JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Current Electricity

Twelve wires each having resistance

2Ω

are joined to form a cube. A battery of 6 V emf is joined across point a and c. The voltage difference between e and f is ______ V.

Answer: 1

Solution

Twelve

2 Ω

wires form a cube; the

6

V cell is connected across a face diagonal (

a

–

c

).

By symmetry of this connection, no current flows through the two wires

e - b

and

g - d

. Reducing the network gives

R_{eq} = \frac{3}{4} R = \frac{3}{4} (2) = 1.5 Ω,

so the total current from the cell is

I = \frac{6}{1.5} = 4

A. This splits at the terminals; the branch containing

e - f

carries

\frac{1}{2}

A through a

2 Ω

arm, giving

V_{e f} = \frac{I _{2}}{2} R = \frac{1}{2} \times 2 = 1 V .

Answer: 1.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Integer answer