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This can be solved by using Kirchhoff's laws.
Using the junction rule, the current coming into a junction is the same as the current leaving it.
At junction a: `i_1 = i_2 + i_3`
Using loop rule in left loop, we get: `6 + 10i_1 +15 i_2 -8 =0`
This can be simplified to: `10i_1 + 15i_2 = 8-6 = 2`
Using the loop rule in right loop, we get: `4 + 12i_3 + 8 -15i_2 = 0`
This can be simplified to: `15 i_2 - 12i_3 = 12`
or, `5i_2-4i_3 = 4`
Using these three equations simultaneously, we obtain the values of currents as:
`i_1 = -0.28A`
`i_2 = 0.32 A`
and `i_3 = -0.6 A`
The negative sign for i1 and i3 indicate that the assumed direction of current is wrong and it flows in the opposite direction. The sign for i2 is positive which means that the assumed direction is correct.
Hope this helps.
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