Figure below shows a portion of an electric circuit with the cur ents in ampreres and their directions. The
magnitude and direction of the cur ent in the portion PQ is :
Asked by Amyra B | 22nd Oct, 2014, 07:55: PM
According to kirchhoff's first law (Current Law or Juction Law):
The algebraic sum of currents meeting at a junction is zero.
i.e. In a circuit, at any juction the sum of the currents entering the junction must be equal to the sum of currents leaving the junction.
We will apply Kirchhoff's junction at the junctions L , M, N and O as shown in the figure below:
At junction L:
The current would be 2 A + 8A = 10 A towards right
At junction M:
The current from junction L (10A) divides into two, one goes upwards towards junction O and other goes downwards towards junction N.
At junction N:
Its given that at junction N the current splits to 4A and 2 A. So the current entering junction N will be the sum of both 4 A and 2 A. So current entering junction N will be 6 A downwards.
At junction O:
So the current reaching junction O will be 10 A - 6 A = 4 A upwards.
This current entering the junction O gets divided into two: 1 A towards right side and hence towards left it will be 4 A - 1A =3A towards left.
At junction Q:
So the current entering junction Q will be the sum 3A +3A = 6 A
Hence we can say that the current in the portion PQ = 6A towards left.
Answered by Jyothi Nair | 27th Oct, 2014, 11:36: AM
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