Question

A $5\mu F$capacitor is charged fully by a $220V$ supply. It is then disconnected from the supply and is connected in series to another uncharged $2.5\mu F$ capacitor. If the energy change during the charge redistribution is $(X/100)J$ then value of $X$ to the nearest integer is _________.

Answer 1

Step 1: Given data

Value of capacitors, $C_1\left(C\right)=5\mu F$

$C_2(C/2)=2.5\mu F$

Value of initial potential, $V_1=220V$

Value of final potential, $V_2=0V$

energy change during the charge redistribution, $∆H=(X/100)J$

Step 2: Find the change in energy,$\mathbf{∆}\mathit{H}$

Heat loss, $\mathrm{\Delta }H=U_i–U_f=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}(V_1-V_2{)}^2$

$=\frac{1}{2}\times \frac{5\times 2.5}{(5+2.5)}(220-0{)}^2\mu J$

$=\frac{5\times 11\times 22}{3}\times {10}^{-4}J$

$=4\times {10}^{-2}J$

Step 3: Compare the value of change in energy, ($\mathbf{∆}\mathit{H}$) with the given value

On comparing we will get, $X=4$

Hence, Value of $X$ to the nearest integer is 4.