Analyse the given circuit in the steady state condition. Charge on the capacitor is `q_(0)=16muC`. If the e.m.f. of the battery is `6k` then find the value of `k`.

Answer

A

Explanation

a. Current inlower branch `=E/8=3A`
Current in uper branch `=E/9=24/9=2.67A`
b. Pd across the capacitor `=E/2-E/3=E/6`
From `q=CV`, we have `16=(4)E/6`
`:. E=24V`
c. After short circuitting the bttery, we will have to find net resistance across capacitor to calculate equivalent value of `tau_C` in dischargein. `3Omega` and `6Omega` are in parallel. Similarly `4Omega` and `4Omega` are in parallel. They are then in series
`:. R_(n et)=4Omega`
`tau_C=CR_(n et)=(4xx4)mus=16mus`
During discharging `q=q_0e^(-t/tau_C)`
or `8=16e^(-t/16)`
Solving this equation we get t=`11.1mus`