Power, Energy & Which Bulb Is Brightest
- Compute power dissipated using the three equivalent expressions
- Choose the appropriate power expression from what is held constant
- Rank bulbs by brightness in series and parallel arrangements
Three forms of the same equation
Power dissipated in a resistor is P = IV, and substituting Ohm's law gives P = I²R and P = V²/R. All three are always true. They are not interchangeable in practice, though, because which one makes a comparison easy depends on what is the same across the resistors being compared — and that is the actual skill being tested.
Brightness in series and in parallel
Bulbs in series carry the same current, so P = I²R makes the higher-resistance bulb brighter. Bulbs in parallel have the same voltage across them, so P = V²/R makes the lower-resistance bulb brighter. Students who memorize one case get the other exactly backward, which is why the exam usually presents both in the same question.
What happens when a bulb is removed
Remove a bulb from a series string and the circuit breaks — everything goes out. Remove one from a parallel branch and the others are unaffected, since each branch still has the full source voltage. The more interesting case is a mixed circuit: removing a parallel branch raises the total resistance, which reduces total current, which reduces the voltage dropped across any series element and therefore dims it. Tracing that chain is a standard free-response task.
Energy and cost
Energy is power times time, and the domestic unit is the kilowatt-hour: one kilowatt for one hour, which is 3.6 × 10⁶ J. A device rated in watts on a stated supply voltage has an implied resistance R = V²/P — which is how a "60 W bulb" question converts into a resistance you can put in a circuit.
A 4.0 Ω and a 12 Ω resistor are connected first in series, then in parallel, across a 24 V source. In each case find the power dissipated in each resistor.
- 1.Series: R_total = 16 Ω, so I = 24/16 = 1.5 A through both.
- 2.P₄ = I²R = (1.5)²(4.0) = 9.0 W. P₁₂ = (1.5)²(12) = 27 W. The larger resistor dissipates more.
- 3.Parallel: each has the full 24 V across it.
- 4.P₄ = V²/R = 576/4.0 = 144 W. P₁₂ = 576/12 = 48 W. Now the smaller resistor dissipates more.
P = V²/R uses the voltage across that resistor, not the source voltage, unless the resistor genuinely has the full source voltage across it. In a series circuit it does not, and using the source voltage overstates the power badly.
Two bulbs of different resistance are connected in series. The brighter bulb is:
Two resistors are connected in parallel across a battery. Compared with the 20 Ω resistor, the 5 Ω resistor dissipates:
A resistor carries 2.0 A with 6.0 V across it. The power dissipated is:
Before ranking power, write down whether the elements share current or share voltage. That one line determines which expression to use and therefore which direction the ranking goes.
Answer the 3 checkpoints as you read.
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