Which of the following are true? Bulb A is brighter than Bulb B in (a) The effec
ID: 2003704 • Letter: W
Question
Which of the following are true?
Bulb A is brighter than Bulb B in (a)
The effective resistance of Bulbs A and B in (a) is greater than the effective resistance of Bulbs A, B, and C in (b)
The resistance of Bulb A is greater than the effective resistance of Bulbs B and C in (b)
In (b) Bulb A is in series with Bulb B
In (b) Bulb A is in neither series or parallel with Bulb B
The effective resistance of Bulbs A and B in (a) is less than the effective resistance of Bulbs A, B, and C in (b)
Bulb A is in parallel with Bulb B in (a
)Bulbs B and C are connected in series in (b)
The effective resistance of Bulbs A and B in (a) is equal to the effective resistance of Bulbs A, B, and C in (b)
Bulb B is in parallel with Bulb C in (b)
In (b) Bulb A is in parallel with Bulb B
Bulb A is in series with Bulb B in (a)
Explanation / Answer
Bulb A is brighter than Bulb B in (a)
True. Because it will get all the battery current and hence will be brighter than B.
The effective resistance of Bulbs A and B in (a) is greater than the effective resistance of Bulbs A, B, and C in (b).
True. Let R be the values of all the resistors A B and C. For (a) R(eq) = R+R = 2R and for (b) R(eq) = R + Rbc = R + R/2 = 3R/2. Hence, R(eq) A > R(eq) B. [ 1/Rbc = 1/R + 1/R = 2/R => Rbc = R/2]
The resistance of Bulb A is greater than the effective resistance of Bulbs B and C in (b).
True. R(A) = R and R(BC) = R/2. [ 1/Rbc = 1/R + 1/R = 2/R => Rbc = R/2]
In (b) Bulb A is in series with Bulb B.
True. This can be claerly seen through the circuit diagram given.
In (b) Bulb A is in neither series or parallel with Bulb B
True. From the circuit diagram (b) A is in series with the equivalent resistance of B and C.
The effective resistance of Bulbs A and B in (a) is less than the effective resistance of Bulbs A, B, and C in (b).
False. Let R be the values of all the resistors A B and C. For (a) R(eq) = R+R = 2R and for (b) R(eq) = R + Rbc = R + R/2 = 3R/2. Hence, R(eq) A > R(eq) B. [ 1/Rbc = 1/R + 1/R = 2/R => Rbc = R/2]
Bulb A is in parallel with Bulb B in (a)
False. Bulb A is in series with B and this can be clearly seen.
Bulbs B and C are connected in series in (b).
False. They are connected in parallel to each other in (b).
The effective resistance of Bulbs A and B in (a) is equal to the effective resistance of Bulbs A, B, and C in (b).
False.Let R be the values of all the resistors A B and C. For (a) R(eq) = R+R = 2R and for (b) R(eq) = R + Rbc = R + R/2 = 3R/2. Hence, R(eq) A > R(eq) B. [ 1/Rbc = 1/R + 1/R = 2/R => Rbc = R/2]
Bulb B is in parallel with Bulb C in (b)
True. Its clearly visible in the circuit diagram.
In (b) Bulb A is in parallel with Bulb B.
False. From the circuit diagram (b) A is in series with the equivalent resistance of B and C
Bulb A is in series with Bulb B in (a)
True. Its clearly visible in the circuit diagram.
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