The potential difference (voltage) of one cell is 1,5 V. Determine the readings on V₁, V₂ and V₃ with reasons.

Since there are two identical cells in series, the total voltage supplied is the sum of their individual voltages.

$$V_{\text{total}} = 1.5 \text{ V} + 1.5 \text{ V} = 3.0 \text{ V}$$

This total voltage is measured by $V_3$, which is across the two cells. So, $V_3 = 3.0 \text{ V}$.

In a parallel circuit, the voltage across each branch (and across the entire parallel combination) is the same as the total voltage supplied by the source.

• $V_1$ is connected across one bulb, which is a parallel branch. Therefore, $V_1 = V_{\text{total}} = 3.0 \text{ V}$.
• $V_2$ is connected across the entire parallel combination of the three bulbs. Therefore, $V_2 = V_{\text{total}} = 3.0 \text{ V}$.

Here is the completed table for Question 1:

$$\boxed{ \begin{array}{|l|l|l|} \hline \textbf{Reading} & \textbf{Value with unit} & \textbf{Reason} \\ \hline V_1 & \textbf{3.0 V} & \text{Voltage across parallel branches is equal to total voltage.} \\ \hline V_2 & \textbf{3.0 V} & \text{Voltage across parallel combination is equal to total voltage.} \\ \hline V_3 & \textbf{3.0 V} & \text{Total voltage of 2 cells in series.} \\ \hline \end{array} }$$

Ammeter $A_4$ is in series with the two resistors in the lower branch. In a series circuit, the current is the same at all points. Since $A_2$ measures the current in this branch, $A_4$ will have the same reading as $A_2$.

$$A_4 = A_2 = 4 \text{ A}$$

Let the resistance of one resistor be $R$.

• The lower branch has two identical resistors in series, so its total resistance is $R_{\text{lower}} = R + R = 2R$.
• The upper branch has one resistor, so its resistance is $R_{\text{upper}} = R$.
• In a parallel circuit, the voltage across both branches is the same. Let this voltage be $V$.
• For the lower branch: $V = A_2 \times R_{\text{lower}} = 4 \text{ A} \times 2R = 8R$.
• For the upper branch: $V = A_1 \times R_{\text{upper}} \implies 8R = A_1 \times R$.
• Solving for $A_1$: $A_1 = \frac{8R}{R} = 8 \text{ A}$.

Ammeter $A_3$ measures the total current flowing from the battery. In a parallel circuit, the total current is the sum of the currents in the individual parallel branches.

$$A_3 = A_1 + A_2 = 8 \text{ A} + 4 \text{ A} = 12 \text{ A}$$

Here is the completed table for Question 2:

$$\boxed{ \begin{array}{|l|l|l|} \hline \textbf{Reading} & \textbf{Value with unit} & \textbf{Reason} \\ \hline A_1 & \textbf{8 A} & \text{Voltage across parallel branches is equal. Current is inversely proportional to resistance (V/R).} \\ \hline A_2 & 4 \text{ A} & \text{Given} \\ \hline A_3 & \textbf{12 A} & \text{Total current is the sum of currents in parallel branches.} \\ \hline A_4 & \textbf{4 A} & \text{Current is the same at all points in a series circuit (within a branch).} \\ \hline \end{array} }$$