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**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 i_{2} 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.