In the circuit shown in the diagram, the current in the solenoid (opening is per
ID: 1556235 • Letter: I
Question
In the circuit shown in the diagram, the current in the solenoid (opening is perpendicular to the plane of the page) is in the direction shown (counterclockwise). The current is INCREASING linearly in time. The battery in the circuit has an emf equal to the emf of a single loop around the solenoid. The bulbs are all identical and the internal resistance of the battery is much less than the resistance of the bulbs. Give the voltage that a voltmeter would read for each bulb assuming an ideal voltmeter connected directly next to the bulb in case. For each of the following parts, give the brightness of each bulb when the indicated & change is made. Take each part SEPARATELY... if bulb #1 is unscrewed in solenoid part b), assume that it is replaced when the change in part c) is made. Explain your answers briefly. a) For the circuit shown. b) If Bulb #1 were unscrewed from its socket. c) If Bulb #2 were unscrewed from its socket. d) The wire is cut at the point labeled C in the circuit. e) The circuit is shorted from A to B by a wire around the outside of the right side of the circuit. f) The circuit is shorted from A to B by a wire around the outside of the left side of the circuit (outside the wire with the DC battery).Explanation / Answer
(a) When the circuit is in normal form, as the bulbs are in series combination so same current will flow through them. As brightness of a bulb is directly proportional to its resistance for same current and here both bulbs are identical, So both bulbs will have same brightness.
(b) If bulb 1 were unscrewed from its socket or bulb 1 is removed from circuit then effective resistance of circuit will go down and so there is a increase in current in circuit. So brightness of bulb 2 will increase.
(c) When bulb 2 is removed from circuit then also same case occurs as of (b). Current in circuit will increase and so brightness of bulb 1 will also increase.
(d) If the wire is cut at point C , then circuit will be an open circuit and no current will flow through circuit. So there will not be any light in any of the bulbs.