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

Options

A. 10 C
B. 20 C
C. 30 C
D. 40 C

Answer

To find the total charge that flows in the first five seconds when the current in the circuit is given by the equation \( i = 1.2t + 3 \), we can follow these steps:

### Step 1: Understand the relationship between charge and current
The total charge \( Q \) that flows through a circuit over a time interval can be calculated by integrating the current \( i(t) \) over that time interval. Mathematically, this is expressed as:
\[ Q = \int_{t_1}^{t_2} i(t) \, dt \]
where \( t_1 \) and \( t_2 \) are the limits of the time interval.

### Step 2: Set up the integral for the given current function
Given the current function:
\[ i(t) = 1.2t + 3 \]
we need to integrate this from \( t = 0 \) to \( t = 5 \) seconds:
\[ Q = \int_{0}^{5} (1.2t + 3) \, dt \]

### Step 3: Perform the integration
We can split the integral into two parts:
\[ Q = \int_{0}^{5} 1.2t \, dt + \int_{0}^{5} 3 \, dt \]

Now we calculate each integral separately:

1. For the first integral:
\[ \int 1.2t \, dt = 1.2 \cdot \frac{t^2}{2} = 0.6t^2 \]
Evaluating from 0 to 5:
\[ 0.6(5^2) - 0.6(0^2) = 0.6 \cdot 25 = 15 \]

2. For the second integral:
\[ \int 3 \, dt = 3t \]
Evaluating from 0 to 5:
\[ 3(5) - 3(0) = 15 \]

### Step 4: Combine the results
Now, we add the results of both integrals:
\[ Q = 15 + 15 = 30 \, \text{Coulombs} \]

### Final Answer
The total charge that will flow in the first five seconds is:
\[ Q = 30 \, \text{C} \]

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