Question

Two capacitors c1=2uf and c2=6uf in series are connected in parallel to a third capacitor 4uf and then connected to a battery of emf 2v.How much energy is lost by the battery in charging the combination

Answer 1

The energy lost by the battery in charging the combination is $11\times10^-^6J$.

Explanation:

The equivalent capacitance of the combination $C_1$ and $C_2$ are in series and connected to the parallel to third capacitor $C_3$.

$C_e_q_u=\frac{C_1C_2}{C_1+C_2}+C_3$

Given $C_1=2\mu F,C_2=6\mu F, C_3=4\mu F$

So,

$C_{equ} =\frac{2\times6}{2+6}+4=5.5\mu F$

Now total charge flown in battery,

$Q=C_{equ}V$

$Q=5.5\mu F\times2V$

$Q=11\mu C$

In charging the energy lost by the battery is

 $U=\frac{1}{2}QV$

substitute the value, we get

$U=\frac{1}{2}\times11\times10^-^6F\times2V$

$U=11\times10^-^6J$.

Thus, the energy lost by the battery in charging the combination is $11\times10^-^6J$.

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