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

Hey Small, good to see you again. Step 1: 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 \, A. Step 2: Solve 3 (b) (ii) Voltmeter reading when key K is open. When the key K is open, the branch with the 20 resistor is still connected directly across the 12 V voltage source. Therefore, the voltage across the 20 resistor is the source voltage. The voltmeter reading will be 12 \, V. Step 3: Solve 3 (c) Combined resistance when key K is closed. The circuit diagram shows two resistors in parallel. One resistor is 20 . The other resistor (in the branch with the switch and ammeter) has no value indicated. We will assume this unlabeled resistor is also 20 \, for the calculation. When the key K is closed, the two 20 \, resistors are in parallel. The combined resistance (R_eq) for two parallel resistors is given by: R_eq = (R_1 × R_2)/(R_1 + R_2) Substituting the values: R_eq = (20 \, × 20 \, )/(20 \, + 20 \, ) = (400 \, ^2)/(40 \, ) = 10 \, The combined resistance of the circuit is 10 \, (assuming the unlabeled resistor is 20 \, ). Step 4: Solve 3 (d) Ammeter reading when key K is closed. Assuming the resistor in the ammeter branch is 20 \, , and the key K is closed, the voltage across this parallel branch is 12 \, V. The current (I) through this branch is given by Ohm's Law: I = (V)/(R) Substituting the values: I = 12 \, V20 \, = 0.6 \, A The ammeter reading will be 0.6 \, A (assuming the unlabeled resistor is 20 \, ). Step 5: Solve 3 (e) (i) Equivalence of 1 kWh in joules. 1 kilowatt-hour (kWh) is a unit of energy. 1 \, kWh = 1 \, kW × 1 \, h Convert kilowatts to watts and hours to seconds: 1 \, kW = 1000 \, W 1 \, h = 3600 \, s 1 \, kWh = 1000 \, W × 3600 \, s = 3,600,000 \, J 1 \, kWh = 3.6 × 10^6 \, J Step 6: Solve 3 (e) (ii) Calculate the cost of electricity. First, calculate the total power consumption: Power of cooker = 1000 \, W = 1 \, kW Power of immersion heater = 300 \, W = 0.3 \, kW Total power = 1 \, kW + 0.3 \, kW = 1.3 \, kW Next, calculate the total energy consumed per day: Energy = Total power × Time Energy = 1.3 \, kW × 10 \, h = 13 \, kWh Finally, calculate the total cost: Cost = Energy consumed × Cost per kWh Cost = 13 \, kWh × 75 \, frs/kWh = 975 \, frs The cost of electricity used is 975 \, frs Step 7: Solve 4 (a) (i) Two properties of plastics useful as engineering materials. Two properties of plastics that make them useful as engineering materials are: • They are lightweight with a high strength-to-weight ratio.