Three 12 V, 24 W lamps are parallel across 12 V. (a) Find current in each and total current. (b) Find equivalent resistance. (c) Explain why parallel is preferable to series. (d) Choose a fuse from 3 A, 5 A and 10 A. (e) One lamp fails open: describe others and total current. (f) Explain live-wire fuse and switch placement in a mains version.

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Each 2.0 A; total 6.0 A; equivalent 2.0 Ω; choose 10 A. One failed branch leaves two normal lamps and 4.0 A total. Each branch has 12 V, so I = P/V = 24/12 = 2.0 A. Junction conservation gives total current 6.0 A. Equivalent resistance is V/Itotal = 12/6.0 = 2.0 Ω. It is below each operating lamp resistance, as expected for parallel paths. Parallel branches each receive rated voltage and operate independently. Series lamps would share voltage, be dimmer and all extinguish if one filament opened. The lowest listed fuse above normal 6.0 A is 10 A. A 5 A fuse would interrupt normal operation. An open branch removes only one path. Two lamps remain at 12 V and normal brightness; total current becomes 4.0 A. Put switch and fuse in live so operation disconnects dangerous live potential. A neutral-only fuse could open while internal components remain live relative to Earth.

Safe parallel lighting circuit

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