To solve the problem in the image, we need to calculate the total current flowing through the circuit based on the resistances and voltage provided. Let's break it down:

1. Circuit Configuration:

   • There is a 12V battery.
   • Two resistors, $R_1 = 6 \, \Omega$ and $R_2 = 4 \, \Omega$, are connected in parallel.
   • The total current through the circuit needs to be calculated.

2. Parallel Resistors Formula: For two resistors $R_1$ and $R_2$ in parallel, the equivalent resistance $R_{eq}$ is given by: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$ Substituting the given values: $\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12}$ Thus, $R_{eq} = \frac{12}{5} = 2.4 \, \Omega$

3. Total Current: Using Ohm's Law, the total current $I$ is: $I = \frac{V}{R_{eq}} = \frac{12 \, \text{V}}{2.4 \, \Omega} = 5 \, \text{A}$

Thus, the current flow through the switch and fuse is 5 A, which corresponds to option b.

Would you like more details or have any further questions?

Here are 5 related questions to deepen your understanding:

1. How do you calculate the equivalent resistance for resistors in series?
2. How does the total current change if the resistors were connected in series instead of parallel?
3. What is the relationship between voltage, current, and resistance in Ohm’s Law?
4. How would you determine the voltage drop across each resistor in this circuit?
5. What happens to the total resistance if another resistor is added in parallel?

Tip: When resistors are in parallel, the total resistance is always less than the smallest resistor in the combination.