Really basic what happens when the switch closed question

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When the switch in the circuit is closed, nearly all current flows through the switch due to its low resistance, leaving lamp C with insufficient current to illuminate. The resistance of lamp C is much higher compared to the switch and parallel wires, resulting in no voltage drop across lamp C. In theoretical problems, the switch and wires are assumed to have zero resistance, which further explains why lamp C does not light up. If a resistor were added in the switch's path, more current would flow to lamp C, allowing it to illuminate. Understanding this concept highlights the importance of current paths in electrical circuits.
sameeralord
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Hello guys,

I got a small question. It says circuit consists of three identical lamps connected to a battery with a switch as shown in the diagram. When the switch is closed what happens.

I did this really quick on paint.

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they are not allowing me to post URL because I don't have 15 posts. I'll do some spaces and stuff please type the w w w thing properly.

It says that lamps C turns off because no current flows through it. But I don't get it. Shouldn't the current divide and each section get some current. Why is C getting no current when the switch is closed. Any help would be appreciated. Thanks:rolleyes:smile:
 
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You are supposed to have both positive and negative current for the lamp to turn on, no?
 
sameeralord said:
Hello guys,
...
It says that lamps C turns off because no current flows through it. But I don't get it. Shouldn't the current divide and each section get some current. Why is C getting no current when the switch is closed. Any help would be appreciated. Thanks:rolleyes:smile:

The switch and wire in parallel with the "C" lamp have extremely low resistance.
The "C" lamp will have much higher resistance in comparison.
When the switch is closed, nearly all of the current will flow through the switch.
The current flowing through the lamp will not be sufficient to cause it to illuminate.

For theoretical schoolwork problems, the resistance in the switch and wires are generally assumed to be zero. So when the switch is closed, there will be no voltage drop across the "switch" leg of the circuit. With no voltage drop across the "C" lamp, there will be no current.
 
OmCheeto said:
The switch and wire in parallel with the "C" lamp have extremely low resistance.
The "C" lamp will have much higher resistance in comparison.
When the switch is closed, nearly all of the current will flow through the switch.
The current flowing through the lamp will not be sufficient to cause it to illuminate.

For theoretical schoolwork problems, the resistance in the switch and wires are generally assumed to be zero. So when the switch is closed, there will be no voltage drop across the "switch" leg of the circuit. With no voltage drop across the "C" lamp, there will be no current.

Thank you for your help:smile:. I think I get it now. The current took the path of least resistance in this circuit. Let's say there was a resistor in the wire with the switch then you would see more current flow to the globe. Am I right!. Thanks a lot for the help. Really appreciate it. :smile:
 
sameeralord said:
Thank you for your help:smile:. I think I get it now. The current took the path of least resistance in this circuit. Let's say there was a resistor in the wire with the switch then you would see more current flow to the globe. Am I right!. Thanks a lot for the help. Really appreciate it. :smile:

Exactly right.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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