Question

Two bulbs $ 40W $ and $ 60W $ and rated voltage $ 240V $ are connected in series across a potential difference of $ 420V $ . Which bulb will work at above its rated voltage?
A. $ 40W\text{ bulb} $
B. $ 60W\text{ bulb} $
C. Both will work
D. None of these

Answer 1

Hint: For the bulbs connected in series, current through both of them would be the same. To check the working voltage of the bulb we need to calculate the resistance of the bulb and current in the circuit. Resistance of bulb can be calculated by viewing the rating of bulb mentioned, that is, voltage drop across bulb and its working power.

Formulae used:
 $ R=\dfrac{{{V}^{2}}}{P} $
 $ I=\dfrac{V}{R} $

Complete step-by-step answer:
Bulbs contain a thin coil of wire called filament. When an electric current passes through the bulb, filament gets heated up and produces light as a result of heating effect. The current through a filament lamp or a bulb is not proportional to the potential difference across its terminals. At high temperature, the atoms in the filament start vibrating more and collide more. This results in the requirement of more energy to pass current through the filament.
Resistance of a bulb is given by:
 $ R=\dfrac{{{V}^{2}}}{P} $
Where $ V $ is the potential difference across the terminals of bulb and $ P $ is the power of the bulb
Calculating resistance of bulb 1 $ (40W,240V) $
 $ {{R}_{1}}=\dfrac{{{V}^{2}}}{P}=\dfrac{240\times 240}{40}=1440 $
 $ {{R}_{1}}=1440\Omega $
Similarly, resistance of bulb 2 $ (60W,240V) $
 $ {{R}_{2}}=\dfrac{{{V}^{2}}}{P}=\dfrac{240\times 240}{60}=960 $
 $ {{R}_{2}}=960\Omega $
As the two bulbs are connected in series, equivalent resistance of the circuit would be:
 $ \begin{align}
  & {{R}_{S}}={{R}_{1}}+{{R}_{2}} \\
 & {{R}_{S}}=1440+960=2400 \\
\end{align} $
 $ {{R}_{S}}=2400\Omega $
Voltage applied in the circuit is gives as $ 420V $
Current through the circuit $ I=\dfrac{V}{{{R}_{S}}}=\dfrac{420}{2400}=0.175 $
 $ I=0.175A $
Now,
Voltage drop across bulb 1, $ {{V}_{1}}=I{{R}_{1}}=\dfrac{420}{2400}\times 1440=252 $
 $ {{V}_{1}}=252V $
Voltage drop across bulb 2, $ {{V}_{2}}=I{{R}_{2}}=\dfrac{420}{2400}\times 960=178 $
 $ {{V}_{2}}=178V $
As the voltage drop across bulb 1 is more than its rated voltage, therefore, bulb 1 $ (40W,240V) $ will work above its rated voltage.
Hence, the correct option is A.

Note: While calculating the resistances of bulbs, take the rated voltage only. While calculating current passing through the bulbs, take the applied voltage. Voltage and Power are always mentioned on the bulb, this is what we call a rating of bulb.