Simple DC Circuits: What Happens to L.B. A & C When B is Removed?

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When lightbulb B is removed from the circuit, the circuit becomes a series connection for lightbulbs A and C. This change results in the total voltage from the battery being divided between A and C, leading to a decrease in brightness for both bulbs. Since they are now in series, the current flowing through them is reduced compared to when B was present. Consequently, both A and C will shine dimmer than they did with B in the circuit. The overall effect is a significant reduction in brightness for both lightbulbs A and C.
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Homework Statement



L.B. = Lightbulb
A,B,C are all identical light bulbs

B = battery

If a circuit looks like this

-----L.B.A----------------
| | |
| | |
| | |
B L.B.B L.B.C
| | |
| | |
| | |
------------------------

What happens to brightness of A and C when B is removed?


Homework Equations






The Attempt at a Solution

 
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wow! its not supposed to look like that! ok let me describe it. There is a rectangle with a line down the middle. On the left side of the rectanlge is a battery. On the top left half of the rectange there is lightbulbA and on the line down the middle of the rectangle there is lightbulbB and on the right side of the rectangle there is lightbulbC.

What happens to the brightness of A and C when B is removed?
 
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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