JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Current Electricity

The energy dissipated by a resistor is 10 mJ in 1 s when an electric current of 2 mA flows through it. The resistance is ____

Ω

. (Round off to the Nearest Integer)

Answer: 2500

Solution

A steady current

I

in a resistance

R

dissipates energy

E = I^{2} R t

in time

t

, so

R = \frac{E}{I ^{2} t} = \frac{10 \times 1 0 ^{- 3}}{( 2 \times 1 0 ^{- 3} ) ^{2} \times 1} = \frac{1 0 ^{- 2}}{4 \times 1 0 ^{- 6}}

The resistance of the resistor is

2500 Ω

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer