Where Do the Missing Watts Go? A Lamp, a Car Battery and a 15 m Extension Lead
At a school science fair Emeka runs a lamp rated 12 V, 24 W directly from a 12 V car battery and it glows at normal brightness. To move the lamp to a display table 15 m away he uses a two-core extension lead of 1.0 mm² copper cable (resistivity of copper 1.7 × 10⁻⁸ Ω m; with two conductors the lead has about 0.034 Ω of resistance per metre of its length). The lamp is now visibly dimmer. A classmate says the battery must be flat. The teacher disagrees and asks Emeka to account for every watt. Using a clamp meter Emeka reads a current of 1.70 A, and a voltmeter across the battery terminals reads 11.1 V while the lamp is running. With the lamp disconnected the battery's emf is 12.0 V. Emeka also has a spare 15 m lead made of 2.5 mm² cable and wants to know whether changing the lead is worth the trouble, or whether he should simply move the battery closer to the table.
1. Calculate the resistance of the extension lead and the current it should carry in series with the lamp, and check whether the 1.70 A reading is consistent with the measured terminal voltage.
Read the worked answer
Resistance of the lead. Resistance per metre of lead (two conductors carrying current in opposite directions) = 2ρL/A. Per metre: 2 × 1.7 × 10⁻⁸ ÷ 1.0 × 10⁻⁶ = 0.034 Ω. For 15 m: R_lead = 0.034 × 15 = 0.51 Ω.
Resistance of the lamp from its rating. R = V²/P = 12² ÷ 24 = 6.0 Ω. (This is the hot resistance at full brightness; it is a reasonable working value and is the figure the rating implies.)
Current in the series circuit. The lamp and lead are in series across the battery terminals, so I = V_terminal ÷ (R_lamp + R_lead) = 11.1 ÷ 6.51 = 1.70 A.
Verdict. The measured 1.70 A agrees with the prediction to within rounding. The lead is not broken, shorted or 'blocking' the current; it simply adds 0.51 Ω in series with a 6 Ω lamp, which is about 8% of the total resistance - enough to be visible in a lamp, which dims noticeably for a small drop in voltage.
Caution. A tungsten filament's resistance rises with temperature, so a dimmer lamp has a slightly lower resistance than 6 Ω and the real current will drift a little above this estimate; the agreement within a few per cent is therefore good.
2. Using a power budget, decide between the classmate's 'flat battery' explanation and the lead as the main cause of the dimming.
Read the worked answer
Voltage drops around the circuit.
- Across the lamp: V = IR = 1.70 × 6.0 = 10.2 V, so P_lamp = 10.2 × 1.70 = 17.3 W (also I²R = 17.3 W).
- Across the lead: V = 1.70 × 0.51 = 0.87 V, so P_lead = 0.87 × 1.70 = 1.5 W.
- Check: 10.2 + 0.87 = 11.07 V ≈ the measured 11.1 V.
Internal resistance of the battery. The emf is 12.0 V but only 11.1 V appears at the terminals at 1.70 A, so the lost 0.9 V is dropped inside the battery: r = 0.9 ÷ 1.70 = 0.53 Ω, dissipating 0.9 × 1.70 = 1.5 W.
Total check. 17.3 + 1.5 + 1.5 = 20.3 W ≈ emf × current = 12.0 × 1.70 = 20.4 W. The budget balances.
What the lamp would do at the battery terminals (no lead). I = 12.0 ÷ (6.0 + 0.53) = 1.84 A, giving P_lamp = 1.84² × 6.0 = 20.3 W.
Judgement. Removing the lead would restore the lamp from 17.3 W to 20.3 W, a gain of about 3.0 W; the lead is genuinely responsible for that. But the lamp can never reach its rated 24 W from this battery, because the internal resistance costs a further 24 − 20.3 ≈ 3.7 W even with the lamp at the terminals. The classmate is wrong to call the battery flat - its emf is still 12.0 V and 0.53 Ω is normal for a car battery - but the teacher is also only partly right: both the battery and the lead take a share.
3. Emeka considers swapping to the 2.5 mm² lead. Calculate the improvement, and advise him on the best practical fix.
Read the worked answer
Resistance of the thicker lead. Resistance per metre = 2 × 1.7 × 10⁻⁸ ÷ 2.5 × 10⁻⁶ = 0.0136 Ω m⁻¹. For 15 m: R = 0.0136 × 15 = 0.204 Ω.
New current and lamp power. Total resistance = 6.0 + 0.204 + 0.53 = 6.73 Ω; I = 12.0 ÷ 6.73 = 1.78 A. Lamp voltage = 1.78 × 6.0 = 10.7 V; P_lamp = 1.78² × 6.0 = 19.0 W.
Comparison. Lamp power rises from 17.3 W to about 19.0 W, roughly a 10% improvement. The lead's share of the voltage drop falls from 0.87 V to about 0.36 V, i.e. 3% of the supply - usually considered acceptable.
Advice. The thicker lead helps but does not restore full brightness, because the battery's own 0.53 Ω still costs about 3.5 W and no cable can remove that. Moving the battery to the table removes the lead loss completely (20.3 W), and it is also the free option if the battery can safely be carried. The complete solution is a stabilised 12 V supply that can hold 12 V at about 2 A, which would let the lamp reach close to its rated 24 W.
Assumption to state. The filament resistance is treated as constant at 6 Ω; in reality it rises with temperature, so the true current will be a little lower than these figures when the lamp is bright. The ranking of the options, however, is unaffected.