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To solve this problem, we need to calculate the amount of electric charge (\( Q \)) that flows through the resistor over the first 5 seconds based on the current (\( I \)) vs. time (\( t \)) graph provided.

### Key formula:
The amount of charge \( Q \) can be found using the relation:
\[
Q = \int_0^t I(t) \, dt
\]
In simpler terms, for a current-time graph, \( Q \) is the area under the curve of \( I(t) \) over the given time interval.

### Analyzing the graph:
The graph shows a piecewise linear current function over time:
1. From \( t = 0 \) to \( t = 3 \), the current is constant at \( I = 4 \, \text{A} \).
2. From \( t = 3 \) to \( t = 5 \), the current linearly decreases from \( I = 4 \, \text{A} \) to \( I = 2 \, \text{A} \).

### Step-by-step calculation:

1. **First interval (from \( t = 0 \) to \( t = 3 \) seconds):**
   - The current is constant at \( 4 \, \text{A} \).
   - The area under the graph for this interval is a rectangle:
     \[
     Q_1 = I \times t = 4 \, \text{A} \times 3 \, \text{s} = 12 \, \text{C}
     \]

2. **Second interval (from \( t = 3 \) to \( t = 5 \) seconds):**
   - The current decreases linearly, forming a trapezoid.
   - The area of this trapezoid is given by:
     \[
     Q_2 = \frac{1}{2} \times (I_1 + I_2) \times t = \frac{1}{2} \times (4 \, \text{A} + 2 \, \text{A}) \times 2 \, \text{s} = \frac{1}{2} \times 6 \times 2 = 6 \, \text{C}
     \]

### Total charge:
\[
Q_{\text{total}} = Q_1 + Q_2 = 12 \, \text{C} + 6 \, \text{C} = 18 \, \text{C}
\]

Thus, the total charge flowing through the resistor in the first 5 seconds is **18 coulombs**.

The correct answer is **d. 18**.

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Would you like any further details or explanations?

Here are 5 related questions to consider:
1. How do you calculate the charge when the current function is not linear?
2. What is the relationship between current, voltage, and resistance in a circuit?
3. How does the shape of the current vs. time graph affect the charge calculation?
4. How can you calculate power dissipated in the resistor using this graph?
5. How would you calculate the charge if the current graph extended beyond 5 seconds?

**Tip:** The area under the curve in a current vs. time graph always represents the total charge transferred in that time interval.

Grafik di bawah menunjukkan kuat arus yang mengalir dalam suatu hambatan R, sebagai fungsi waktu. Banyaknya muatan listrik yang mengalir dalam hambatan tersebut selama 5 sekon pertama adalah... (coulomb)

Solve the problem of calculating the electric charge that flows through a resistor over 5 seconds by analyzing a current-time graph. This problem covers key physics concepts, including electric charge, current, and graphical integration. Suitable for students in Grades 10-12.

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