A circuit consists of three identical lamps connected to a battery. Battery has some internal resistance. Switch S, open at first, is closed.
1) What happens to brightness of lamps B and C?
2) What happens to current in battery?
3) What happens to potential across lamps A and C?
4) What happens to total power delivered to lamps by battery?
See attached figure.
I = emf/R+r
P = I*deltaV
The Attempt at a Solution
This is my attempt at the solution; I am almost certain that my answers are correct but I would appreciate if someone could verify, and explain if possible, if I am incorrect.
1) B brightness remains the same. C brightness increases since the circuit is complete.
2) Current in battery decreases because of increased resistance.
3) Potential is unchanged for lamp A but increases for C because circuit is now closed.
4) Total power increases, since there is an increase in potential due to the closed circuit.
I would appreciate any assistance, thank you!
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