Three capacitors each having capacitance `C=2 muF` are connected with a battery of emf`30 V` as shown in. when the switch S is closed, find
a. the amount of charge flowing through the battery ,
b. the heat generated in the circuit,
c. the energy supplied by the battery,
d. the amount of charge flowing through the switch S.
.

Answer

a. `20 muC` b. `0.3mJ` c. `0.6 mJ` d. 60 muC`

Explanation

Before closing the switch
`C_(eq)=(2CxxC)/(2C+C)=2/3C=2/3xx2xx4/3 muF`
`:. q_(i)=C_(eq)V=4/3xx30=40 muC`
The energy is `U_(i)=1/2C_(eq)V^(2)=6xx10^(-4) J`
After closing the switch
`C'_(eq)=C` {Since left capacitors are short-circuited}
`:.` `V=CV=2xx30=60 muC`
and `U_(f)=1/2C'_(eq)V^(2)=1/2CV^(2)=9xx10^(-9) J`
a. Deltaq=q_(f)-q_(i)=60-40=20 muC`
b. `W_(cell)=DeltaU+DeltaH or VDeltaq=U_(f)U_(i)+DeltaH`
:. DeltaH=6xx10^(-4)-3xx10^(-4)=3xx10^-4) J=0.3 mJ`
c. `W_(cell)=VDeltaq=30xx20xx10^(-6)=6xx10^(-4)=0.6 mJ`
d. Their charge flowon through switch`=`charge supplied by battery+charges flowing for neutralization of leftward capacitors
`=20+20+20=60 muC`
{Since charges on each leftward capacitors are `20muC` before short circuiting}`