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In the electric network shown in the Fig., use Kirchhoff's rules to calculate the power consumed by the resistanceR = 4 Omega |
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Answer» <P> Solution :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` |
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