QUESTION 8 (Start on a new page.) A battery with an internal resistance of 0,5 Ω and an unknown emf (ε) is connected to three resistors, a high resistance voltmeter and an ammeter of negligible resistance, as shown in the circuit diagram below. The resistance of the connecting wires must be ignored.

When S is open, only $R_1$ is in the external circuit.

$$R_{ext, open} = R_1 = 4 \, \Omega$$

The total resistance of the circuit is $R_{total, open} = R_{ext, open} + r = 4 \, \Omega + 0.5 \, \Omega = 4.5 \, \Omega$.
The current flowing is $I_{open} = \frac{\mathcal{E}}{R_{total, open}} = \frac{\mathcal{E}}{4.5}$.
The voltmeter measures the voltage across $R_1$:

$$V_{R1, open} = I_{open} R_1 = \frac{\mathcal{E}}{4.5} \times 4 = \frac{8}{9}\mathcal{E}$$

When S is closed, $R_2$ and $R_3$ are in parallel.

$$R_p = \left(\frac{1}{R_2} + \frac{1}{R_3}\right)^{-1} = \left(\frac{1}{25 \, \Omega} + \frac{1}{15 \, \Omega}\right)^{-1}$$ $$R_p = \left(\frac{3+5}{75}\right)^{-1} = \left(\frac{8}{75}\right)^{-1} = \frac{75}{8} \, \Omega = 9.375 \, \Omega$$

The total external resistance is $R_{ext, closed} = R_1 + R_p = 4 \, \Omega + 9.375 \, \Omega = 13.375 \, \Omega$.
The total resistance of the circuit is $R_{total, closed} = R_{ext, closed} + r = 13.375 \, \Omega + 0.5 \, \Omega = 13.875 \, \Omega$.
The current flowing is $I_{closed} = \frac{\mathcal{E}}{R_{total, closed}} = \frac{\mathcal{E}}{13.875}$.
The voltmeter measures the voltage across $R_1$:

$$V_{R1, closed} = I_{closed} R_1 = \frac{\mathcal{E}}{13.875} \times 4 = \frac{32}{111}\mathcal{E}$$

$$V_{R1, open} - V_{R1, closed} = 1.5 \, V$$ $$\frac{8}{9}\mathcal{E} - \frac{32}{111}\mathcal{E} = 1.5$$ $$\mathcal{E} \left(\frac{8}{9} - \frac{32}{111}\right) = 1.5$$ $$\mathcal{E} \left(\frac{8 \times 37 - 32 \times 3}{333}\right) = 1.5$$ $$\mathcal{E} \left(\frac{296 - 96}{333}\right) = 1.5$$ $$\mathcal{E} \left(\frac{200}{333}\right) = 1.5$$ $$\mathcal{E} = 1.5 \times \frac{333}{200} = \frac{3}{2} \times \frac{333}{200} = \frac{999}{400}$$ $$\mathcal{E} = 2.4975 \, V$$

The EMF of the battery is $\boxed{\text{2.50 V}}$.

8.3.2 Total external resistance of the circuit

$$R_{ext, closed} = R_1 + R_p$$ $$R_{ext, closed} = 4 \, \Omega + 9.375 \, \Omega$$ $$R_{ext, closed} = 13.375 \, \Omega$$

The total external resistance of the circuit is $\boxed{\text{13.38 \Omega}}$.

8.3.1 Reading on the ammeter

The ammeter measures the total current $I_{closed}$.

$$I_{closed} = \frac{\mathcal{E}}{R_{ext, closed} + r}$$ $$I_{closed} = \frac{2.4975 \, V}{13.375 \, \Omega + 0.5 \, \Omega}$$ $$I_{closed} = \frac{2.4975 \, V}{13.875 \, \Omega}$$ $$I_{closed} \approx 0.180 \, A$$

The reading on the ammeter is $\boxed{\text{0.18 A}}$.

8.4 A learner makes the following statement: The current through resistor $R_3$ is larger than the current through resistor $R_2$. Is this statement CORRECT? Choose from YES or NO. Explain the answer.

• YES
• Explanation: Resistors $R_2$ and $R_3$ are connected in parallel, meaning the potential difference across them is the same. According to Ohm's Law ($I = V/R$), for a constant voltage, the current is inversely proportional to the resistance. Since $R_3$ ($15 \, \Omega$) is smaller than $R_2$ ($25 \, \Omega$), the current through $R_3$ will be larger than the current through $R_2$.

8.5 The 4 $\Omega$ resistor is now removed from the circuit. How will this affect the emf of the battery? Choose from INCREASES, DECREASES or REMAINS THE SAME.

• REMAINS THE SAME
• Explanation: The electromotive force (EMF) of a battery is an intrinsic property of the battery itself, determined by its chemical composition. It does not depend on the external circuit connected to it. Removing a resistor from the external circuit will change the external resistance, current, and terminal potential difference, but not the EMF.