A circuit has two resistors in parallel, R1 = 4 ohms and R2 = 12 ohms, connected to a 9 V battery. Calculate the equivalent resistance, supply current, and current through each resistor.

Step 1: Identify the given values.
Two resistors are connected in parallel: $R_1 = 4 \Omega$ and $R_2 = 12 \Omega$.
The battery voltage is $V = 9 \text{ V}$.

a) Determine the equivalent circuit resistance.
For resistors connected in parallel, the equivalent resistance $R_{eq}$ is given by the formula:

$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$

Substitute the given values:

$$\frac{1}{R_{eq}} = \frac{1}{4 \Omega} + \frac{1}{12 \Omega}$$

Find a common denominator:

$$\frac{1}{R_{eq}} = \frac{3}{12 \Omega} + \frac{1}{12 \Omega}$$ $$\frac{1}{R_{eq}} = \frac{3+1}{12 \Omega}$$ $$\frac{1}{R_{eq}} = \frac{4}{12 \Omega}$$ $$\frac{1}{R_{eq}} = \frac{1}{3 \Omega}$$

Invert both sides to find $R_{eq}$:

$$R_{eq} = 3 \Omega$$

The equivalent circuit resistance is $\boxed{3 \Omega}$.

b) Determine the supply current.
The supply current $I_{supply}$ can be found using Ohm's Law, $I = \frac{V}{R_{eq}}$, where $V$ is the battery voltage and $R_{eq}$ is the equivalent resistance of the parallel combination.

$$I_{supply} = \frac{9 \text{ V}}{3 \Omega}$$ $$I_{supply} = 3 \text{ A}$$

The supply current is $\boxed{3 \text{ A}}$.

c) Determine the current in each resistor.
In a parallel circuit, the voltage across each resistor is the same as the supply voltage.
For resistor $R_1$:

$$I_1 = \frac{V}{R_1} = \frac{9 \text{ V}}{4 \Omega}$$ $$I_1 = 2.25 \text{ A}$$

For resistor $R_2$:

$$I_2 = \frac{V}{R_2} = \frac{9 \text{ V}}{12 \Omega}$$ $$I_2 = 0.75 \text{ A}$$

The current in the $4 \Omega$ resistor is $\boxed{2.25 \text{ A}}$.
The current in the $12 \Omega$ resistor is $\boxed{0.75 \text{ A}}$.