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in the cirmuit given beiiow, lind the camem through 100bog302+15 |
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Answer» let the voltage at the centre be V and current in 10Ω = i1 , 20Ω = i2 and 30Ω = i3 let i1 = i2+i3 so, according to Kirchoff rules => (10-v)/10 = (v-15)/30 + (v-6)/20=> 10-v = v/3 -5 +v/2 -3=> 11v/6 = 18=> v = 108/11 so, Ii = (10-108/11)/10 = 2/110 = 1/55 A |
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