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)

Solution

Physics sir · 3 steps

Energy in one second is the power, and power is I²R. Convert milli-units to SI before you divide.

1. Find the power. Power is energy per unit time: P = (10 × 10⁻³ J)/(1 s) = 1 × 10⁻² W.
2. Link power to resistance. For a resistor carrying current I, the heat produced per second is P = I²R. Here I = 2 × 10⁻³ A, so I² = 4 × 10⁻⁶ A².

   R = P/I² = (1 × 10⁻²)/(4 × 10⁻⁶) Ω = 2.5 × 10³ Ω

3. So. The resistance is 2500 Ω.

The common slip: 5. Forgetting to square the current and using R = P/I gives 10⁻²/(2 × 10⁻³) = 5 Ω, not 2500.

The idea to keep. When you know the current and the heat, use P = I²R. Change mJ to J and mA to A first, so the powers of ten work out correctly.

Answer from the NTA final answer key. Solution by PrepWiser.