The circuit shown contains five identical light bulbs, labeled A to E. a)Rank th
ID: 1351182 • Letter: T
Question
The circuit shown contains five identical light bulbs, labeled A to E. a)Rank the bulbs in order from brightest to least bright. Justify your rankings carefully and rigorously. b) How would your answer to (a) change if the orientation of the battery were reversed? c) Indicate whether each of the other bulbs gets brighter, dimmer, or stays the same (corn pa red to situation a) if bulb C is removed from its socket, leaving a gap in the circuit there. Justify. d) Indicate what happens to the brightness of each bulb (compared to situation a) if bulb C is shorted out (i.e., replaced by a wire). Justify.Explanation / Answer
a)
resistance of each bulb = r
Voltage across two branches is same. = V
total resistance of branch containing D and E = r + r = 2r
total resistance of branch containing A, B and C = r + (r x r/(r + r)) = 1.5 r
Current in DE branch = V/2r = 0.5 V/r
Current in ABC branch = V/1.5r = 0.67 V/r
A> E=D>C=B
b)
answer would be same
c)
Bulb C removed
total resistance of branch containing D and E = r + r = 2r
total resistance of branch containing A, B = r + r = 2 r
Current in DE branch = V/2r = 0.5 V/r
Current in AB branch = V/2r = 0.5 V/r
Since Current is same in both branched through each bulb, all bulbs will have same brightness.
Current in AB has reduced from situation 1. hence bulb A gets dimmer but bulb B gets brighter as current through it has increased
d)
Bulb C is short circuited
total resistance of branch containing D and E = r + r = 2r
total resistance of branch containing A, B = r
Current in DE branch = V/2r = 0.5 V/r
Current in AB branch = V/r = V/r
Since Current is same in both branched through each bulb, all bulbs will have same brightness.
Current in AB has increased from situation a. hence bulb A gets brighter. bulb B gets turned off as it is short ciruited.
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