Question

A capacitor of capacitance $2\mu F$ is charged to a potential difference of 5 V. Now the charging battery is disconnected and the capacitor is connected in parallel to a $5\Omega$ and R resistance as shown in figure. If total heat produced in $5\Omega$ resistance is $10\mu J$, then R is

• A. $10\Omega$
• B. $7.5\Omega$
• C. $3.33\Omega$
• D. $2.5\Omega$

Answer 1

The correct option is C $3.33\Omega$

Total Energy = $\frac{1}{2}C{V}^2=\frac{1}{2}\times 2\times 5^2=25\mu J$
Let's call the $5\Omega$ resistor as '$r$'.
Heat across $r$ is = 10 $\mu$J,
So heat across R is = 15 $\mu$J
Ratio of heat across r and R is $\frac{i_r^{2}r}{i_R^{2}R}$
and $\frac{i_r}{i_R}=\frac{R}{r}$
Threfore,
$\frac{R^{2}r}{r^{2}R}=\frac{10}{15}$
$\Rightarrow \frac{R}{r}=\frac{2}{3}$
$\Rightarrow R=r\frac{2}{3}=3.33\Omega$