Electric Power: Study Guide (Physics 2 11.4)

Power as a rate of energy transfer

In Topic 11.3 you saw that a resistor converts electrical energy into thermal energy. This topic is about how fast it does so. Power is the rate at which energy is transferred, converted or dissipated:

P = ΔE / Δt

Its unit is the watt (W), and 1 W = 1 J/s.

Deriving P = IΔV

Two definitions you already know give the electrical form.

• Potential difference is energy per unit charge (Topic 10.5): when a charge q passes through an element with potential difference ΔV across it, the energy transferred is ΔE = qΔV.
• Current is charge per unit time (Topic 11.1): I = q/Δt.

Divide the energy by the time:

P = ΔE/Δt = qΔV/Δt = (q/Δt)ΔV, so P = IΔV

This holds for any circuit element: a resistor, a bulb, a motor or a battery. Units check: A × V = (C/s)(J/C) = J/s = W.

Example: 2.0 C passes through a lamp with 6.0 V across it in 0.50 s. The energy transferred is 2.0 C × 6.0 V = 12 J, so P = 12 J ÷ 0.50 s = 24 W. Using the formula: I = 2.0 C ÷ 0.50 s = 4.0 A and P = 4.0 A × 6.0 V = 24 W.

Two forms for a resistor

For a resistor, ΔV = IR. Substituting into P = IΔV gives two derived forms:

• replace ΔV: P = I(IR), so P = I²R
• replace I: P = (ΔV/R)ΔV, so P = ΔV²/R

All three forms give the same number for the same resistor. Choose the one whose quantities you know, or the one that keeps the shared quantity in view when you compare elements.

Energy into, out of and within a circuit

Follow the energy around a single loop.

• Into the circuit. The battery converts chemical energy into electrical energy. Charges gain potential energy as they pass through it. For an ideal battery of emf ℰ carrying current I, the rate is P = Iℰ.
• Out of the circuit. In each resistor, charges lose potential energy and the energy becomes thermal energy, which spreads into the surroundings. A bulb also emits light. We say the energy is dissipated: it leaves the circuit and cannot be recovered as electrical energy.
• Within the circuit. Some elements store energy instead. A charging capacitor (Topic 10.6) stores it in its electric field, and can return it later.

Energy is conserved. With an ideal battery and ideal wires, the rate at which energy enters equals the total rate at which it is dissipated and stored. This is a powerful check on any calculation.

How power depends on current, potential difference and resistance

For an ohmic resistor (constant R):

• P ∝ I² and P ∝ ΔV². Double the current through a resistor and its power becomes 4 times as large. Triple the potential difference and the power becomes 9 times as large.
• What happens when R changes depends on what stays fixed:
  • Same ΔV (each resistor connected directly across the same battery): P = ΔV²/R, so the smaller resistance has the greater power. With 9.0 V across it, a 3.0 Ω resistor dissipates 27 W but a 6.0 Ω resistor only 13.5 W.
  • Same I (resistors one after another in a single loop, which share one current): P = I²R, so the larger resistance has the greater power. With 2.0 A in each, the 3.0 Ω resistor dissipates 12 W and the 6.0 Ω resistor 24 W.

Neither rule is wrong. They answer different questions. Always ask “what is the same for these elements?” before choosing.

Bulb brightness

The brightness of a bulb increases with its power. So to decide which of several bulbs is brightest, compare their powers. You often do not need numbers:

1. Decide what the bulbs share: the same ΔV, or the same I.
2. Use P = ΔV²/R or P = I²R to compare.
3. A bulb that carries no current (in an open loop or short-circuited, Topic 11.2) has P = 0 and is off.

Bulbs are often labelled with a rating, such as “6.0 V, 3.0 W”. That means 3.0 W when 6.0 V is across it. With a different ΔV the power is different. In this course, treat a bulb’s resistance as constant unless the question says otherwise, even though a real filament’s resistance rises as it heats (Topic 11.3).

Worked example 1: a heater three ways

Question. A heater has a resistance of 24 Ω and is connected across a 120 V supply. Find (a) the current, (b) the power, using all three forms, and (c) the energy transferred in 3.0 minutes.

1. (a) I = ΔV/R = 120 V ÷ 24 Ω = 5.0 A.
2. (b) P = IΔV = (5.0 A)(120 V) = 600 W.
3. Check: P = I²R = (5.0 A)²(24 Ω) = 600 W, and P = ΔV²/R = (120 V)² ÷ 24 Ω = 600 W.
4. (c) Δt = 3.0 × 60 s = 180 s. ΔE = PΔt = (600 W)(180 s) = 1.08 × 10⁵ J.

Answer. (a) 5.0 A; (b) 600 W; (c) 1.1 × 10⁵ J.

Check. All three forms agree, which confirms the arithmetic. If you forget to convert minutes, you get 1800 J, sixty times too small.

Worked example 2: which bulb is brighter?

Question. Bulb X is rated 6.0 V, 3.0 W and bulb Y is rated 6.0 V, 12 W. Assume each has constant resistance. (a) Each bulb in turn is connected alone across a 6.0 V battery. Which is brighter? (b) Both bulbs are now connected one after the other in a single loop with the 6.0 V battery, as in Figure 2. An ammeter in the loop reads 0.40 A. Find the power of each bulb and say which is brighter. (c) Show that energy is conserved.

1. Resistances from the ratings, R = ΔV²/P: R_X = (6.0 V)² ÷ 3.0 W = 12 Ω; R_Y = (6.0 V)² ÷ 12 W = 3.0 Ω.
2. (a) Alone across 6.0 V, each bulb has its rated ΔV, so P_X = 3.0 W and P_Y = 12 W. Y is brighter. (Same ΔV: smaller R, more power.)
3. (b) In one loop the current is the same in both: I = 0.40 A. Use P = I²R.
   • P_X = (0.40 A)²(12 Ω) = 1.92 W.
   • P_Y = (0.40 A)²(3.0 Ω) = 0.48 W.
   • X is brighter, four times the power of Y. (Same I: larger R, more power.)
4. (c) Power delivered by the battery: Iℰ = (0.40 A)(6.0 V) = 2.4 W. Total dissipated: 1.92 W + 0.48 W = 2.4 W. They match.

Answer. (a) Y is brighter; (b) P_X = 1.9 W, P_Y = 0.48 W, so X is brighter; (c) 2.4 W in and 2.4 W out.

Interpretation. The “stronger” 12 W bulb is the dimmer one in the single loop. The rating describes the bulb at 6.0 V, not in every circuit. Both bulbs are also dimmer than when each was alone, because they now carry a smaller current.

Worked example 3: a symbolic derivation and a factor of change

Question. A heater element is a wire of resistivity ρ, length L and diameter d, connected across a fixed potential difference ΔV. (a) Derive an expression for its power in terms of these quantities. (b) The original heater has a power of 900 W. A new element is made from the same material, with 1.5 times the diameter and twice the length, and is connected to the same supply. Find its power.

1. (a) Resistance (Topic 11.3): R = ρL/A with A = πd²/4, so R = 4ρL/(πd²).
2. Fixed ΔV, so use P = ΔV²/R = ΔV² ÷ [4ρL/(πd²)].
3. P = πd²ΔV² / (4ρL).
4. (b) Only d and L change, so P ∝ d²/L. Factor = (1.5)² ÷ 2 = 2.25 ÷ 2 = 1.125.
5. New power = 1.125 × 900 W = 1012.5 W.

Answer. (a) P = πd²ΔV²/(4ρL); (b) about 1.0 × 10³ W (1010 W).

Check. Units: (m²)(V²) ÷ [(Ω·m)(m)] = V²/Ω = W. Sense: a thicker wire has less resistance, so more power at fixed ΔV; a longer wire has more resistance, so less. The thicker wire wins slightly. Note also that halving ΔV would cut the power to a quarter, because P ∝ ΔV².

Common misconceptions

• “A bigger resistance always means more power.” Only when the current is the same. At the same ΔV the smaller resistance has more power.
• “The higher-rated bulb is always brighter.” A rating only applies at the stated ΔV. Worked example 2 shows the opposite in a single loop.
• “Current is used up in a bulb.” The current entering and leaving a bulb is the same. Energy is what the bulb transfers out of the circuit.
• “Power is energy.” Power is the rate of energy transfer. Energy = power × time, with time in seconds for joules.
• “Doubling the current doubles the power.” For a resistor, P ∝ I², so doubling I makes P four times as large.
• Using P = IΔV with the battery’s emf for one bulb. Use the potential difference across that particular element.

Where this leads

Power is the energy side of every circuit you will analyse. Next, Topic 11.5 combines resistors into compound circuits, where you will predict bulb brightness in branches using these rules: continue with Topic 11.5, Compound Direct Current (DC) Circuits. First test yourself with the practice questions, then use the revision notes and the topic checklist to consolidate.