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Question
Solve the following question.
Using Kirchhoff’s rules, calculate the current through the 40 Ω and 20 Ω resistors in the following circuit.
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Solution
Apply KVL through ABCDA
`80 - 20i_1 - 40(i_1 - i_2) = 0`
`80 - 60i_1 + 40i_2 = 0` ....(1)
Apply KVL through FEDCF
`40 + 40(i_1 - i_2) - 10i_2 = 0`
`40 + 40i_1 - 50i_2 = 0` ......(2)
`4i_2 - 6i_1 = -8 (1)` .....[ from (1)]
`-5i_2 + 4i_1 = -4 (2)` .....[ from (2)]
Multiply equation (2) by `6/4` and add with equation (1)
`4i_2 - 6i_1 = -8`
`(-30i_2)/4 + 6i_1 = -6`
__________________________
`4i_2 - 30/4i_2 = -14`
`(-14)/4i_2 = -14`
`i_2 = (-14)/7 xx 2`
`i_2 = 4 A`
Put the value of i2 in equation (1)
`4 xx (4) - 6i_1 = -8`
`16 - 6i_1 = -8`
`6i_1 = 16 + 8 = 24`
`i_1 = 4A`
So, current through 40 Ω resistor = `i_1 - i_2`
= 4 - 4
= 0 A
Hence, current through 20 Ω resistor = 4A.
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