In the electric network shown in the Fig., use Kirchhoff's rules to calculate the power consumed by the resistance `R = 4 Omega`

Explanation

Applying Kirchhoff.s second rule for loop ABCDA, we have
`-12 +I_1 xx 2 + (I_1 + I_2) xx 4 = 0`
`rArr 6I_1 + 4I_2 = 12 ` ....(i)
Again for loop ADFEA, we have
` - (I_1 + I_2) xx 4 + 6 = 0`
` rArr 4I_1 + 4I_2 = 6` ...(ii)
From (i) and (ii), we get `I_1 = 3A , I_2 = - 1.5 A`
Hence current flowing through the resistance `R=4Omega` will be `I_1 + I_2 = 3 - 1.5 = 1.5 A`
` therefore ` Power consumed by the resistor `P = I^2 R = (1.5)^2 xx 4= 9W`