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    Part C
    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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