Halogen Lamp Inrush and Fuse Selection
A 12 V, 60 W halogen lamp is installed at the end of a two-core copper cable whose two conductors together have a total length of 16 m and cross-sectional area 1.5 mm². It is supplied by a 12.6 V lead-acid battery of internal resistance 0.080 Ω and protected by a 5 A fast-blow fuse. Copper resistivity is 1.7×10−8 Ω m. When switched on, the lamp sometimes blows the fuse immediately; on other occasions it works, but the cable becomes noticeably warm and the light is dimmer than expected. A clamp meter shows a steady current of about 4.7 A. The cold resistance of the tungsten filament is about one-twelfth of its hot resistance. The installer suggests replacing the 5 A fuse with a 10 A fuse because “the lamp is 5 A, so 10 A gives margin”.
1. Calculate the steady lamp voltage, the cable power dissipation, and explain the warm cable and reduced brightness.
Read the worked answer
Hot lamp resistance: R_L = V²/P = 12²/60 = 2.40 Ω. Cable resistance: R_c = ρL/A = 1.7×10−8 × 16 / 1.5×10−6 = 0.181 Ω.
Total circuit resistance = 2.40 + 0.181 + 0.080 = 2.661 Ω, so I = 12.6/2.661 = 4.73 A.
Lamp voltage = 4.73 × 2.40 = 11.4 V; cable drop = 4.73 × 0.181 = 0.86 V; battery internal drop = 4.73 × 0.080 = 0.38 V. Cable dissipation = I²R_c = 4.73² × 0.181 ≈ 4.1 W. This is enough to make a small cable noticeably warm and to explain the drop from 60 W to about 54 W at the lamp. The lamp is dimmer because it is not receiving its rated 12 V.
2. Estimate the cold-filament inrush current and decide whether a 5 A fast-blow fuse is appropriate. Evaluate the installer’s 10 A suggestion.
Read the worked answer
Cold filament resistance ≈ 2.40/12 = 0.20 Ω. Total cold resistance = 0.20 + 0.181 + 0.080 = 0.461 Ω, so inrush current ≈ 12.6/0.461 ≈ 27 A.
That is about 5.5 times the fuse rating. A 5 A fast-blow fuse can nuisance-blow because the filament heats and its resistance rises within milliseconds to tens of milliseconds. The correct response is not to fit a 10 A fuse: that reduces protection of the 1.5 mm² cable and lampholder and may allow a fault current to persist. A 5 A time-delay (T) fuse, or a 6.3 A time-delay fuse with verified cable rating, is more appropriate. The choice should be checked against the fuse’s I²t let-through and the cable’s short-circuit withstand, not just the steady 5 A.
3. The lamp is still dim after a correct fuse is fitted. Describe measurements that would distinguish a high battery internal resistance from excessive cable resistance, and calculate the expected battery terminal voltage under the steady 4.73 A load.
Read the worked answer
Expected battery terminal voltage = 12.6 − (4.73 × 0.080) = 12.22 V. If the battery posts read close to 12.22 V under load while the lamp reads 11.4 V, the 0.86 V cable drop is normal and the cable is the main cause of dimness. If the battery posts read well below 12.22 V, the battery internal resistance is higher than 0.080 Ω or the battery is discharged.
To measure internal resistance, record open-circuit voltage E, connect a known load resistor, and measure terminal voltage V and current I; then R_int = (E − V)/I. To check the cable, measure the voltage between the two ends of one conductor under load and compare with I × R_c. A four-wire resistance measurement of each core would confirm whether the cable is damaged or whether the connectors are resistive.