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.

$I=\frac{2v}{3R}\Rightarrow I^{\prime }=\frac{I^{\prime }}{2}=\frac{V}{3R}$

$\therefore PowderdevelopedinAandB$

= $(I^{\prime }{)}^{2}R=\frac{V^2}{9R^2}\times R=\frac{V^2}{9R}$

$\therefore PowderdevelopedinC=I^{2}R=\frac{\left(4V^2\right)}{9R^2}\times R=\frac{\left(4V^2\right)}{9R}$

$WhenAisburnt,Circuitis$

$I=\frac{V}{2R}$

$\therefore Powerdevelopedin$

$BorC=I^{2}R=\frac{V^2}{4R^2}\times R=\frac{V^2}{4R}$

$\therefore PowerofBincreases$

$PowerofCdecreases$