Solve 3 (b) (i) Ammeter reading when key K is open.

When the key K is open, the circuit branch containing the ammeter is an open circuit. No current can flow through an open circuit.
The ammeter reading will be $0 \, \text{A}$.

When the key K is open, the branch with the 20 $\Omega$ resistor is still connected directly across the 12 V voltage source. Therefore, the voltage across the 20 $\Omega$ resistor is the source voltage.
The voltmeter reading will be $12 \, \text{V}$.

The circuit diagram shows two resistors in parallel. One resistor is 20 $\Omega$. The other resistor (in the branch with the switch and ammeter) has no value indicated. We will assume this unlabeled resistor is also $20 \, \Omega$ for the calculation.
When the key K is closed, the two $20 \, \Omega$ resistors are in parallel.
The combined resistance ($R_{eq}$) for two parallel resistors is given by:

Substituting the values:

The combined resistance of the circuit is $\boxed{\text{}10 \, \Omega\text{}}$ (assuming the unlabeled resistor is $20 \, \Omega$).

Assuming the resistor in the ammeter branch is $20 \, \Omega$, and the key K is closed, the voltage across this parallel branch is $12 \, \text{V}$.
The current ($I$) through this branch is given by Ohm's Law:

Substituting the values:

The ammeter reading will be $\boxed{\text{}0.6 \, \text{A}\text{}}$ (assuming the unlabeled resistor is $20 \, \Omega$).

1 kilowatt-hour (kWh) is a unit of energy.

Convert kilowatts to watts and hours to seconds:

First, calculate the total power consumption:
Power of cooker = $1000 \, \text{W} = 1 \, \text{kW}$
Power of immersion heater = $300 \, \text{W} = 0.3 \, \text{kW}$
Total power = $1 \, \text{kW} + 0.3 \, \text{kW} = 1.3 \, \text{kW}$
Next, calculate the total energy consumed per day:
Energy = Total power $\times$ Time
Energy = $1.3 \, \text{kW} \times 10 \, \text{h} = 13 \, \text{kWh}$
Finally, calculate the total cost:
Cost = Energy consumed $\times$ Cost per kWh
Cost = $13 \, \text{kWh} \times 75 \, \text{frs/kWh} = 975 \, \text{frs}$
The cost of electricity used is $\boxed{\text{}975 \, \text{frs}\text{}}$

Two properties of plastics that make them useful as engineering materials are:

• They are lightweight with a high strength-to-weight ratio.