2=0.135)$

• A. $4.7\mu \text{J}$
• B. $3.7\mu \text{J}$
• C. $5.7\mu \text{J}$
• D. $2.7\mu \text{J}$

Answer 1

The correct option is A $4.7\mu \text{J}$

Given,
$C=5\mu \text{F}$, $Q_0=20\mu \text{C}$, $R=5\Omega$ , $t_1=25\mu \text{s}$ and $t_2=50\mu \text{s}$

While discharging, charge on the capacitor after time $t$ is,$$Q=Q_0{e}^{\frac{-t}{RC}}$$Substituting the values,

$Q=\left[20e^{\frac{-t}{5\times 5}}\right]$

So, $t_1=25\mu \text{s}$, charge

$Q_{25}=20e^{-\frac{25}{25}}=\frac{20}{e}\mu \text{C}$

and at $t_2=50\mu \text{s}$, charge

$Q_{50}=\frac{20}{e^2}\mu \text{C}$

As we know,
$\text{Heat dissipated = [Initial energy-Final energy] of capacitor}$

$\Rightarrow H=\frac{1}{2C}[Q_{25}^2-Q_{50}^2]$

$\Rightarrow H=\frac{400}{2\times 5\times e^2}[1-\frac{1}{e^2}]\mu \text{J}$

$\therefore H\approx 4.7\mu \text{J}$

Hence, option $\left(a\right)$ is correct.