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

Answer

A, D

Explanation

Applying Kirchoff's law in ABFGA
`6-(i_1 + i_2)4 = 0 `……….(i)
Applying Kirchoff's law in BCDEFB
`i_2 xx 3 - 3- 2 + 2i_2 + (i_2 + i_1)4 = 0 ` ………………(ii)
Putting the value of `4(i_1 + i_2) = 6` in (ii) ,brgt `3i_2 - 5 + 2i_2 + 6 = 0`
`:. i_2 = -1/5 A`
Substituting this value in (i) , we get
`i_1 = 1.5 - (-1/5)= 1.7A` .
Thereforecurrent in `R_3`
`= i_1 + i_2 = 1.7 - 0.2 = 1.5A`
To find the p.d. across the capacitor
`V_(E) - 2 - 0.2 xx 2 = V_(G)`
`:. V_(E) - V_(G) = 2.4V`
or `V = 24V`
`:. Energy stored in capacitor = 1/2 CV^2`
`= 1/2 xx 5 xx 10^(-6) xx (2.4)^2 = 1.44 xx 10^(-5)J` .