A) (i) Name the meters A and V. (ii) Calculate the readings of meters A and V when the key K is open. (iii) Calculate the reading of meters A and V when K is closed.

Since the circuit is open, no current flows through the ammeter A.

$$\text{Reading of A} = \boxed{\text{}0 \text{ A}\text{} }$$

With no current flowing through the $2 \Omega$ resistor, there is no potential difference across it ($V=IR$).

$$\text{Reading of V} = \boxed{\text{}0 \text{ V}\text{} }$$

a) (iii) Calculate the reading of meters A and V when K is closed.
When the key K is closed, the $10 \Omega$ and $5 \Omega$ resistors are in parallel, and this combination is in series with the $2 \Omega$ resistor. The total voltage supplied by the battery is $12 \text{ V}$.

$$\frac{1}{R_p} = \frac{1}{10 \Omega} + \frac{1}{5 \Omega}$$ $$\frac{1}{R_p} = \frac{1}{10 \Omega} + \frac{2}{10 \Omega} = \frac{3}{10 \Omega}$$ $$R_p = \frac{10}{3} \Omega \approx 3.33 \Omega$$

The parallel combination is in series with the $2 \Omega$ resistor.

$$R_{total} = R_p + 2 \Omega$$ $$R_{total} = \frac{10}{3} \Omega + 2 \Omega = \frac{10}{3} \Omega + \frac{6}{3} \Omega = \frac{16}{3} \Omega \approx 5.33 \Omega$$

$$I_A = \frac{12 \text{ V}}{\frac{16}{3} \Omega} = \frac{12 \times 3}{16} \text{ A} = \frac{36}{16} \text{ A} = \frac{9}{4} \text{ A}$$ $$I_A = 2.25 \text{ A}$$

The reading of meter A is $\boxed{\text{}2.25 \text{ A}\text{}}$.

The total current $I_A$ flows through the $2 \Omega$ resistor.

$$V_V = I_A \times 2 \Omega$$ $$V_V = \frac{9}{4} \text{ A} \times 2 \Omega = \frac{18}{4} \text{ V} = \frac{9}{2} \text{ V}$$ $$V_V = 4.5 \text{ V}$$

The reading of meter V is $\boxed{\text{}4.5 \text{ V}\text{}}$.

b) In a modern power socket, there are three cable terminals. One of the cables carries a switch and fuse. Name the three cables and specify the one which each of fuse and switch is connected to.

• The three cables are: Live wire, Neutral wire, and Earth wire.
• The fuse and the switch are both connected to the Live wire.

c) (i) What is the equivalence of $1 \text{ kWh}$ in joules?

$$1 \text{ kWh} = 1 \text{ kilowatt} \times 1 \text{ hour}$$ $$1 \text{ kW} = 1000 \text{ W}$$ $$1 \text{ hour} = 3600 \text{ seconds}$$

$$1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s}$$ $$1 \text{ kWh} = 3,600,000 \text{ J}$$ $$1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$$

The equivalence of $1 \text{ kWh}$ in joules is $\boxed{\text{}3.6 \times 10^6 \text{ J}\text{}}$.

c) (ii) On an electric cooker is marked $4000 \text{ W}$. Find the cost of using this cooker 5 hours daily for 30 days, if electricity cost 55 Frs. per unit.

$$E = P \times \text{Total time}$$ $$E = 4 \text{ kW} \times 150 \text{ hours} = 600 \text{ kWh}$$

The cost per unit is 55 Frs. and 1 unit is $1 \text{ kWh}$.

$$\text{Total cost} = \text{Energy consumed} \times \text{Cost per unit}$$ $$\text{Total cost} = 600 \text{ kWh} \times 55 \text{ Frs./kWh} = 33,000 \text{ Frs.}$$

The cost of using the cooker is $\boxed{\text{}33,000 \text{ Frs.}\text{}}$.