In the circuit shown in figure, find the current through the branch BD
5A
0A
3A
4A
As we have learnt,
Kirchhoff's Law -
-
The current in the circuit are assumed as shown in the fig.
Applying KVL along the loop ABDA, we get
– 6i1 – 3 i2 + 15 = 0 or 2i1 + i2 = 5 …..(i)
Applying KVL along the loop BCDB, we get
– 3(i1 – i2) – 30 + 3i2 = 0 or – i1 + 2i2 = 10 …..(ii)
Solving equation (i) and (ii) for i2, we get i2 = 5 A.
Option 1)
5A
Option 2)
0A
Option 3)
3A
Option 4)
4A
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