1. Six identical light bulbs are placed in two different circuits as shown above

Question

1. Six identical light bulbs are placed in two different circuits as shown above. The first circuit (at the top) has a battery with a small internal resistance, r, while the second has a power supply. The power supply is set to = 9.0 V, the same as the battery is rated.

(a) Initially, how do the brightnesses of bulbs A and C compare? Please explain your answer.

(b) When switch 2 is closed, how do the brightnesses of A and B compare? Is A as bright, brighter than or less bright than it was before switch 2 was closed? Please explain your answer.

(c) When switch 3 is closed, how do the brightnesses of C and D compare? Is C as bright, brighter than or less bright than it was before switch 3 was closed? Please explain your answer.

(d) Now both switches 3 and 4 are closed. How do the brightnesses of E and F compare? How do the brightnesses of C and F compare? Please explain your answer.

(e) If we close all of the switches 3, 4 and 5, compare the brightnesses of bulbs C, D, E and F. Please explain your answer.

(f) If we close switch 1, what will happen to bulb A? What will happen to the internal resistor of the battery? Please explain your answer.

(g) Why are short circuits dangerous?

Explanation / Answer

Let resistance of all bulbs = R

a) as in this case, IA = E/(r + R)

Ic = E/R

so current in C is larger than current in A, C will be more bright.

b) IA = IB = E/2(r + R/2) = E/(2r + R)

A and B both will be equally bright.

current in A is less than previous case, so it will be less brigher after closing switch 2

c) Ic = ID = E/2(R/2) = E/R

C and D both will be equally bright.

current in C is same as previous case, so it will be equally brigher after closing switch 3 as in previous case.

d) as E and F both are in series, current through E will be same as current in F and will be half than current in C

so E will be less bright than C and same as bright as F.

e) in this case F is short circuited and we can remove this from the circuit which will not affect the circuit current,

also C, D and E all are in parallel with equal current and so will be equally brighter.

f) if we close switch 1, A and B both will be short circuited, and no current will flow through them. current in circuit will be E/r.

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