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In the circuit shown , the emf of each b...

In the circuit shown , the emf of each battery is `60 V` and `C_(1)=2 muF` and `C_(2)=3 muF`. Find the charges that will flow through the sections `1 , 2` and `3` after the Key is closed.
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Text Solution

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Before closing the switch , both capacitors are in series. Their equivalent capacitance is
`C_(eq)=(2xx3)/(2+3)=6/5 muF`
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So charge on each capacitor is
`q=C_(eq)xx120=6/5xx120=144 muC`
After closing the switch, the potential difference across each capacitor is `60 V`. So charge on them are
`q_(1)=C_(1)xx60=2xx60=120muC`
`q_(2)=C_(2)xx60=3xx60=180 mu C`
Charge flowing through section `1` is
`Deltaq_(1)=q-q_(1)=144-120=24 muC`
Charge flowing through section `2` is
`Deltaq_(2)=q_(2)-q=180-144=36 muC` Charge flowing though dection `3` is
`Deltaq_(3)=Deltaq_(1)+Deltaq_(2)=24+36=60 mu C`.
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