IB — physicsFoundational2 min read

Electrical Power and Energy

P = IV = I²R = V²/R; energy E = Pt; cost of electricity; kWh

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# Electrical Power and Energy — IB Physics

1. Electrical Power

P=IV=I2R=V2R\boxed{P = IV = I^2R = \frac{V^2}{R}}

Power = rate of energy transfer. Units: watts (W).

2. Electrical Energy

E=Pt=IVtE = Pt = IVt

Units: joules (J) or kilowatt-hours (kWh). 11 kWh =3.6×106= 3.6 \times 10^6 J.

3. Cost of Electricity

Cost=energy (kWh)×price per kWh\text{Cost} = \text{energy (kWh)} \times \text{price per kWh}

4. Power Transmission

Power loss in cables: Ploss=I2RP_{\text{loss}} = I^2R

To reduce loss: use high voltage (lower current for same power). This is why electricity is transmitted at high voltage and stepped down by transformers.

Worked Example: Example 1

Problem

A 60W bulb operates for 8 hours. Cost at 0.25/kWh.0.25/kWh. E = 0.060 \times 8 = 0.48kWh.Cost=kWh. Cost =0.48 \times 0.25 = $0.12$.

Solution

Worked Example: Example 2

Problem

A heater draws 13A from 230V. Find power and energy in 2 hours. P=230×13=2990P = 230 \times 13 = 2990 W ≈ 3 kW. E=3×2=6E = 3 \times 2 = 6 kWh = 2.16×1072.16 \times 10^7 J.

Solution

Worked Example: Transmission

Problem

100 kW transmitted through 5Ω cables. Compare losses at 1000V vs 100,000V. At 1000V: I=100000/1000=100I = 100000/1000 = 100 A. Ploss=1002×5=50,000P_{\text{loss}} = 100^2 \times 5 = 50{,}000 W (50% lost!) At 100,000V: I=1I = 1 A. Ploss=12×5=5P_{\text{loss}} = 1^2 \times 5 = 5 W (negligible)

Solution

6. Practice Questions

    1. A 2kW kettle boils water for 3 minutes. Find energy used in kJ and kWh. (2 marks)
    1. Explain why the National Grid uses high voltages. (3 marks)
    1. A 12V battery powers a motor at 3A for 1 minute. The motor lifts 2kg by 5m. Find the efficiency. (4 marks)

    Answers

    1. E=2000×180=360,000E = 2000 \times 180 = 360{,}000 J = 360 kJ. =360000/3.6×106=0.1= 360000/3.6 \times 10^6 = 0.1 kWh.
    1. Power loss = I2RI^2R. At higher voltage, the same power can be transmitted with lower current (since P=IVP = IV). Lower current means less power lost as heat in the cables.

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Summary

  • P=IV=I2R=V2/RP = IV = I^2R = V^2/R
  • E=PtE = Pt; 1 kWh = 3.6 MJ
  • Transmission: high V → low I → low I2RI^2R losses
Tags:
#ib#physics#electricity#power#energy#kwh#efficiency

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