Heat, Current and Fuse Choice in a Station Generator Room
A police station backup room has a 12 V battery bank feeding three identical 24 W emergency lamps through a common fuse. The wiring from the battery to the fuse box is 6 m of copper cable with a total round-trip resistance of 0.10 ohm. The fuse is rated 5 A. On a hot afternoon all three lamps are switched on, and a constable notices the cable near the fuse box feels warm and the lamps look slightly dim. He wants to know whether the setup is safe, whether the dimness is expected, and what would happen if a fourth identical lamp were added. He has a multimeter but no clamp meter, and he must decide before nightfall whether to change the fuse.
1. Calculate the current drawn by three lamps and the voltage drop along the cable. Is the 5 A fuse adequate?
Read the worked answer
Each lamp: P = 24 W at 12 V, so I = P / V = 24 / 12 = 2 A per lamp.
Three lamps: 3 × 2 = 6 A total.
Cable drop: V = I × R = 6 × 0.10 = 0.60 V.
Voltage reaching the lamps: 12 − 0.60 = 11.4 V, about 5 percent low.
Fuse adequacy: a 5 A fuse carrying 6 A is not adequate - it is being run above its rating and will either blow or run hot. The correct action is to fit a fuse rated above the expected current with a sensible margin, for example a 10 A fuse, provided the cable itself can carry that current. A fuse protects the cable, not the appliance, so the cable's current rating must be checked before simply going bigger. The constable should not fit a fuse far above the load, because then the cable becomes the weakest link.
2. Explain why the lamps look dim and the cable feels warm, using the numbers. Then explain why adding a fourth lamp is a bad idea even if the fuse were upgraded.
Read the worked answer
Dimness: the lamps receive 11.4 V instead of 12 V. Lamp power falls roughly with the square of voltage for a resistive load, so about (11.4/12)² ≈ 0.90, roughly a 10 percent drop in power - visible as slight dimming. The missing power is dissipated in the cable.
Warmth: the cable carries 6 A through 0.10 ohm, so it dissipates I²R = 36 × 0.10 = 3.6 W along its length. That is small in absolute terms but concentrated in a short run near a connection, and a poor or loose terminal raises local resistance, which concentrates heat exactly where the constable felt it. Warmth at a joint is a warning sign, not normal operation.
Fourth lamp: total current becomes 8 A. Cable drop becomes 8 × 0.10 = 0.80 V, leaving 11.2 V, so dimming worsens. Cable dissipation becomes 64 × 0.10 = 6.4 W, nearly double, and any weak joint gets hotter. Upgrading the fuse alone removes the protection without fixing the cause. The right fix is thicker or shorter cable and clean, tight terminals, then size the fuse to the cable.
3. The constable has a multimeter but no clamp meter. Describe how he can measure the current safely and what reading would confirm his diagnosis.
Read the worked answer
A multimeter measures current only in series, so he must not simply clip it across the circuit. Safe approach:
- Switch off the lamps and isolate the battery.
- Set the multimeter to its DC current range, choosing a range above the expected 6 A, and confirm the meter and leads are rated for that current.
- Break the circuit at the fuse holder and connect the meter in series, observing polarity.
- Switch on and read the current, then switch off before disconnecting.
Expected confirmation: a reading near 6 A with three lamps on supports the calculation. A reading well above 6 A, or a cable drop measured in parallel across the cable ends that is much more than 0.6 V, points to excess resistance, likely at a joint.
A safer alternative that avoids series current measurement entirely: measure voltage across the battery and across the lamp terminals with the lamps on. A difference of about 0.6 V confirms the drop, and a much larger difference localises the fault. Measuring voltage in parallel is lower-risk than breaking a live circuit, and for a constable without a clamp meter it is the better first step.