The problem you've uploaded asks to calculate the total current in a circuit containing three resistors, each with a resistance of 9 ohms, connected in parallel to a power supply with a voltage of 12 V.

Step-by-step solution:

1. Parallel Circuit Formula: When resistors are connected in parallel, the total resistance $R_{\text{total}}$ can be calculated using the following formula: $\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$ In this case, $R_1 = R_2 = R_3 = 9$ ohms. Thus, the formula simplifies to: $\frac{1}{R_{\text{total}}} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9}$ $\frac{1}{R_{\text{total}}} = \frac{3}{9} = \frac{1}{3}$ So, $R_{\text{total}} = 3 \, \Omega$

2. Ohm's Law: Ohm's Law relates the voltage ($V$), current ($I$), and resistance ($R$) in a circuit: $V = I R$ To find the total current ($I_{\text{total}}$) in the circuit, we rearrange the formula: $I_{\text{total}} = \frac{V}{R_{\text{total}}}$ Substituting the known values: $I_{\text{total}} = \frac{12}{3} = 4 \, \text{A}$

Thus, the total current in the circuit is 4 A.

Let me know if you'd like further details on the process or if you have any questions.

Follow-up Questions:

1. What happens to the total resistance if the resistors were connected in series instead of parallel?
2. How would the total current change if the voltage were doubled to 24 V?
3. How does the total current change if one of the resistors is removed from the parallel circuit?
4. How would the power consumed by each resistor be calculated in this circuit?
5. What is the current through each resistor individually in this parallel setup?

Tip:

In a parallel circuit, the total resistance is always less than the smallest individual resistance. This is because each parallel path provides an additional way for current to flow.