More practice questions

4. Q4. Assertion: an ac current of rms value 2 A heats a resistor at the same rate as a steady 2 A dc current. Reason: the rms current is defined as the dc current that gives the same average power loss. Are both true, and does the reason explain the assertion?

   Show answerHide answer

   Both are true, and the reason explains the assertion (NCERT page 180).

5. Q5. An electric iron is rated 1000 W at 220 V. Find its resistance, the rms current and the peak current.

   Show answerHide answer

   R = V²/P = 220²/1000 = 48.4 Ω; I = P/V = 1000/220 ≈ 4.55 A; i_m = √2 × 4.55 ≈ 6.43 A (NCERT Eqs. (7.6)–(7.9), page 180).

6. Q6. Show from sin² ωt = ½(1 − cos 2ωt) that the average power in a resistor is half the peak power.

   Show answerHide answer

   The peak power is i_m²R. Over a cycle cos 2ωt averages to zero, so sin² ωt averages to ½, and the average power is ½i_m²R, half the peak (NCERT page 179).

7. Q7. A source v = 200 sin(314t) V is connected to a 40 Ω resistor. Find the frequency, the rms voltage and the average power.

   Show answerHide answer

   ω = 314 rad/s, so ν = 314/2π ≈ 50 Hz. V = 200/√2 ≈ 141 V, and P = V²/R = 141.4²/40 ≈ 500 W (NCERT Eqs. (7.1), (7.7)–(7.8), pages 178–180).