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

Solution

Physics sir · 3 steps

Current is the rate at which charge flows, so add it up over time: integrate I(t) from 1 s to 2 s.

1. Charge from current. Since I = dq/dt, the charge in a small time dt is dq = I dt. Because the current changes with time, you integrate instead of multiplying.

   q = ∫₁² (0.02t + 0.01) dt = [0.01t² + 0.01t]₁²

2. Apply the limits. At t = 2: 0.01 × 4 + 0.01 × 2 = 0.06. At t = 1: 0.01 × 1 + 0.01 × 1 = 0.02. Subtract: 0.06 - 0.02 = 0.04.
3. So. The charge that flows from t = 1 s to t = 2 s is 0.04 C, option (d).

The idea to keep. Charge is the area under the I versus t graph. When the current varies, integrate between the given times, and always subtract the value at the lower limit.

Answer from the NTA final answer key. Solution by PrepWiser.