In the circuit below, two resistors in parallel are connected with a resistor in series. The volt drop over the 8 Ω resistor in parallel is 12 V.

Let's solve the questions based on the information provided.

### 4.1 The current through the 8 Ω resistor
The voltage drop across the 8 Ω resistor is given as 12 V. We can use Ohm's Law, \( V = I \cdot R \), to find the current \( I \):

\[
I = \frac{V}{R} = \frac{12 V}{8 \, \Omega} = 1.5 \, A
\]

**Answer 1**: 1.5 A

### 4.2 The current through the 4 Ω resistor
Since the 8 Ω and 4 Ω resistors are in parallel, they will have the same voltage drop across them. Thus, the voltage across the 4 Ω resistor is also 12 V. Again, using Ohm's Law:

\[
I = \frac{V}{R} = \frac{12 V}{4 \, \Omega} = 3 \, A
\]

**Answer 2**: 3 A

### 4.3 The total current in the ammeter
Since the resistors are in parallel, the total current \( I_{total} \) is the sum of the currents through each resistor:

\[
I_{total} = I_{8 \, \Omega} + I_{4 \, \Omega} = 1.5 \, A + 3 \, A = 4.5 \, A
\]

**Answer 3**: 4.5 A

### 4.4 The value of resistor 3
To find the value of resistor 3, we need to know the voltage drop across it. Assuming the total voltage supplied is 12 V (the same voltage across the parallel combination), the voltage drop across resistor 3 would be:

\[
V_{3} = V_{total} - V_{parallel} = V_{total} - 12V
\]

However, if the total voltage supplied by the source was not given, we cannot calculate the resistance of resistor 3 directly without more information. Assuming the total voltage is greater than 12 V, if we denote the supply voltage as \( V_{s} \), then:

If the total voltage is not provided, we cannot calculate.

If we assume that the voltage drop across the entire circuit equals the total voltage supplied, we would have more specific calculations based on the power requirements or other specified conditions.

So, without additional information about the supply voltage, we cannot definitively answer this. If you could provide the voltage source or other parameters, we can calculate it.

### 4.5 The equivalent resistance
To find the equivalent resistance (\( R_{eq} \)), you first calculate the equivalent resistance of the parallel combination and then add the series resistance:

1. For the parallel resistors \( R_1 = 8 \, \Omega \) and \( R_2 = 4 \, \Omega \):

\[
\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{8} + \frac{1}{4} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}
\]

So,

\[
R_{parallel} = \frac{8}{3} \, \Omega
\]

2. If the value of resistor 3 (let's denote it as \( R_3 \)) is known or assumed, the total equivalent resistance in series with it would be:

\[
R_{eq} = R_{parallel} + R_3 = \frac{8}{3} + R_3 \, \Omega
\]

Again, without the value of resistor 3 or the specifications, we need additional values to compute this.

#### Summary:
- **Answer 1**: 1.5 A
- **Answer 2**: 3 A
- **Answer 3**: 4.5 A
- **Answer 4**: (need supply voltage or specific conditions to answer)
- **Answer 5**: \( R_{eq} = \frac{8}{3} + R_3 \, \Omega \) (need resistor 3 value)