The circuits shown below are all constructed using identical bulbs of resistance

Question

The circuits shown below are all constructed using identical bulbs of resistance R and ideal batteries of emf epsilon. Assume for simplicity that the resistance of a bulb does not depend on the temperature of its filament. (a) Rank the circuits in order of increasing power supplied by the battery, defending your ranking with physical reasoning and a calculation. (b) In each of the four circuits, a bulb has been labeled with an 'X.' Rank the circuits above according to how bright bulb 'X' is, in order of increasing brightness. Assume that the brightness of the bulb is directly proportional to the power it dissipates. Again, defend your ranking based on a clear statement of your physical reasoning and a calculation which supports your claim.

Explanation / Answer

Given

all are identical bulbs of resistance R and Batteries are identical fo emf E

from the given circuits

a) increasing power supplied by the battery, is P = E*I = I^2*R = E^2/R

given E is same so the power supplied depends on resistance on if the resistance is more less power in the circuit

Now calculating the total resistance in all the circuits

A .

both the bulbs are in series so R = R+R = 2R

B. both the bulbs are in parallel so R = R*R/(R+R) = R/2 = 0.5R

C. below two bulbs are in parallel their resistance is R/2,and the top bulb is also in parallel to the lower section so the net resistance is RC = R/3= 0.33R

D. Lower section bulbs in parallel so they have resistance R/2 and the upper bulb is in series with R, so resistance is (R/2+R) = 3R/2 = 1.5R

the rank of increasing power supply is C > B > D > A

b)

Brightness of the bulb depends on the power dissipation at the bulb , here all bulbs are of same resistance so power dissipation depends on the potential drop acroos each resistor

according to the formula P = E^2/R if v across resistor is more it will be more brighter

A. The most of the potential will be dropped across the first bulb say E

B. Bulbs are in parallel combination so the potential difference will be same across each bulb so the potential drop will be E/2

C. all are in parallel combination so the potential drop across the bulb X is E/3

D. here the first the potential will be dropped at first bulb remaining will be equally dropped across X and below one so the potential drop across X will be less when compared with other three cases

so the ranking of brightness is A > B > C > D

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