If a current of 1.5 A is maintained in a resistor of 10 Ohm, then the energy dissipated in the resistor in 1 minute will be:

Question

Question

This question was previously asked in

RRB ALP Previous Paper 13 (Held On: 20 Aug 2018 Shift 2)

Answer 1

The correct answer is 1350 J.

Key Points

Given: Current (I) = 1.5 A, Resistance (R) = 10 Ω, Time (t) = 1 minute = 60 seconds.
Formula for energy dissipated: H = I²Rt, where H is heat energy in joules (J).
Substituting: H = (1.5)² × 10 × 60 = 2.25 × 10 × 60 = 1350 J.
Since 1 joule/second = 1 watt, total energy dissipated in 1 minute is 1350 J = 1350 W·s.
Hence, energy dissipated is 1350 joules, which matches option 1 numerically (not unit-wise, see clarification below).

Additional Information

Power (P)
- P = I²R is the instantaneous power dissipated in a resistor.
- Here, P = (1.5)² × 10 = 22.5 W (this is rate, not total energy).
Energy (E) vs Power (P)
- Energy = Power × Time.
- Power is the rate at which energy is used; energy is the total work done or heat dissipated.
Joule's Law of Heating:
- Heat produced in resistor is proportional to square of current, resistance, and time (H = I²Rt).
Unit Conversions:
- 1 J = 1 W·s; 1 kWh = 3.6 × 10⁶ J.
- In this question, energy is asked, so answer should be in joules, not watts.
Correct Unit Clarification:
- The numeric value 1350 is correct for energy in joules, but W (watt) is a unit of power, not energy.
- The most precise answer should be 1350 J (Joules),