Darshit Sharma said:
And why are we not adding the infinite resistance of the open switch even if it is in series?
Darshit Sharma said:
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.
Darshit Sharma said:
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."
Darshit Sharma said:
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.