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Analyzing Electrical Circuits -- Which lamps will have current flowing through them?

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Homework Statement


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State and explain whether or not current will flow through the lamps in these circuits
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The Attempt at a Solution


I am confused with figure 6 as my explanation is different from my textbook's answer. My textbook says:
"Figure 6 – the lamp is short circuited due to the switch arranged in parallel across it. The current will flow through the switch rather than the lamp due to its far lower resistance".

My explanation would be different. I would say that the LDR will have some significant resistance (which is unknown). The combined resistance of the piece of wire and lamp would be close to zero if you add them reciprocally as they are in parallel and the switch/piece of wire would have a very low resistance. Therefore, the LDR will get all the potential difference and the lamp/wire parallel combo will get none, i.e. no short circuit.

Also, could someone check if my explanation of what would happen if there was no LDR please. If the LDR wasn't there then would the lamp actually work because the lamp is in parallel with the switch and since these would be the only components they would get the full p.d. of the battery? I know in reality the short circuit would cause a huge surge in current and the battery's internal resistance would eat up the p.d. but if we assume there is no internal resistance would the lamp work if the LDR was missing in figure 6?

Thanks.
 

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  • #2
gneill
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It is the lamp itself that is short circuited by the switch. So long as a zero resistance is placed across it, current will bypass the lamp.
Also, could someone check if my explanation of what would happen if there was no LDR please. If the LDR wasn't there then would the lamp actually work because the lamp is in parallel with the switch and since these would be the only components they would get the full p.d. of the battery?
No, taking the components as ideal that would be an "illegal" circuit, one where our clever modelling of real circuits as ideal components creates an unphysical scenario. Another example might be to place two different ideal current generators in the same wire.
I know in reality the short circuit would cause a huge surge in current and the battery's internal resistance would eat up the p.d. but if we assume there is no internal resistance would the lamp work if the LDR was missing in figure 6?
Still no. If you assume no internal resistance for the battery (ideal component) then why not have the wires ideal too? That short circuit on the lamp will still bypass any current. This is the problem with unphysical setups, you can't get logical results from them.
 
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Thanks for the information.

It is the lamp itself that is short circuited by the switch. So long as a zero resistance is placed across it, current will bypass the lamp.
But does this not mean we are now using an "illegal" circuit as the switch in figure 6 of the OP would not have zero resistance? Let me give an example that I would be grateful if you could help me understand and I will use no "illegal" bits.

Imagine a circuit with a 12V battery which just has a 24W, 12V bulb connected to it therefore drawing 2A from the battery assuming the cells have little internal resistance. Let's say that you now place 10cm piece of copper wire with a diamter of 2mm in parallel across the bulb. In reality this piece of wire has a definite resistance given by R = pl/A. Using the resistivity of copper, a length of 10cm and a diameter of 2mm gives a resistance of 0.0006Ω. Since the 6Ω of the bulb is so much larger then the combined parallel resistance will be roughly 0.0006Ω. This would draw a current of 20,000A if no fail safe kicked in. If a battery could (and I doubt one exists) maintain 12V (i.e. none lost to internal resistance) then according to the physics the bulb should stay on as it has 12V across it still and a resistance of 6Ω which it hasn't lost therefore according to V=IR it must get 2A and operate at normal brightness albeit for a very short amount of time before the battery is drained?

My point is that in reality with a normal 12V battery the bulb will go out because the battery (even with a very low internal resistance) will take most of the available 12V of p.d therefore very little is left for the bulb so the bulb gets very little current because it gets very little p.d. However, my book says that the current will bypass the bulb as it flows through the least resistive path and some other sources use this terminology for short circuiting. This really makes no sense to me after reading other sources online which I think seem well presented. These sources say that it is completely wrong to think about current bypassing a component since there are already electrons in the component and shorted wire before it is switched on. So it is not like some large current is waiting to come along and chooses to go down the shorted wire but rather there is almost no p.d. across the bulb so almost no current flows through it whereas even a small bit of p.d. left after going through the internal resistance will cause a large current to flow in a piece of wire as it only has 0.0006Ω of resistance.
 
  • #4
gneill
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Thanks for the information.


But does this not mean we are now using an "illegal" circuit as the switch in figure 6 of the OP would not have zero resistance? Let me give an example that I would be grateful if you could help me understand and I will use no "illegal" bits.

Imagine a circuit with a 12V battery which just has a 24W, 12V bulb connected to it therefore drawing 2A from the battery assuming the cells have little internal resistance. Let's say that you now place 10cm piece of copper wire with a diamter of 2mm in parallel across the bulb. In reality this piece of wire has a definite resistance given by R = pl/A. Using the resistivity of copper, a length of 10cm and a diameter of 2mm gives a resistance of 0.0006Ω. Since the 6Ω of the bulb is so much larger then the combined parallel resistance will be roughly 0.0006Ω. This would draw a current of 20,000A if no fail safe kicked in. If a battery could (and I doubt one exists) maintain 12V (i.e. none lost to internal resistance) then according to the physics the bulb should stay on as it has 12V across it still and a resistance of 6Ω which it hasn't lost therefore according to V=IR it must get 2A and operate at normal brightness albeit for a very short amount of time before the battery is drained?
since you've assumed no ideal components, yes the lamp might light for a brief instant (I say might because it's not clear what will burn out in what order -- some part of the wiring is going to melt through like a fuse pretty quickly).
My point is that in reality with a normal 12V battery the bulb will go out because the battery (even with a very low internal resistance) will take most of the available 12V of p.d therefore very little is left for the bulb so the bulb gets very little current because it gets very little p.d. However, my book says that the current will bypass the bulb as it flows through the least resistive path and some other sources use this terminology for short circuiting. This really makes no sense to me after reading other sources online which I think seem well presented. These sources say that it is completely wrong to think about current bypassing a component since there are already electrons in the component and shorted wire before it is switched on. So it is not like some large current is waiting to come along and chooses to go down the shorted wire but rather there is almost no p.d. across the bulb so almost no current flows through it whereas even a small bit of p.d. left after going through the internal resistance will cause a large current to flow in a piece of wire as it only has 0.0006Ω of resistance.
Yes, that statement about current taking the least resistive path is nonsense in the real world unless you are dealing with superconductors.
 
  • #5
CWatters
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By illegal gniell means the electrical equivalent of this situation...

"An irresistible force is applied to an immovable object. Which one wins?"

For example you cannot short circuit an ideal voltage source nor open circuit an ideal current source. If you find yourself doing that you have to replace the ideal components with real world components.

Most of the time (including solving problem 6) you can use ideal components. It makes very little difference to the answer if you assume the switch has zero resistance or a real small resistance. Both are likely to be much smaller than the resistance of a light dependant resistance (probably by >5 orders of magnitude).
 
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  • #6
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By illegal gniell means the electrical equivalent of this situation...

"An irresistible force is applied to an immovable object. Which one wins?"

For example you cannot short circuit an ideal voltage source nor open circuit an ideal current source. If you find yourself doing that you have to replace the ideal components with real world components.

Most of the time (including solving problem 6) you can use ideal components. It makes very little difference to the answer if you assume the switch has zero resistance or a real small resistance. Both are likely to be much smaller than the resistance of a light dependant resistance (probably by >5 orders of magnitude).
Thanks. Are you both in agreement that the textbook answer is not really great: "Figure 6 – the lamp is short circuited due to the switch arranged in parallel across it. The current will flow through the switch rather than the lamp due to its far lower resistance". The last sentence implies the current is choosing to go down the path of least resistance. Whereas I think from reading around a better answer would be to discuss the fact the switch/lamp parallel combo has almost no resistance and therefore receives no voltage as most of it will go to the LDR. I think the fact that a large current flows through the switch has nothing to do in directly explaining why the lamp goes out.
 
  • #8
CWatters
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Imagine a circuit with a 12V battery which just has a 24W, 12V bulb connected to it therefore drawing 2A from the battery assuming the cells have little internal resistance. Let's say that you now place 10cm piece of copper wire with a diamter of 2mm in parallel across the bulb. In reality this piece of wire has a definite resistance given by R = pl/A. Using the resistivity of copper, a length of 10cm and a diameter of 2mm gives a resistance of 0.0006Ω. Since the 6Ω of the bulb is so much larger then the combined parallel resistance will be roughly 0.0006Ω. This would draw a current of 20,000A if no fail safe kicked in. If a battery could (and I doubt one exists) maintain 12V (i.e. none lost to internal resistance) then according to the physics the bulb should stay on as it has 12V across it still and a resistance of 6Ω which it hasn't lost therefore according to V=IR it must get 2A and operate at normal brightness albeit for a very short amount of time before the battery is drained?
It is valid to consider such cases but if you are going to do that you may have to go much further. Why did you only treat the wire in parallel with the bulb as non-ideal? What about the rest of the wire in the circuit not to mention the battery?

Typically problems like this (eg number 6) do not include sufficient information to do a real world analysis. So you are forced to treat everything as ideal.
 
  • #9
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It is valid to consider such cases but if you are going to do that you may have to go much further. Why did you only treat the wire in parallel with the bulb as non-ideal? What about the rest of the wire in the circuit not to mention the battery?

Typically problems like this (eg number 6) do not include sufficient information to do a real world analysis. So you are forced to treat everything as ideal.
Whether or not you treat components as ideal still makes the textbook answer bad as it implies the current is choosing the path of least resistance. If you treat the switch as ideal then surely the parallel combo of the switch and lamp is exactly zero and therefore they will get none of the p.d. as it will all go to the LDR which is now a completely different answer to the book. Or am I missing what you mean by ideal.
 
  • #10
CWatters
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Thanks. Are you both in agreement that the textbook answer is not really great: "Figure 6 – the lamp is short circuited due to the switch arranged in parallel across it. The current will flow through the switch rather than the lamp due to its far lower resistance". The last sentence implies the current is choosing to go down the path of least resistance. Whereas I think from reading around a better answer would be to discuss the fact the switch/lamp parallel combo has almost no resistance and therefore receives no voltage as most of it will go to the LDR. I think the fact that a large current flows through the switch has nothing to do in directly explaining why the lamp goes out.
I think both answers (no current or no voltage) are fine.

I would word the no voltage answer like this....

"An ideal switch has zero resistance so there can be no voltage across the bulb so it does not light up".
 
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