Search any question & find its solution
Question:
Answered & Verified by Expert
A $6.0$ volt battery is connected to two light bulbs as shown in figure. Light bulb 1 has resistance 3 ohm while light bulb 2 has resistance $6 \mathrm{ohm}$. Battery has negligible internal resistance. Which bulb will glow brighter?
Options:
Solution:
2195 Upvotes
Verified Answer
The correct answer is:
Bulb 1
Bulb 1
Total resistance $=\frac{6 \times 3}{6+3}=2 \Omega$ Current in circuit $=\frac{6}{2}=3 A$
Therefore current through bulb 1 is $2 \mathrm{~A}$ and bulb 2 is $1 A$. So bulb 1 will glow more.
Therefore current through bulb 1 is $2 \mathrm{~A}$ and bulb 2 is $1 A$. So bulb 1 will glow more.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.