Q 12-03-277JEE MainJEE Main 2019 (12 Jan, Shift 2)Medium
In the given circuit diagram, the currents, $I_1 = -0.3$ A, $I_4 = 0.8$ A and $I_5 = 0.4$ A, are flowing as shown. The currents $I_2$, $I_3$ and $I_6$, respectively, are:
Answer: (C) $1.1$ A, $0.4$ A, $0.4$ A
Apply the junction rule:
At P: $I_6 = I_5 = 0.4$ A.
At S: $I_4 = I_5 + I_3 \Rightarrow I_3 = 0.8 - 0.4 = 0.4$ A.
At R: $I_1 + I_2 = I_4 \Rightarrow I_2 = 0.8 - (-0.3) = 1.1$ A.
(Check at Q: $I_6 + I_3 = 0.8 = I_1 + I_2$.)
Solution by Sreeraj P, M.Sc Physics