they're pretty easy i guess but i dont quite understand,
thank u :)!!!!!!! 10 pnts best answer !
http://tinypic.com/r/34grr01/5
thank u :)!!!!!!! 10 pnts best answer !
http://tinypic.com/r/34grr01/5
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Q1:
A) When the switch is closed, all 3 light bulbs will light up.
B) When ANY light bulb is unscrewed, the circuit will become open and therefore disconnected and all bulbs will go out.
C) with one load removed from a series circuit, the measured voltage across the other 2 circuit element will increase.
Q2
A) Circuit is disconnected, it will be as if the switch never closed.
B) light bulbs 1, 2, 3 and 5 will remain operational, circuit is now like a series circuit as oppose to parallel
C) this will create a short circuit across bulb 4, since current take the path of least resistance and therefore bypassing bulb 4, bulb 1, 2 and 5 will remain operational, bulb 4 will not.
Q3
A) all bulb will be operational except for bulb 2, because it is in series with 4 and that part of the circuit is not closed
B) same situation as A) all bulb operational except for 5, it is in series with 3, and it is not an open circuit at that part
C) entire circuit is now open, all bulbs shut down
Q4
A) entire circuit is open, all bulbs down
B) same as A)
C) all bulb is operational except for 3, since it is a parallel connection only bulb 3 is affected
D) all bulb operational except 5 and 6, same concept, 5 and 6 are connected in series, but they are a parallel branch and therefore will not affect the rest of the circuit.
A) When the switch is closed, all 3 light bulbs will light up.
B) When ANY light bulb is unscrewed, the circuit will become open and therefore disconnected and all bulbs will go out.
C) with one load removed from a series circuit, the measured voltage across the other 2 circuit element will increase.
Q2
A) Circuit is disconnected, it will be as if the switch never closed.
B) light bulbs 1, 2, 3 and 5 will remain operational, circuit is now like a series circuit as oppose to parallel
C) this will create a short circuit across bulb 4, since current take the path of least resistance and therefore bypassing bulb 4, bulb 1, 2 and 5 will remain operational, bulb 4 will not.
Q3
A) all bulb will be operational except for bulb 2, because it is in series with 4 and that part of the circuit is not closed
B) same situation as A) all bulb operational except for 5, it is in series with 3, and it is not an open circuit at that part
C) entire circuit is now open, all bulbs shut down
Q4
A) entire circuit is open, all bulbs down
B) same as A)
C) all bulb is operational except for 3, since it is a parallel connection only bulb 3 is affected
D) all bulb operational except 5 and 6, same concept, 5 and 6 are connected in series, but they are a parallel branch and therefore will not affect the rest of the circuit.
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This a poor way to show resistors (as an inductor). However if you assume the values by the coils are resistance values, then just use Ohm's law to solve the problems. In some cases you may need to put the resistprs in series or parallel.