State Ohm's law. 2) Calculate the reading of the instruments labeled A and V. 3) What is the cost of using resistor R for 6 hour a day for 30 days if power costs 15p per kWh?

Using Ohm's law, $I = \frac{V}{R}$.

$$I = \frac{12 \text{ V}}{5.5 \text{ } \Omega}$$ $$I \approx 2.1818 \text{ A}$$

The reading of instrument A is approximately 2.18 A.

Since the resistor is connected directly across the 12 V supply in a simple series circuit, the voltage across it is equal to the supply voltage.

$$V_{resistor} = V_{supply}$$ $$V_{resistor} = 12 \text{ V}$$

The reading of instrument V is 12 V.

3) What is the cost of using resistor R for 6 hours a day for 30 days if power costs 15N per kWh?

Power $P = \frac{V^2}{R}$

$$P = \frac{(12 \text{ V})^2}{5.5 \text{ } \Omega}$$ $$P = \frac{144 \text{ V}^2}{5.5 \text{ } \Omega}$$ $$P \approx 26.1818 \text{ W}$$

Total time $t = 6 \text{ hours/day} \times 30 \text{ days}$

$$t = 180 \text{ hours}$$

Energy $E = P \times t$

$$E = 26.1818 \text{ W} \times 180 \text{ hours}$$ $$E = 4712.724 \text{ Wh}$$

Convert Watt-hours to kilowatt-hours:

$$E = \frac{4712.724 \text{ Wh}}{1000 \text{ Wh/kWh}}$$ $$E \approx 4.7127 \text{ kWh}$$

Cost = Energy consumed $\times$ Cost per kWh

$$Cost = 4.7127 \text{ kWh} \times 15 \text{ N/kWh}$$ $$Cost \approx 70.6905 \text{ N}$$

The total cost is approximately 70.69 N.

Final Answers:

1. Ohm's law: $V = IR$
2. Ammeter reading (A): $\boxed{\text{2.18 A}}$ Voltmeter reading (V): $\boxed{\text{12 V}}$
3. Cost of using resistor R: $\boxed{\text{70.69 N}}$