JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Current Electricity

The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of

R

is:

Answer: (A)

Solution

Reducing the network to a single equivalent resistance gives

R_{eq} = 8 R

. The power drawn from the source is

P = \frac{V ^{2}}{R _{eq}}

, so

4 = \frac{1 6 ^{2}}{8 R} = \frac{256}{8 R}

. Solving,

8 R = 64

and

R = 8 Ω

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option