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3a) A 2A electrical heater was connected to a 110V supply for 1 hour. Calculate the cost of running the electricity electric heater. b) State one limited limitation of the ohms Ohion's law. c) Name the two cables on a pin plug.

Physics

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a) To calculate the cost of running the electrical heater, we first need to find the power consumed and then the total energy consumed. The cost of electricity per unit is not provided, so we will assume a common rate of $0.15 per kilowatt-hour (kWh).

  1. Calculate the power (PP) of the heater.

    P=V×IP = V \times I P=110 V×2 AP = 110 \text{ V} \times 2 \text{ A} P=220 WP = 220 \text{ W}
  2. Convert the power from watts to kilowatts (kW).

    P=220 W1000P = \frac{220 \text{ W}}{1000} P=0.22 kWP = 0.22 \text{ kW}
  3. Calculate the total energy (EE) consumed in kilowatt-hours (kWh).

    E=P×tE = P \times t E=0.22 kW×1 hourE = 0.22 \text{ kW} \times 1 \text{ hour} E=0.22 kWhE = 0.22 \text{ kWh}
  4. Calculate the cost of running the heater.

    (Assuming electricity cost is $0.15 per kWh).

    Cost=E×Cost per kWh\text{Cost} = E \times \text{Cost per kWh}

    Cost=0.22 kWh×$0.15/kWh\text{Cost} = 0.22 \text{ kWh} \times \$0.15/\text{kWh} Cost=$0.033\text{Cost} = \$0.033 The cost of running the electric heater is \boxed{\text{}\0.033}$

    b) One limitation of Ohm's Law is that it only applies to ohmic materials (like most metals) where resistance remains constant regardless of the voltage or current. It does not apply to non-ohmic devices (e.g., diodes, transistors, or thermistors) where the resistance changes with voltage or current, or if physical conditions like temperature change significantly.

    c) The two cables on a 2-pin plug are the:

    • Live wire
    • Neutral wire
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