Question

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

Answer 1

Given C=5.0$\mu F$, charge q=20$\mu C$ and resistance R=5.0$\Omega$, time interval t=25-5=20$\mu s$

Heat dissipated in the wire in time period from 5 to 25 $\mu$s

$H={\int }_{t_1}^{t_2}PRdt$--------------(1)

Also power $P=\left(\frac{V_0}{R}\right)e^{\frac{-2t}{RC}}$----------(2)

Now 2 in 1 we get,

$H={\int }_{t_1}^{t_2}\left(\frac{V_0}{R}\right)e^{\frac{-2t}{RC}}Rdt$

$=V_0^{2}R|{(-\frac{1}{2RC})e^{\frac{-2t}{RC}}|}_5^{25}$

$V_0^{2}R|(-\frac{1}{2RC})e^{\frac{-2t_2}{RC}}-(-\frac{1}{2RC})e^{\frac{-2t_1}{RC}}|$

$=5V_0^2|(-\frac{1}{50})e^{\frac{-2\times 25}{25}}-(-\frac{1}{50})e^{\frac{-2\times 5}{25}}|$

$=5V_0^2|-\frac{0.135}{50}+\frac{0.670}{50}|$

$H=0.0535V_0^2$-----------(3)

Now, current $I=\frac{q}{t}$

$I=\frac{20}{20}=1A$

Again V=IR=1$\times$ 5=5V----------(4)

Putting 4 in 3

$H=0.0535\left(5^2\right)$

$H=1.3375J$