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find current supplied by each battery and power dissipated in 1ohm resistor
Asked by snehashiragannavar773 | 21 Oct, 2023, 02:38: PM
Let us apply Kirchoff's voltage law to the loop ABEFA . We get

-0.25 I2  +12 -12 +0.20 I1 = 0

From above circuit , we get ,  I1 = 1.25 I2 .................. (1)

Let us apply Kirchoff's voltage law to the loop BCDEB

......................... (2)

If we apply Kirchoff's curremt law to the juntion B , wegt ,

I3 = I1 + I2 .................... (3)

By using eqn.(1) and eqn.(3) , we rewrite eqn.(2) as

Hence I2 = 4.8 A ;

From eqn.(1) , we get , I1 = 6 A ;

Hence current passing through 1 Ω resistor is I3 = ( 6 + 4.8 ) A \ = 10.8 A

Power diispation in 1 Ω resitor = I32 R = (10.8 × 10.8 × 1 ) W = 117 W

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Answered by Thiyagarajan K | 21 Oct, 2023, 04:23: PM

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