A $0.25\mu F$capacitor is charged to$50V$It is then connected in series with a $25\Omega$ resistor and a $100\Omega$ resistor and allowed to discharge completely. How much energy is dissipated by the $25\Omega$ resistor?

We need to find how much energy is dissipated by the $25\Omega$resistor.

When the capacitor is fully charged, the energy is stored due to it's capacitance and stored energy is related to the voltage across the capacitor in form of,

$E=\frac{1}{2}C{V}^2$ $equation\left(1\right)$

Now, Let us plug the values for C and V into equation(1) to get E

$$\begin{gathered} E=\frac{1}{2}C{V}^2 \\ =\frac{1}{2}(0.25\mu F){\left(50V\right)}^2=312.5\mu J \end{gathered}$$

When the current flows through a resistor, it dissipates energy, the rate of the dissipated energy is the power. this rate where the energy is transferred from the current to resistor is

$P_R=I^{2}R$ $equation\left(2\right)$

As shown by equation (1), the dissipated power increases as the value of the resistance increases in this case, the resistor$100\Omega$dissipates more energy than $25\Omega$.the capacitor is fully discharged which means all its stored energy dissipates through the two resistors

$E_{100\Omega }+E_{25\Omega }=312.5\mu J$

The dissipated energy by resistor $100\Omega$is four times the dissipated energy by $25\Omega$because the resistor is four times and current is the same.

$E_{100\Omega }=4E_{25\Omega }$

Therefore, Let us use this expression into equation $\left(3\right)$to get$E_{25\Omega }$

$$\begin{gathered} E_{100\Omega +}{E}_{25\Omega }=312.5\mu J \\ 4E_{25\Omega +}{E}_{25\Omega }=312.5\mu J \\ {E}_{25\Omega }=62.5\mu J \end{gathered}$$.