JEE Main 2025 Apr 4 Shift 1, Physics Q13: Introduction And Ohm’s Law

JEE Main2025Apr 4, Shift 1Physics

Current passing through a wire as function of time is given as $I (t) = 0.02 t + 0.01$ A. The charge that will flow through the wire from $t = 1$ s to $t = 2$ s is :

A
0.06 C
B
0.02 C
C
0.07 C
D
0.04 C

Solution

$q = \int_{1}^{2} I (t) d t = \int_{1}^{2} (0.02 t + 0.01) d t = [0.01 t^{2} + 0.01 t]_{1}^{2} = (0.04 + 0.02) - (0.01 + 0.01) = 0.04$ C.

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JEE Main 2025 Apr 4 Shift 1, Physics Q13: Introduction And Ohm’s Law

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