A $6.0$ volt battery is connected to two light bulbs as shown in figure. Light bulb 1 has resistance 3 ohm…
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?
Bulb 1 will glow more first and then its brightness will become less than bulb 2
Bulb 1
Bulb 2
Both glow equally
Solution
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.
