Take the faulty
headlamp as lamp A, and the lamp which is affected by this as lamp B.
lamp B 'takes' a greater portion of the voltage in order to overcome its greater resistance. Now taking a greater share of the voltage, it is brighter (P= V^2 / R).
On the other hand, now having the greater resistance, less current flows through B. This means it now has less power (P= I^2 *R).
HOW SHOULD I GO ABOUTT?
PLEASE HELP SOMEONEE
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- Thread Starter
- 28-04-2016 23:47