Question

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

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

Hint:

Integrate the current function to find the charge.

Answer 1

The Correct Option is D

1. Charge calculation: \[ q = \int I(t) dt \] \[ q = \int_{1}^{2} (0.02t + 0.01) dt \] \[ q = \left[ 0.02 \frac{t^2}{2} + 0.01t \right]_{1}^{2} \] \[ q = \left[ 0.01(4) + 0.01(2) \right] - \left[ 0.01(1) + 0.01(1) \right] \] \[ q = 0.04 \mathrm{~C} \] Therefore, the correct answer is (4) 0.04 C.

Answer 2

We are given the current as a function of time:

\[ I(t) = 0.02t + 0.01 \, \text{A} \]

We are asked to find the total charge that flows through the wire between \( t = 1 \, \text{s} \) and \( t = 2 \, \text{s} \).

Concept Used:

The total charge \( Q \) passing through a conductor in a time interval is obtained by integrating the current over that interval:

\[ Q = \int_{t_1}^{t_2} I(t) \, dt \]

Step-by-Step Solution:

Step 1: Substitute the expression for \( I(t) \) into the formula for \( Q \).

\[ Q = \int_{1}^{2} (0.02t + 0.01) \, dt \]

Step 2: Integrate each term separately.

\[ Q = 0.02 \int_{1}^{2} t \, dt + 0.01 \int_{1}^{2} dt \]

Step 3: Compute each integral.

\[ \int t \, dt = \frac{t^2}{2}, \quad \int dt = t \]

Substitute these results:

\[ Q = 0.02 \left[\frac{t^2}{2}\right]_{1}^{2} + 0.01 [t]_{1}^{2} \]

Step 4: Evaluate between the limits.

\[ Q = 0.02 \left(\frac{4 - 1}{2}\right) + 0.01 (2 - 1) \] \[ Q = 0.02 \times 1.5 + 0.01 \times 1 \] \[ Q = 0.03 + 0.01 = 0.04 \, \text{C} \]

Final Computation & Result:

Hence, the total charge that flows through the wire between \( t = 1 \, \text{s} \) and \( t = 2 \, \text{s} \) is:

\[ \boxed{Q = 0.04 \, \text{C}} \]

Final Answer: \( 0.04 \, \text{C} \)