Question

The current in a wire varies with time according to the relation $i = 4t + 3t^2$. [here t is in seconds and i is in amperes]. How much charge passes through a cross-section of the wire in the time interval between $t = 5s$ and $t = 10s$?

• A. 1200 C
• B. 875 C
• C. 200 C
• D. 1025 C

Answer 1

Answer: (D)

1025 C

Answer 2

The problem asks to calculate the total charge passing through a cross-section of a wire when the current varies with time according to the relation $i = 4t + 3t^2$. The time interval given is from $t = 5s$ to $t = 10s$.

The relationship between current ($i$), charge ($q$), and time ($t$) is given by: $i = \frac{dq}{dt}$

To find the total charge ($Q$) that passes through the cross-section in a given time interval, we need to integrate the current with respect to time over that interval: $dq = i \ dt$ Integrating both sides from $t_1 = 5s$ to $t_2 = 10s$: $Q = \int_{t_1}^{t_2} i \ dt$

Substitute the given expression for $i$: $Q = \int_{5}^{10} (4t + 3t^2) \ dt$

Now, perform the integration: $\int (4t + 3t^2) \ dt = 4 \int t \ dt + 3 \int t^2 \ dt$ $= 4 \left( \frac{t^2}{2} \right) + 3 \left( \frac{t^3}{3} \right)$ $= 2t^2 + t^3$

Now, evaluate the definite integral using the limits from $t=5$ to $t=10$: $Q = [2t^2 + t^3]_{5}^{10}$ $Q = (2(10)^2 + (10)^3) - (2(5)^2 + (5)^3)$

Calculate the value at the upper limit ($t=10$): $2(10)^2 + (10)^3 = 2(100) + 1000 = 200 + 1000 = 1200$

Calculate the value at the lower limit ($t=5$): $2(5)^2 + (5)^3 = 2(25) + 125 = 50 + 125 = 175$

Subtract the value at the lower limit from the value at the upper limit: $Q = 1200 - 175$ $Q = 1025 \text{ C}$

The total charge that passes through the cross-section of the wire in the given time interval is 1025 C.