Light Bulbs in a parallel circuit

AI Thread Summary
When two bulbs are connected in parallel and one is unscrewed, the brightness of the remaining bulb will double, as the current that was shared between the two bulbs will now flow entirely through the remaining bulb. The voltage across both bulbs remains constant, but the total current drawn from the battery increases, potentially affecting battery performance due to finite output impedance. In practical scenarios, the output voltage may decrease slightly under increased load, but this nuance is not captured in the multiple-choice answers provided. The discussion emphasizes the relationship between voltage, current, and resistance in parallel circuits. Overall, the consensus is that the best answer to the question is that the brightness of the remaining bulb will double.
chemistrykid
Messages
2
Reaction score
0
Q: Two bulbs x and y are connected in parallel to a new dry cell. The switch is closed. If bulb x is unscrewed, the brightness of bulb y will
a) double
b) halve
c) remain the same
d) become 0

A: a? The current from the first branch would go into the second branch?
 
Physics news on Phys.org
chemistrykid said:
Q: Two bulbs x and y are connected in parallel to a new dry cell. The switch is closed. If bulb x is unscrewed, the brightness of bulb y will
a) double
b) halve
c) remain the same
d) become 0

A: a? The current from the first branch would go into the second branch?
The voltage remains the same. What happens when bulb is removed? Think about the resistance of each bulb, and the relationship between voltage, current and resistance.
 
chemistrykid said:
Q: Two bulbs x and y are connected in parallel to a new dry cell. The switch is closed. If bulb x is unscrewed, the brightness of bulb y will
a) double
b) halve
c) remain the same
d) become 0

A: a? The current from the first branch would go into the second branch?
It depends on the battery chemistry. But as with many subtle multiple-choice questions, there is only one best answer in the choices given (as Astro has indicated).

In the real world, however, batteries have a finite output impedance, so even a new battery's output voltage will droop some as it is loaded more. So the output voltage will be a bit less when it is outputting twice the current (to drive both bulbs). So if there were an answer e) that said "increase slightly", that would be the correct choice. Since that isn't listed, what is the next best choice?
 
The light bulbs in your home are wired in parallel. If you unscrew one, how does it affect the others?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top