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Hello students, in this question we have given circuit.
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Now let's take node v1, v2 and v3 and this is reaction for i1 and this is for i2.
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Now here we have given 5 ampere.
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Now as we have given i2.
00:29
Now as we have given i1 plus i2 is 5 ampere.
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So at node v2 we have v2 minus v1 upon 5 plus v2 minus v3 upon 10 is equal to 5.
00:55
Taking this equation 1, now at node v1 we have v2 minus v1 upon 5 is equal to v1 minus v3 upon 20 plus v1 upon 10.
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Taking equation 2.
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Next at node v3 this will be v2 minus v3 upon 10 is equal to v1 minus v3 upon 20 is equal to v3 upon 5.
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Taking this equation 3.
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Now from equation 1 we have 3 v2 minus 2 v1 minus v3 is equal to 50.
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We get this equation.
01:45
From equation 2 further solving we get 4 v2 minus 4 v1 is equal to v1 minus v3 minus v3 plus 2 v2.
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So this will gives equation 7 v1 minus 4 v2 minus v3 is equal to 0.
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From equation 3 we get equation 2 v2 minus 2 v3 plus v1 minus v3.
02:18
So we get this that is equal to 4 v3.
02:23
So we get v1 plus 2 v2 minus 7 v3 will be equal to 0.
02:30
Now from above equations we get v1 is equal to 250 upon 11.
02:38
So this will gives value 22 .72 and v2 we get 400 upon 11.
02:48
So this gives 36 .36 and for v3 we have 150 upon 11.
02:56
So we get 13 .63 v.
03:00
Now we can find vd that is v3 minus v2.
03:06
So this will gives 250 upon 11...