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In the circuit shown in figure. epsi(1)=...



In the circuit shown in figure. `epsi_(1)=3,epsi_(2)=2,epsi_(3)=6V,R_(1)=2,R_(4)=6Omega,R_(3)=2,R_(2)=4Omega and C=5muF`. Find the current in the resistor `R_(3)` and the electrical energy stored in the capacitor C.

Text Solution

Verified by Experts

The correct Answer is:
(I) 1.5
(II) 14.4
In steady state no current will flow through `R_1=6 Omega` Potential difference across `R_3` or `4Omega` is `E_1` or 6V
Current through it will be `6/4=1.5A` from right to left.Because left hand side of this resistance is at higher potential .Now,suppose this 1.5A distributes in `i_2` as shown applying Kichhoff's second law in loop dhfed
`3-3i_1-4xx1.5-2i_1+2=0 therefore i_1=-1/5A=-0.2A`
To find energy stored in capacitor we will have to find potential difference across it or `V_(ad)`
Now, `V_a-2i_1+2=V_d or V_a-V_d=2i_1-2=-2.4 V`
or `V_d-V_a=-2.4V=V_(da)`
Energy stored in capacitor .`U=1/2CV_(da)^(2)=1/2(5xx10^(-6))(2.4)^(2)=1.44xx10^(-5)J=14.4 mu J`
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