We need to determine whether the circuit is a series, parallel, or combination type of circuit. We do this by looking at how current flows through a circuit.

Recall that current is the movement of electric charge from a higher potential to a lower potential.

We assume that the flowing charge is positive (conventional current). This means that the charges start at the positive terminal of our power source (battery). Here the charges have lots of electrical potential energy.

The charges then move towards the negative terminal through the path that is created by the components and wires of the circuit. During this process, the electrical potential energy is converted into thermal energy by the resistors. Therefore, the charges have less electrical potential energy when they reach the negative terminal of the power source (battery).

The flow of current is indicated by the arrows in the diagram below:

From the diagram it is clear that there is only one path for the current to flow, since the circuit does not split into two or more paths. This means that the current flows through all the components, one after the other (in series). We will label this current \(I_A\).

The circuit shown in the diagram is a series circuit.

\(0\)

We know that the circuit in the diagram is a series type circuit.

Recall that in a series circuit the equivalent resistance can be calculated by summing the resistance values of the individual resistors:

\[R_{eq,S} = R_1 + R_2 + R_3 + \ldots ~~~~~\text{(1)}\]

We can rewrite \(\text{(1)}\) for the circuit in question as:

\[R_{eq} = R_1 + R_2 + R_3 + R_4 ~~~~~\text{(2)}\]

We can represent this by drawing an equivalent circuit:

We are given the following information:

• the resistance of \(R_1 = \text{4,7}\text{ Ω}\)
• the resistance of \(R_2 = \text{6,9}\text{ Ω}\)
• the resistance of \(R_3 = \text{4,9}\text{ Ω}\)
• the resistance of \(R_4 = \text{4,3}\text{ Ω}\)

Substituting the above values in \(\text{(2)}\), we get:

\begin{align*} R_{eq} & = R_1 + R_2 + R_3 + R_4\\ & = \text{4,7} + \text{6,9} + \text{4,9} + \text{4,3}\\ & = \text{20,8}\text{ Ω} \end{align*}

Therefore, the total equivalent resistance in the circuit, or \(R_{eq} = \text{20,8}\text{ Ω}\)

\(20.8\)

Potential difference, or voltage is a way to describe the difference in electrical potential energy across a component in a circuit. Remember that a resistor converts electrical potential energy into thermal energy, so the electrical potential is higher on the "in" side compared to the "out" side of a resistor.

Ohm's law describes the relationship between the total current \(~I~\) through an ohmic conductor, its resistance \(R\), and the potential difference \(V~\) across it:

\[I = \frac{V}{R}~~~~~\text{(3)}\]

For the equivalent circuit, we can rewrite \(\text{(3)}\) as:

\[I_A = \frac{V_A}{R_{eq}}~~~~~\text{(4)}\]

We are asked to calculate the potential difference \(V_{1}\) over \(R_{1}\), as shown in the circuit diagram above, so we use Ohm's law and rewrite \(\text{(3)}\) to get:

\[V_1 = I_1R_1~~~~~\text{(5)}\]

Since the circuit in question is a series circuit, we know that the current \(I_{1}\) flowing through \(R_{1}\), is the same as the current \(I_{A}\). This is represented as:

\[I_1 = I_A~~~~~\text{(6)}\]

Substituting \(\text{(6)}\) into \(\text{(5)}\) we get:

\[V_1 = I_AR_1~~~~~\text{(7)}\]

Substituting \(\text{(4)}\) into \(\text{(7)}\) we get:

\[V_{1} = I_AR_1 = \left(\frac{V_A}{R_{eq}}\right)R_1 = \frac{V_AR_1}{R_{eq}}~~~~~\text{(8)}\]

Recall from Question 2:

\[R_{eq} = R_1 + R_2 + R_3 + R_4 ~~~~~\text{(2)}\]

Substituting \(\text{(2)}\) into \(\text{(8)}\) we get:

\[V_{1} = \frac{V_AR_1}{R_1 + R_2 + R_3 + R_4}~~~~~\text{(9)}\]

We are given the following information:

• the resistance of \(R_1 = \text{4,7}\text{ Ω}\)
• the resistance of \(R_2 = \text{6,9}\text{ Ω}\)
• the resistance of \(R_3 = \text{4,9}\text{ Ω}\)
• the resistance of \(R_4 = \text{4,3}\text{ Ω}\)
• the potential difference (voltage) over the battery \(V_A = \text{7,6}\text{ V}\)

Substituting the above values in \(\text{(9)}\), we get:

\begin{align*} V_{1} & = \frac{V_AR_1}{R_1 + R_2 + R_3 + R_4}\\ & = \frac {(\text{7,6})(\text{4,7})} {\text{4,7} + \text{6,9} + \text{4,9} + \text{4,3}}\\ & = 1,71731 \ldots~~~~~\text{(calculated)}\\ & \approx \text{1,717}\text{ V}~~~~~\text{(rounded)} \end{align*}

Therefore, the potential difference (voltage) across \(R_1\), or \(V_1 \approx \text{1,717}\text{ V}\).

\(1.717\)

We are asked to calculate the three remaining potential differences, \(V_2\), \(V_3\), and \(V_4\). This is shown in the diagram below:

We will use the same method to obtain three expression for \(V_2\), \(V_3\), and \(V_4\) as was used in Question 3 for \(V_1\). We get:

\[V_{2} = \frac{V_AR_2}{R_1 + R_2 + R_3 + R_4}~~~~~\text{(10)}\]

\[V_{3} = \frac{V_AR_3}{R_1 + R_2 + R_3 + R_4}~~~~~\text{(11)}\]

\[V_{4} = \frac{V_AR_4}{R_1 + R_2 + R_3 + R_4}~~~~~\text{(12)}\]

We are given the following information:

• the resistance of \(R_1 = \text{4,7}\text{ Ω}\)
• the resistance of \(R_2 = \text{6,9}\text{ Ω}\)
• the resistance of \(R_3 = \text{4,9}\text{ Ω}\)
• the resistance of \(R_4 = \text{4,3}\text{ Ω}\)
• the potential difference (voltage) over the battery \(V_A = \text{7,6}\text{ V}\)

Substituting the above values in \(\text{(10)}\), \(\text{(11)}\), and \(\text{(12)}\) we get:

\begin{align*} V_{2} & = \frac{V_AR_2}{R_1 + R_2 + R_3 + R_4}\\ & = \frac {(\text{7,6})(\text{6,9})} {\text{4,7} + \text{6,9} + \text{4,9} + \text{4,3}}\\ & = 2,52115 \ldots~~~~~\text{(calculated)}\\ & \approx \text{2,521}\text{ V}~~~~~\text{(rounded)} \end{align*}

\begin{align*} V_{3} & = \frac{V_AR_3}{R_1 + R_2 + R_3 + R_4}\\ & = \frac {(\text{7,6})(\text{4,9})} {\text{4,7} + \text{6,9} + \text{4,9} + \text{4,3}}\\ & = 1,79038 \ldots~~~~~\text{(calculated)}\\ & \approx \text{1,790}\text{ V}~~~~~\text{(rounded)} \end{align*}

\begin{align*} V_{4} & = \frac{V_AR_4}{R_1 + R_2 + R_3 + R_4}\\ & = \frac {(\text{7,6})(\text{4,3})} {\text{4,7} + \text{6,9} + \text{4,9} + \text{4,3}}\\ & = 1,57115 \ldots~~~~~\text{(calculated)}\\ & \approx \text{1,571}\text{ V}~~~~~\text{(rounded)} \end{align*}

Therefore, the potential difference (voltage) across \(R_2\), or \(V_2 \approx \text{2,521}\text{ V}\).

The potential difference (voltage) across \(R_3\), or \(V_3 \approx \text{1,790}\text{ V}\).

The potential difference (voltage) across \(R_4\), or \(V_4 \approx \text{1,571}\text{ V}\).

\(2.521\) \(1.790\) \(1.571\)