In the electron network. Kirchhoff's rule to calculate the power consumed by the resistance `R= 4 Omega`

Answer

` 4 = 9W`

Explanation

Let the current in the varies arms of the network be the

Using kirchhoff's cecond rate in the closed loop `ABCDA` we have
`+ epsilon_(1) - l_(1) r_(1) - (l_(1) + l_(2)) R = 0`
or `+ 12 - l_(1) xx 2 - (l_(1) + l_(2)) 4 = 0 `
or `6 l_(1) + 4l_(2) = + 12`
or `3l_(1) + 2 l_(2) = + 6`.......(i)
In the closed loop `ADEFA` we have
`(l_(1) + l_(2)) R - epsilon_(2) = 0 or (l_(1) + l_(2)) 4 = 6`
or `2l_(1) + 2l_(2) = 3 `......(ii)
On solving (i) and (ii) we get
`l_(1) = 3A, l_(2) = - (3)/(2) A`
`:. l_(1) + l_(2) 3 - (3)/(2) = (3)/(2) A`
power connected by resistance `R( = 4 Omega)` is
`= (l_(1) + l_(2))^(2) R = ((3)/(2))^(2) xx 4 = 9W`