Question

Consider a simple circuit containing a battery and three light bulbs. Bulb A is wired in parallel with bulb B and this combination is wired in series with bulb C. What would happen to the brightness of the other two bulbs if bulb A were to burn out?

(A) There would be no change in the brightness of either bulb B or bulb C.

(B) Both would get brighter.

(C) Bulb B would get brighter and bulb C would get dimmer.

(D) Bulb B would get dimmer and bulb C would get brighter.

(E) Only bulb B would get brighter

Answer 1

The correct option is: (C) Bulb B would get brighter and bulb C would get dimmer.

Explanation:

If A were to burn out, the total resistance of the parallel part of the circuit increases, causing less current from the battery and less current through bulb A. However, A and B split the voltage from the battery in a loop and with less current through bulb A, A will have a smaller share of voltage, increasing the potential difference (and the current) through bulb B.