Electric Energy and Power

When electric charge flows through a conductor, electrical energy is converted into other forms — heat, light, or mechanical work. This concept explains how much energy is transferred and at what rate, which is essential for understanding household appliances, fuse ratings, and circuit safety.

Real-Life Analogy: Think of electric power like the flow rate of water through a pipe. Voltage is like water pressure, current is the flow rate, and power is how much "work" that flowing water can do per second — the higher the pressure and flow, the more work gets done.

Step 1: Start with potential difference

• Potential difference (V) is the work done per unit charge:
  V = \[\frac {W}{Q}\]​, which rearranges to W = QV.

Step 2: Bring in current

• Since charge Q = It (current × time), substitute into the energy equation: W = VIt
  This gives energy in terms of voltage, current, and time.

Step 3: Apply Ohm's Law

• Using V = IR, substitute into W = VIt: W = (IR)It = I²Rt
• Or substitute I = V/R instead: W = V(\[\frac {V}{R}\])t = \[\frac {V^2t}{R}\]

Step 4: Convert energy to power
Power is energy per unit time, P = W/t. Divide each energy expression by t:

• P = VI
• P = I²R
• P = \[\frac {V^2}{R}\]

A frequently confused concept — presented here as a direct comparison.

Given: A bulb is rated 500 W, 100 V. It is connected to a 200 V supply. Find the resistance that must be connected in series to allow the bulb to operate safely at its rated power.

Solution:

1. Rated current of bulb: I = P/V = 500/100 = 5 A
2. Resistance of bulb: R_bulb = V/I = 100/5 = 20 Ω
3. Extra voltage to be dropped across series resistor: 200 − 100 = 100 V
4. Series resistance needed: R = V/I = 100/5 = 20 Ω

Answer: A series resistance of 20 Ω is required.