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:

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),