Answer:
B. The other bulb stays lit.
Explanation:
You did not provide the picture of the circuit but here are some suggestions.
Based on the choices given:
A)The other bulb goes out.
B)The other bulb stays lit.
C)Current stops flowing in the circuit.
D)The current path in the circuit is broken.
Choices A, C, and D all happen when a light bulb is unscrewed in a series circuit. A series circuit is a circuit where all the resistors are lined up in a single path. In order for current to flow, the resistors need to be all functional. When you take one out, the flow of current stops because the one path it flows through is broken. As a result, other resistors in the circuit will no longer work.
As for choice B, this occurs in a parallel circuit. In a parallel circuit the resistors are arranged in parallel paths. The paths branch off and thus so does the flow of current. If you take out one of the light bulbs in a parallel circuit, the other light bulbss would remain lit because the flow of current has an alternative path to move through.
Attached is a diagram of a series and a parallel circuit.
Answer:
0.026 V
Explanation:
Given that,
Inductance of the coil, L = 6 mH
The current changes from 0.2 A to 1.5 A in a time interval of 0.3 s
We need to find the magnitude of the average induced emf in the coil during this time interval. The formula for the induced emf is given by :
So, the magnitude of induced emf is 0.026 volts.