A part of circuit in a steady state along with the currents flowing in the branches, the values of resistance etc., is shown in the figure. Calculate the energy stored in the capacitor C (4muF)`

Answer

A, D

Explanation

Applying Kirchoff's first law at junction M, we get the current
`I _1 = 3A`

Applying Kirchoff's first law at junction P, we get current `i_2 = 1A`
Note : No current flows through capacitor at steady state.
Moving the loop along MNO to P
`:. V_(M) - 5 xx i_1 - 1 xx I_1 - 2 xx i_2 = V_p`
`:. V_(M) -V_(P) = 6i_1 + 2i_2 = 6 xx 3 + 2 xx 1 = 20 V`
`:. V = 6 xx 3 + 2 xx 1 = 20 V`
Energy stored in the capacitor
`= 1/2 CV^2 = 1/2 xx 4 xx 10^(-6) xx 20 xx 20 = 8 xx 10^(-4) J` .