Question

use the information and image to answer the following question.
a student is modeling an electric circuit containing one battery and five identical light bulbs. the model created is shown.
image of the circuit with bulb a, bulb b, bulb c, bulb d, bulb e
if the student unscrewed bulb d from the circuit, how would the other bulbs be affected?
a. only bulb c would go out while all the others remain lit. b. bulbs c and b would go out while all the others remain lit. c. bulbs b, d, and e would go out while all the others remain lit. d. all of the bulbs will go out.

Explanation:

1. First, analyze the circuit connections: Bulb C and Bulb D are in series with each other, and this series combination is in parallel with another bulb (let's say Bulb X, but from the diagram, it's the middle bulb), and this parallel group is in series with Bulbs A, B, and E? Wait, no, re - examining: Wait, the circuit: Bulb A and Bulb E are in series with a parallel combination. The parallel combination has three branches? Wait, no, looking at the diagram: Bulb C and D are in series (so they form a series branch), then there's another branch (the middle bulb), and then Bulbs A, B, E are in series? Wait, no, actually, when Bulb D is unscrewed, since Bulb C and D are in series, removing D breaks the series for C and D. But the other branches (Bulb A, B, E, and the middle bulb) are in parallel or series? Wait, no, in a parallel circuit, if one branch is broken (but here C and D are in series in a branch). Wait, no, the key is: Bulb C and D are in series (so they are dependent on each other). So if D is removed, the branch with C and D is open. But the other bulbs (A, B, E, and the middle bulb) are in a different branch? Wait, no, maybe the correct analysis is: The circuit has a main path with A, B, E, and a parallel section with three branches? No, looking at the options, option A says only C goes out. Wait, no, if C and D are in series, then removing D would make C go out (since the series path is broken), but the other bulbs (A, B, E, and the middle bulb) are in a different path (parallel or series - parallel). Wait, maybe the circuit is: Bulbs A and E are in series with a parallel network. The parallel network has three branches: one with the middle bulb, one with C and D in series. So when D is removed, the branch with C and D is open, so C goes out, but the other branches (middle bulb, A, B, E) remain. So only C goes out, others (A, B, E, middle bulb) stay lit. So option A is correct.

Answer:

A. Only Bulb C would go out while all the others remain lit.