Identify the type of circuit.

Here are the solutions for the electrical circuit problem.

Step 1: Identify the type of circuit.
The circuit consists of two branches connected in parallel across the voltage supply, with resistors arranged in series within each branch. Therefore, it is a parallel circuit with series components.

Step 2: Calculate readings when Switch A is closed and Switch B is open.
When switch A is closed and switch B is open, only the top branch ($R_1$ and $R_2$) is active.
The total resistance in the top branch is:

$$R_{top} = R_1 + R_2 = 2 \, \Omega + 4 \, \Omega = 6 \, \Omega$$

The current flowing through the top branch (and the ammeter) is:

$$I_A = \frac{V}{R_{top}} = \frac{12 \, \text{V}}{6 \, \Omega} = 2 \, \text{A}$$

The reading on voltmeter $V_1$ (across $R_1$) is:

$$V_1 = I_A \times R_1 = 2 \, \text{A} \times 2 \, \Omega = 4 \, \text{V}$$

Since switch B is open, no current flows through the bottom branch, so the reading on voltmeter $V_2$ is:

$$V_2 = 0 \, \text{V}$$

Ammeter reading: $\boxed{\text{2 A}}$
**Voltmeter }V_1 reading: $\boxed{\text{4 V}}$
**Voltmeter }V_2 reading: $\boxed{\text{0 V}}$

Step 3: Calculate readings when Switch B is closed and Switch A is open.
When switch B is closed and switch A is open, only the bottom branch ($R_3$ and $R_4$) is active.
The total resistance in the bottom branch is:

$$R_{bottom} = R_3 + R_4 = 12 \, \Omega + 6 \, \Omega = 18 \, \Omega$$

The current flowing through the bottom branch (and the ammeter) is:

$$I_B = \frac{V}{R_{bottom}} = \frac{12 \, \text{V}}{18 \, \Omega} = \frac{2}{3} \, \text{A} \approx 0.667 \, \text{A}$$

Since switch A is open, no current flows through the top branch, so the reading on voltmeter $V_1$ is:

$$V_1 = 0 \, \text{V}$$

The reading on voltmeter $V_2$ (across $R_4$) is:

$$V_2 = I_B \times R_4 = \frac{2}{3} \, \text{A} \times 6 \, \Omega = 4 \, \text{V}$$

Ammeter reading: $\boxed{\text{0.667 A}}$
**Voltmeter }V_1 reading: $\boxed{\text{0 V}}$
**Voltmeter }V_2 reading: $\boxed{\text{4 V}}$

Step 4: Calculate readings when Both switches are closed.
When both switches are closed, both branches are active and connected in parallel.
The resistance of the top branch is $R_{top} = 6 \, \Omega$.
The resistance of the bottom branch is $R_{bottom} = 18 \, \Omega$.
The current through the top branch is:

$$I_{top} = \frac{V}{R_{top}} = \frac{12 \, \text{V}}{6 \, \Omega} = 2 \, \text{A}$$

The current through the bottom branch is:

$$I_{bottom} = \frac{V}{R_{bottom}} = \frac{12 \, \text{V}}{18 \, \Omega} = \frac{2}{3} \, \text{A} \approx 0.667 \, \text{A}$$

The total current (ammeter reading) is the sum of the currents in the parallel branches:

$$I_{total} = I_{top} + I_{bottom} = 2 \, \text{A} + \frac{2}{3} \, \text{A} = \frac{6}{3} \, \text{A} + \frac{2}{3} \, \text{A} = \frac{8}{3} \, \text{A} \approx 2.667 \, \text{A}$$

The reading on voltmeter $V_1$ (across $R_1$) is:

$$V_1 = I_{top} \times R_1 = 2 \, \text{A} \times 2 \, \Omega = 4 \, \text{V}$$

The reading on voltmeter $V_2$ (across $R_4$) is:

$$V_2 = I_{bottom} \times R_4 = \frac{2}{3} \, \text{A} \times 6 \, \Omega = 4 \, \text{V}$$

Ammeter reading: $\boxed{\text{2.667 A}}$
**Voltmeter }V_1 reading: $\boxed{\text{4 V}}$
**Voltmeter }V_2 reading: $\boxed{\text{4 V}}$

Step 5: Calculate the resistance of the full circuit.
The resistance of the full circuit refers to the equivalent resistance when both switches are closed.
The equivalent resistance $R_{eq}$ for parallel branches is given by:

$$\frac{1}{R_{eq}} = \frac{1}{R_{top}} + \frac{1}{R_{bottom}}$$ $$\frac{1}{R_{eq}} = \frac{1}{6 \, \Omega} + \frac{1}{18 \, \Omega}$$

To add these fractions, find a common denominator, which is 18:

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