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Consider thr network as shown in fig. Current is supplied to the network by two batteries as shown. Find the values of currents I_1 , I_2, I_3 . The direction of the currents are as indicated by the arrows. |
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Answer» Solution :Applying Kirchhoff.s 1st LAW to junction C, we get `I_1 + I_2 + I_3 = 0` Applying Kirchhoff.s Iind law to the closed meshes ACDA and BCDB, we get `5 I_1 + 2I_3 = 12""…..(2)` `3I_2 + 2I_3 + 6 ""….(3)` SUBTRACTING eq (3) from eq (2) we get `5I_1 - 3I_2 = 6 "".....(4)` Multiplying eq.(1) by 2 adding with eq. (2) we get `7I_2 + 2I_2 = 12 ""......(5)` Multiplying eq (4) by 2 and eq. (5) by 3 and adding them we get `31I_1 = 48 impliesI_1 = 1.548 A.`  Putting value of `I_1 ` in eq. (5) we get `I_2 =0.58 A`and from eq. 1 we get `I_3 = I_1 + I_2 = 2.128 A` |
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