JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Current Electricity

In the given circuit diagram, a wire is joining points

B

and

D

. The current in this wire is:

(A)zero
(B)2 A✓ Correct
(C)0.4 A
(D)4 A

Answer: (B)

Solution

Reducing the network with Kirchhoff's rules, the branches combine to an effective resistance of

\frac{4}{5} + \frac{6}{5} = 2 Ω

, so the

20

V source drives a total current

i = \frac{20}{2} = 10

A. When this current splits through the two arms and the junction rule is applied at B, the wire joining B and D carries the difference of the two branch currents,

\frac{4 i}{5} - \frac{3 i}{5} = \frac{i}{5} = \frac{10}{5} = 2

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option

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