Question

A source of e.m.f. $E = 15$ V and having negligible internal resistance is connected to a variable resistance so that the current in the circuit increases with time $(t)$ at $I=1.2t+3$. Then, the total charge that will flow in first 6 second will be

• A. 38 C
• B. 43 C
• C. 39.6 C
• D. 40.6 C

Answer 1

Answer: (C)

39.6 C

Answer 2

The relationship between current ($I$) and charge ($Q$) is given by $I = \frac{dQ}{dt}$. To find the total charge that flows in a given time interval, we integrate the current with respect to time over that interval.

Given the current as a function of time: $I = 1.2t + 3$

We need to find the total charge ($Q$) that flows in the first 6 seconds, which means from $t=0$ s to $t=6$ s. The total charge $Q$ is given by the definite integral of $I$ with respect to $t$: $Q = \int_{t_1}^{t_2} I \ dt$

Substitute the given expression for $I$ and the limits of integration: $Q = \int_{0}^{6} (1.2t + 3) \ dt$

Now, perform the integration: $Q = \left[ \frac{1.2t^2}{2} + 3t \right]_{0}^{6}$ $Q = \left[ 0.6t^2 + 3t \right]_{0}^{6}$

Evaluate the expression at the upper limit ($t=6$) and subtract the value at the lower limit ($t=0$): $Q = (0.6(6)^2 + 3(6)) - (0.6(0)^2 + 3(0))$ $Q = (0.6 \times 36 + 18) - (0 + 0)$ $Q = (21.6 + 18) - 0$ $Q = 39.6 \text{ C}$

The information about the e.m.f. and internal resistance is not required for this calculation as the current is directly given as a function of time.