Question

Consider a simple circuit containing a battery and three identical incandescent bulbs A, B and C. 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. Only bulb B would get brighter
• B. Both A and B would get brighter
• C. Bulb B would get brighter and bulb C would get dimmer
• D. There would be no change in the brightness of either bulb B or bulb C.

Answer 1

The correct option is C Bulb B would get brighter and bulb C would get dimmer

When bulb didnot burn out, Bulb A is wired in parallel with bulb B and this combination is wired in series with bulb C with equal resistance r.
Effective resistance will be $\frac{r\times r}{r+r}+r=\frac{3r}{2}$.
Current through C is $\frac{v}{\frac{3r}{2}}=\frac{2v}{3r}$ where v is voltage.
Current through A and B is $\frac{2v}{2\times 3r}=\frac{v}{3r}$
When bulb A burns there are only two bulbs in series connection and the effective resistance become 2r with the current flowing though B and C being Case 1 :$\frac{v}{2r}$ .
Now brightness depends on the power which in turn is proportional to current.
Current in B increased to $\frac{v}{2r}$ from $\frac{v}{3r}$ and in C decreased to $\frac{v}{2r}$ from $\frac{2v}{3r}$.
So Bulb B would get brighter and bulb C would get dimmer.