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In the given circuit E1 = 3E2 = 2E3 = ...

In the given circuit
`E_1 = 3E_2 = 2E_3 =` 6 volts `R_1 = 2R_4 = 6Omega`
`R_3 = 2R_2 = 4Omega` `C = 5 muf.`
Find the current in R and the energy stored in the capacitor.

Text Solution

Verified by Experts

Applying Kirchhoff's law in ABFGA
`6 - (i_1+i_2) 4 =0 (i)`
Applying Kirchhoff's law in BCDEFB
`i_2xx3-3-2+2i_2+(i_2+i_1) 4 =0 (ii)`

Putting the value of `4(i_1+i_2) = 6 "in" Eq. (ii)`, we get
`3i_2 -5+2i_2+6 = 0 or i_2 = -(1)/(5)`
Substituting this value in Eq. (i) we get
`i_1 =1.5 -(-(1)/(5)) = 1.7A`
Therefore, current in `R_3 is i_1 + i_2 =1.5A.` To find the potentail
differece across the capacitor, we have
`V_E -2 - 0.2 xx2 =V_G`
or ` V_E - V_G = 2.4 V`
Therefore, energy stored in capacitor is
`(1)/(2)CV^2= (1)/(2)xx5xx10^(-6)xx(2.4)^2 = 1.44xx10^(-5)J`
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