A steady direct current \( I \) produces the same heat in a resistor as a sinusoidal alternating current of peak value \( i_m \) produces in the same resistor over one complete cycle. The ratio \( I/i_m \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1/\sqrt{2} \)
The heat produced in a resistor depends on \( i^{2}R \). So an alternating current heats a resistor, even though its average value over a cycle is zero. A steady current that gives the same heat in the same time has to produce the same average power. That steady current is the rms value of the alternating current.
For the ac \( i=i_m\sin\omega t \) the average power over a cycle is:
\( \displaystyle \bar{p}=\frac{1}{2}i_m^{2}R \)
The direct current \( I \) in the same resistor gives the power:
\( P=I^{2}R \)
For equal heat in equal time the two powers are equal:
\( \begin{aligned} I^{2}R &= \frac{1}{2}i_m^{2}R \\ I &= \frac{i_m}{\sqrt{2}} \end{aligned} \)
Therefore, \( I/i_m=1/\sqrt{2} \).
NCERT (2026-27):
AC Voltage Applied to a Resistor – “To express ac power in the same form as dc power (P = I2R), a special value of current is defined and used. It is called, root mean square (rms) or effective current … and is denoted by Irms or I. … It is defined by”
\( \displaystyle I=\sqrt{\overline{i^{2}}}=\sqrt{\frac{1}{2}i_m^{2}}= \)\( \displaystyle \frac{i_m}{\sqrt{2}} \)\( \displaystyle =0.707\,i_m \qquad (7.6) \)
AC Voltage Applied to a Resistor – “In fact, the I or rms current is the equivalent dc current that would produce the same average power loss as the alternating current.”