In the circuit shown in figure find a. the equivalent capacitance and b. the charge stored in each capacitor.
Text Solution
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a. the capacitors are in parallel. Hence, the equivalent capacitance is `C=C_1+C_2+C_3` or `C=(1+2+3)=6muF` b. total charge drawn from the battery `K` `q=CV-6xx100muC=600muC` This charge will be distributed in the ratio of their capacities. Hence, `q_1:q_2:q_3=C_1:C_2:C_3=1:2:3` `:. q_1=(1/(1+2+3))xx600=100muC` `q_2=(2/(1+2+3))xx600=200muC` and `q_3=(3/(1+2+3))xx600=300muC` Alternate solution: Since the capacitors are in parallel the `PD` across each of them in `100V`. Therefore from `q=CV`, the charge stored in `1muF` capacitor is `100muC`, in `2muF` capacitor is `200muC` and that in `3muF` capacitor is `300muC`
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