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:

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RRB ALP Previous Paper 13 (Held On: 20 Aug 2018 Shift 2)
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  1. 1350 J
  2. 15 J
  3. 135 J
  4. 22.5 J

Answer (Detailed Solution Below)

Option 1 : 1350 J
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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),
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