Current through a circuit when the key is open

[Mister T]

And why are we not adding the infinite resistance of the open switch even if it is in series?

But that 2.4 ohms should be in series with that infinite resistance open switch....so infinite resistance + 2.4 = infinite resistance

You usually get in trouble with infinities. Here's the correct way to think about the relation $R=V/I$. We measure the current $I$ through a circuit element or branch, and we measure the voltage $V$ across that same element or branch. We define the ratio $V/I$ to be the resistance $R$. If we take several measurements of $V$ and $I$ and calculate the ratio $V/I$ for each measurement we can get get some interesting insight. If that ratio is the same for each measurement then we say the circuit element is ohmic. (Ohm's Law is the assertion that the ratio is constant, a law which, like all laws, has limits of validity. In other words, it's not always valid.) If that ratio is not constant, we say the device is nonohmic. A resistor (you know, the thing with the color bands on it) is ohmic over a wide range of values of $V$ and $I$. A good example of a nonohmic device is an incandescent light bulb. The ratio is not constant, that is, the resistance varies. This is because as you raise $V$ you increase $I$ but you also raise the temperature of the bulb's filament, causing the resistance $R$ to also increase. In other words, the ratio $V/I$ is not constant.

A switch is an entirely different type of circuit element. I've never heard of anyone even trying to measure the resistance of a switch. When the switch is closed, ideally it has zero resistance. When the switch is open I suppose you could raise $V$ so high that a spark jumps across the switch, which would a very high current for a very short time. So forget about the resistance of switches and the notion that they have an infinite resistance when open. The only time the resistance of a switch becomes significant is when it's closed and it offers some resistance that is not negligible in comparison to the other circuit elements. The same is true of the connecting wires in a circuit. We use the term ideal to mean the resistance is so small in comparison to the other circuit elements that we can ignore it. It's said to be negligible.

It's easy when taking a physics course to become concerned with the variables and their values. That's of course important, especially when trying to get the right answers to questions. And it's also important when doing experiments. But don't forget that at it's heart, physics is about the study of phenomena. In other words, it's phenomenology. In this case the phenomenon we're studying is the behavior of a switch.

My friends are saying 0.6 and 2.4, the answer key says 2.4 and you all are also saying 0.6 and 2.4, so which one should be correct and why? And if this is an incorrect-framing-of-the-question issue, please instruct me about the different resistances we can get if we consider different interpretations and different "points."

Okay, sir. But how do we reach the final 2.4 ohms or 0.6 ohm answer? I am still not able to judge even which one is correct 2.4 or 0.6 or maybe infinite resistance.

Remember, we are trying to interpret a poorly-worded question, so there will be different interpretations of what the question means or what the question is asking.

If we interpret Part i) of the question as asking about the parallel combination of the $6\:\Omega$ and the $4\:\Omega$ resistors we proceed as follows. The equivalent resistance $R$ can found using the familiar formula $\frac{1}{R}=\frac{1}{6\:\Omega+4\:\Omega}$, which gives us $R=2.4\:\Omega$. That's the correct answer. The justification for this, I suppose, is that the $0.6\:\Omega$ is supposed to be the internal resistance of the battery, and with with the switch (key) open no current flows through the battery hence the internal resistance is zero.

I don't know how your friends came up with $0.6\:\Omega$. It was only mentioned once AFAIK in this thread by @PhDeezNutz in Post #20, so we'd have to ask him how he came up with it. Again, it would just be a different interpretation of this poorly-worded question.

Now, on to Part ii). My interpretation is that the $0.6\:\Omega$ is the internal resistance of the battery when the switch is closed (and thus a specific amount of current is flowing through the battery). In that case the $0.6\:\Omega$ is in series with the above-mentioned $2.4\:\Omega$, so we add them to get a total of $3.0\:\Omega$.

Part iii). The $3.0\:\Omega$ is the effective resistance of the circuit when the switch (key) is closed. The battery voltage is $6.0\:\mathrm{V}$. So $I=V/R=\frac{6.0\:\mathrm{V}}{3.0\:\Omega}=2.0\:\mathrm{A}$. (This is the specific amount of current mentioned above that gives rise to the internal resistance of $0.6\:\Omega$).

Part iv). To find the current through the $4\:\Omega$ resistor, we need to first find the voltage across it. Note that it's the same as the voltage across the parallel combination that has an equivalent resistance of $2.4\:\Omega$. Thus $V=IR=(2.0\:\mathrm{A})(2.4\:\Omega)=4.8\:\mathrm{V}$. So we have $4.8\:\mathrm{V}$ across the $4\:\Omega$ resistor. Therefore, $I=V/R=\frac{4.8 V}{4.0\:\Omega}=1.2\:\mathrm{A}$.

Another way to find the voltage across the $4\:\Omega$ resistor is to first find the voltage across the $0.6\:\Omega$ resistor. $V=IR=(2.0\:\mathrm{A})(0.6 \Omega)=1.2\:\mathrm{V}$. So we take the $6.0\:\mathrm{V}$ across the battery and subtract the $1.2\:\mathrm{V}$, leaving us with $4.8\:\mathrm{V}$ across the $4\:\Omega$ resistor. The first part of what I see in the answer key for Part iv) appears to be gibberish, and the second part has a typo.

 

[Steve4Physics]

I've never heard of anyone even trying to measure the resistance of a switch.

For information, a switch’s ‘contact resistance’ (resistance when closed) and its ‘insulation resistance’ (resistance when open) are recognised parameters.

In many situations they can be taken as 0 and ‘infinity’. But this is not always the case – e.g. in high current or high voltage applications.

E.g. the contact and insulation resistances for a switch are both given on this datasheet: https://www.farnell.com/datasheets/33031.pdf

 

[jbriggs444]

Rather than dividing voltage by current to get resistance, a useful approach is to take the a derivative. By how much do you need to increase voltage to get a small fixed increment in current?

i.e. what is the slope of the voltage versus current graph?

If the graph is a straight line intersecting the origin then you have an ohmic circuit element. If it is curved, broken or squiggled then one may still calculate a resistance at a particular loading.

This idea becomes relevant if one wants to calculate the resistance (or equivalent resistance) of a network of circuit elements that contains not just resistors but also fixed voltage sources or fixed current sources.
The circuit presented in the OP in this thread is such a network. It contains a voltage source along with three resistors.

We could imagine slicing the circuit open at any chosen point (for instance, the point labelled "A"). We could then measure how much current passes through our slice as a function of any forward voltage we apply through our slice. The slope of the current versus voltage graph (actually the inverse of that slope -- voltage over current) tells us the equivalent resistance of the circuit.

This approach works even if the current versus voltage graph does not intersect with the origin -- even if a current is already flowing when we are applying zero voltage across our slice.

 

[Darshit Sharma]

I'm not! See Post #9.

Yeah, I am in support of infinity as the answer too.

 

[Darshit Sharma]

Rather than dividing voltage by current to get resistance, a useful approach is to take the a derivative. By how much do you need to increase voltage to get a small fixed increment in current?

i.e. what is the slope of the voltage versus current graph?

If the graph is a straight line intersecting the origin then you have an ohmic circuit element. If it is curved, broken or squiggled then one may still calculate a resistance at a particular loading.

This idea becomes relevant if one wants to calculate the resistance (or equivalent resistance) of a network of circuit elements that contains not just resistors but also fixed voltage sources or fixed current sources.
The circuit presented in the OP in this thread is such a network. It contains a voltage source along with three resistors.

We could imagine slicing the circuit open at any chosen point (for instance, the point labelled "A"). We could then measure how much current passes through our slice as a function of any forward voltage we apply through our slice. The slope of the current versus voltage graph (actually the inverse of that slope -- voltage over current) tells us the equivalent resistance of the circuit.

This approach works even if the current versus voltage graph does not intersect with the origin -- even if a current is already flowing when we are applying zero voltage across our slice.

That's ok but won't it give the resistance of the circuit when it is closed (not open)? We are already able to find it without any problem.

 

[Darshit Sharma]

When the switch is open I suppose you could raise $V$ so high that a spark jumps across the switch, which would be a very high current for a very short time. So forget about the resistance of switches and the notion that they have an infinite resistance when open.

And......maybe......if we don't raise the voltage that high then is it near infinity?

I don't know how your friends came up with $0.6\:\Omega$. It was only mentioned once AFAIK in this thread by @PhDeezNutz in Post #20, so we'd have to ask him how he came up with it. Again, it would just be a different interpretation of this poorly-worded question.

Let's see what @PhDeezNutz has to say.

The first part of what I see in the answer key for Part iv) appears to be gibberish, and the second part has a typo.

It is, indeed. This question is just a bit of a time-wasting sort of thing.


Thanks for the efforts.....I got the essence of how we don't take into account the internal resistance (maybe just a wording interpretation but yes---certainly useful for some 2 3 further tests). Not accurate as you said but yes it'll still get the work done for some months.

Thanks @Mister T @berkeman @jbriggs444 @kuruman @Steve4Physics for all your efforts.
(And yeah, thanks to anyone who is left out.)

 

[Mister T]

And......maybe......if we don't raise the voltage that high then is it near infinity?

No. There's no such thing as near infinity AFAIK. Certainly not in this context.

 

[Darshit Sharma]

No. There's no such thing as near infinity AFAIK. Certainly not in this context.

okay sir!!