An ac ammeter reads 15 A. The peak value of the current is

The correct answer is Option (1) → \( 15\sqrt{2}\,\mathrm{A} \)

An ac ammeter reads the rms (effective) value of the current, \( I=15\,\mathrm{A} \). The rms current is related to the peak current by \( I=i_m/\sqrt{2} \).

    \( \begin{aligned} i_m &= \sqrt{2}\,I \\ &= \sqrt{2}\times15\,\mathrm{A} \\ &= 15\sqrt{2}\,\mathrm{A} \end{aligned} \)

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 I_rms or I. … It is defined by

    \( \displaystyle I=\sqrt{\overline{i^{2}}}=\sqrt{\frac{1}{2}\,i_m^{2}}=\frac{i_m}{\sqrt{2}}=0.707\,i_m \qquad (7.6) \)”

AC Voltage Applied to a Resistor – “It is customary to measure and specify rms values for ac quantities.”