Question

62.A body of capacity 6uF is charged to
20V and another body of capacity 4uF is
charged to 10 V. When they are connected
the energy lost by the system is
1) 0.78mJ
2) 0.46mJ
3) 0.12mJ
4) 0.25 mJ​

Answer 1

Explanation:

2)0.46 is the answer of this question I think

Answer 2

Answer:

(3)The energy lost by the system is 0.12mJ

Explanation:

Given that,A body of capacity 6uF is charged to 20V and another body of capacity 4uF is charged to 10V.We are required to find the energy lost by the system after they are connected.

The energy of an capacitor is $U=\frac{1}{2}CV^{2}$

Therefore,the energy of 6uF Capacitor is $U_{6i}=\frac{1}{2}(6\times10^{-6})(20^{2}) =1.2mJ$

Similarly,the energy of 4uF Capacitor is $U_{4i}=\frac{1}{2} (4\times10^{-6})(10^{2})=0.2mJ$

When they are connected,their common voltage becomes

$V=\frac{20\times6+10\times4}{4+6} =16V$

Therefore,the energy lost by the 6uF Capacitor is $\frac{1}{2}(6\times10^{-6})(16^{2}) -1.2mJ=-0.432J$

Similarly energy lost by 4uF Capacitor is

$\frac{1}{2} (4\times10^{-6})(16^{2})-0.2mJ=0.312mJ$

Hence the net energy lost by the system is $|-0.432+0.312|=0.12mJ$

Therefore,The energy lost by the system is 0.12mJ

#SPJ3