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Analyze the given circuit in the steady ...

Analyze the given circuit in the steady state condition. Charge on the capacitor is`q_0=16muC`

(a) Find the current in each branch
(b) Find the emf of the battery.
(c) If now the battery is removed and the points `A` and `C` are shorted. Find the time during which charge on the capacitor becomes
`8muC`

Text Solution

Verified by Experts

Potential difference across the capacitor is
`V_(B)-0=(q_(@))/(C )=(16)/(4)=4"volt"`
`9I_(1)=8I_(2)`
`rArr I_(2)=(9//8)I_(1)`
Applying Kirchoff's voltage law along `ABDA`
`3I_(1)+4=((9)/(8)I_(1))4`
`rArr I_(2)=3A`.
`V_(A)=12V`, `V_(B)=4V`,
as `V_(D)=0`
E.M.F. of the battery `=V_(A)-V_(C )=24V`
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