A lightning stroke carries an average current of \(3\times10^{4}\ \mathrm{A}\) for \(40\ \mu\mathrm{s}\). A room heater draws a steady current of 4 A. For how long must the heater run so that the same charge flows through it?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.3 s
Current is the charge that flows per unit time, so \( q=It \). The charge carried by the stroke is found first. The heater must pass the same charge at its own current.
Given:
\( I_1=3\times10^{4}\ \mathrm{A} \)
\( t_1=40\ \mu\mathrm{s}=40\times10^{-6}\ \mathrm{s} \)
\( I_2=4\ \mathrm{A} \)
\( \begin{aligned} q &= I_1t_1 \\ &= 3\times10^{4}\times40\times10^{-6} \\ &= 1.2\ \mathrm{C} \end{aligned} \)
\( \begin{aligned} t_2 &= \frac{q}{I_2} \\ &= \frac{1.2}{4} \\ &= 0.3\ \mathrm{s} \end{aligned} \)
Therefore, the heater must run for 0.3 s to pass the charge of one lightning stroke.
NCERT (2026-27):
Electric Current – “… The net amount of charge flowing across the area in the forward direction in the time interval t, then, is q = q+– q–. This is proportional to t for steady current”
\( \displaystyle I=\frac{q}{t} \)\( \qquad (3.1) \)
“is defined to be the current across the area in the forward direction. …”