Question

If two bulbs of \[60\;watt\] and \[40\;watt\] are connected in parallel, then
a. Both bulbs with shine equally
b. Both bulbs will get fused
c. \[60\;watt\] bulb will shine more brightly
d. \[40\;watt\] bulb will shine more brightly

Answer 1

Hint: In this question we will use the concept of parallel combination of resistance across which electric potential difference is applied and use the concept of how the voltage distributed in between the appliances connected in parallel across the battery and then we will find current pass through in both the bulbs and compare them to get the desired answer.

Formula Used: -
\[P = VI\]
Where \[P,\;V,\;I\]are power, voltage, current in the conductor respectively.
Given Data:
\[{P_1}\]=60 watt
\[{P_2}\]=40 watt

Complete solution: -
If the bulbs are connected in parallel combination then they will have same potential difference across it and different electric current passed through it so
\[{I_1} = \dfrac{{{P_1}}}{V}\] and \[{I_2} = \dfrac{{{P_2}}}{V}\]
\[{I_1} = \dfrac{{60\,watt}}{V}\] and \[{I_2} = \dfrac{{40\,watt}}{V}\]
Here, \[{I_1} > {I_2}\]
Here, the current in the first bulb is more than the current in the second bulb since the power output in the first bulb is more than that of the second bulb.
The brightness of the bulb depends upon the output power delivered by the bulb and which depends upon the current flow on through the bulb. So, the bulb of 60 watt is brighter than that of 40 watt.
Hence option \[(C)\]is correct.

Note: - In this question we should know the concept of combination of resistances, the voltage distribution and current distribution in series and parallel combinations we must be aware of the relation between power, electric potential difference and resistance of the bulbs used.