Electrical Power and Energy Transfer

$\boxed{P = I \times V}$

Where:

• $P$ = power (watts, W)
• $I$ = current (amperes, A)
• $V$ = potential difference (volts, V)

Alternative Equations

Combining $P = IV$ with $V = IR$:

$\boxed{P = I^2 \times R}$

Combining $P = IV$ with $I = V/R$:

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

When to Use Each

• $P = IV$ — when you know current and voltage
• $P = I^2R$ — when you know current and resistance (no voltage)
• $P = V^2/R$ — when you know voltage and resistance (no current)

$\boxed{E = P \times t}$

Also: $E = I \times V \times t$ and $E = Q \times V$

Where:

• $E$ = energy (joules, J)
• $P$ = power (watts, W)
• $t$ = time (seconds, s)

Electricity companies charge per kilowatt-hour (kWh):

$\text{Energy (kWh)} = \text{Power (kW)} \times \text{Time (hours)}$

$\text{Cost} = \text{Energy (kWh)} \times \text{price per kWh}$

Remember: Convert W to kW (÷1000) and minutes/seconds to hours (÷60 or ÷3600).

Fuses protect circuits from excessive current. To choose a fuse:

1. Calculate the normal operating current: $I = P/V$
2. Choose the next fuse rating above this value

Standard fuse ratings: 3 A, 5 A, 13 A

Example: A 920 W appliance at 230 V: $I = 920/230 = 4$ A → use a 5 A fuse.

• $P = IV$, $P = I^2R$, $P = V^2/R$
• $E = Pt = IVt$
• Energy in kWh = power (kW) × time (hours)
• Cost = energy (kWh) × price per kWh
• Fuse rating: calculate $I = P/V$, choose next fuse above