StreakPeaked· Practice

ExamsJEE AdvancedPhysics

An electric heater operates as follows: a switch charges a capacitor C to EMF Ue through the source's internal resistance Ri. Then the switch discharges the capacitor fully through the heater element of resistance R. This cycle repeats at frequency f. Given C = 100 micro-F, R = 20 ohm, Ue = 90 V, Ri = 80 ohm, f = 2 per second. Which of the following statements are correct? (A) Heat generated in the heating element during capacitor charging is 81 mJ. (B) Heat generated in the heating element during capacitor discharging is 405 mJ. (C) The average power of the heater is 4.86 W. (D) The efficiency of the heater is 3/5.

  1. Heat generated in the heating element during capacitor charging is 81 mJ.
  2. Heat generated in the heating element during capacitor discharging is 405 mJ.
  3. The average power of the heater is 4.86 W.
  4. The efficiency of the heater is 3/5.

Correct answer: Heat generated in the heating element during capacitor charging is 81 mJ.

Solution

During charging: total dissipated = stored energy = 0.405 J, shared as R/(R+Ri) = 20/100 = 1/5 to heater. During discharging: all 0.405 J goes to heater. Efficiency = total heater power / source power.

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →