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.
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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?
 
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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?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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