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Sagot :
The rank of the bulbs based on their brightness when their resistances remain constant are C>A=B>D=E and C is the brightest among other bulbs
The power that occurs across any resistor is equal to the square of the current through that resistor and resistance. The power in general defines the brightness of the bulb.
It is given that the equivalent resistance of A and B bulb R(AB) = Rx R/R+R = R/2 and the equivalent resistance of bulbs C, D, and E,
R(CDE) = R x (R+R)/ R+R+R =2R/3
Therefore, the power of bub
A= P(A) = I(A)² R(A) = V²/R
As the bulbs A and B have the same resistance,
P(A )=P(B) =V²/R.............(1)
Then the power through bulb C,
P(C) = (2/3 x I(CDE)) x R = (2/3 x (3V/2R)² x R = 3/2 x V²/R..............(2)
Again, the power through D and E bulbs are equal, therefore
P(D) =P (E) = (1/3 x I(CDE)) x R = (2/3 x (3V/2R)² x R = 3/4 x V²/R.............(3)
On Comparing equations (1), (2), and (3), the brightness of the bulbs is ranked as follows:
P(C) > P(A) = P(B) > P(D) = P(E)
Therefore, the rank of the bulbs based on their brightness is C>A=B>D=E
The brightness of bulb C is the brightest among all other bulbs placed in the circuit.
To Learn more about power, click below:
brainly.com/question/2933971
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