A sinusoidal voltage of frequency 50 Hz is applied across a resistor R, and the current is \( i=i_m\sin\omega t \). Consider the following statements:
(A) The instantaneous power \( i^{2}R \) is positive or zero at every instant.
(B) The instantaneous power varies periodically with a frequency of 100 Hz.
(C) The average power over one complete cycle is \( i_m^{2}R \).
(D) The current and the voltage reach their maximum values at the same instant.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
For a current \( i=i_m\sin\omega t \) in a resistor, the instantaneous power is \( p=i^{2}R \), that is, \( i_m^{2}R\sin^{2}\omega t \). With the identity \( \sin^{2}\omega t=\frac{1}{2}(1-\cos2\omega t) \) it can be written as:
\( \displaystyle p=\frac{1}{2}\,i_m^{2}R\,(1-\cos2\omega t) \)
(A) The instantaneous power \( i^{2}R \) is positive or zero at every instant. — True. The power depends on \( i^{2} \), which cannot be negative whichever way the current flows. The resistor takes energy from the source throughout the cycle.
(B) The instantaneous power varies periodically with a frequency of 100 Hz. — True. The power contains \( \cos2\omega t \). So it repeats at twice the frequency of the current, \( 2\times50\,\mathrm{Hz} \), which is 100 Hz. In every cycle of the current the power rises to its peak twice and falls to zero twice.
(C) The average power over one complete cycle is \( i_m^{2}R \). — False. The average of \( \cos2\omega t \) over a cycle is zero, so the average power is:
\( \displaystyle \bar{p}=\frac{1}{2}\,i_m^{2}R \)
This is half of the value given in the statement.
(D) The current and the voltage reach their maximum values at the same instant. — True. In a pure resistor the voltage and the current are in phase, so they pass through zero and reach their maxima together.
NCERT (2026-27):
AC Voltage Applied to a Resistor – “The instantaneous power dissipated in the resistor is”
\( p=i^{2}R \)\( =i_m^{2}R\sin^{2}\omega t \qquad (7.4) \)
“… Using the trigonometric identity, sin2 ωt = 1/2 (1– cos 2ωt), we have < sin2 ωt > = (1/2) (1– < cos 2ωt >) and since < cos2ωt > = 0…, we have,”
\( \displaystyle \langle\sin^{2}\omega t\rangle=\frac{1}{2} \)
“Thus,”
\( \displaystyle \bar{p}=\frac{1}{2}\,i_m^{2}R \)\( \qquad [7.5(c)] \)
AC Voltage Applied to a Resistor – “… Note, in particular that both v and i reach zero, minimum and maximum values at the same time. Clearly, the voltage and current are in phase with each other.”