# Conceptual Questions about Voltage, Current, Resistance and their relationship

First off, mad props to berkeman for suggesting Art of Electronics. It's the first textbook that has actually connected with me. I'll be buying myself a copy.

## Homework Statement

Please refer to the images that I've attached below. I had copied down the images from the textbook, since it was only available as a reserve from the library. I hope my handwriting is readable.

www.rangepla.net/pixels/x/1.jpg[/URL]

[h2]Homework Equations[/h2]

Ohm's law, V = IR.

[h2]The Attempt at a Solution[/h2]

[i][u]A[/u][/i]
Why doesn't current flow like how the arrow shows? If the wires are assumed to be of zero resistance (the intro course I'm taking does assume so), then wouldn't the first two resistors get shorted out? I figured that current would want to follow the path that has free resistance, hence the thought. I forgot to draw the ground, but it's on the bottom of the circuit. Say, below the first resistor. The two ends are connected to a voltage source.

[i][u]B[/u][/i]
R1 = R2 = 10k and V = 30V. I'm supposed to find output voltage (V across R2) when (a) R_load = 0 and (b) R_load is 10k. I get 15V for (a) and 10V for (b). I can get the answer mathematically, which really doesn't mean that I [I]understood[/I] what the problem was asking. From Ohm's law, it's clear that the function V=IR is linear. Given a constant I, V should decrease when R decreases. In this given example, net I is not the same in (a) and (b). In fact, change in I is less than the change in R (net values). Why? I assumed that since a function is linear, the drops must be the same too. Maybe I don't even understand the meaning of a linear function! :surprised

[i][u]C[/u][/i]
I came to two conclusions about this one. The textbook claims that they're equal, and even after thinking about it, I just don't get it.

(i) If I consider the 15V source to be perfect (in picture B), then I can accept the fact that the voltage across the two terminals to be 15V. I already know that the V_out in A is 15V. So, what's the point of the 5k resistor (R_th)? Why should it be 5k? Can't it be something else, like, 3k?

(ii) Current in A is 1.5mA. Voltage drop across R2 is 15V.
Current in B is 3mA, Voltage drop across R_th is 15V. If this were to be true, then the V_out would be 0. I'm certain that it's not 0. I can also instinctively say that this way of thinking is definitely wrong. Isn't Thevenin voltage an open circuit voltage?

Okay, that's about it for now. I'll read the rest of the fundamentals today, and I'll most likely be back with more questions. Thank you in advance to anyone who can help me.

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Mapes
Homework Helper
Gold Member
Why doesn't current flow like how the arrow shows? If the wires are assumed to be of zero resistance (the intro course I'm taking does assume so), then wouldn't the first two resistors get shorted out? I figured that current would want to follow the path that has free resistance, hence the thought.
Unfortunately, it's not possible to look only at where you've drawn your arrows and predict anything about the current. Since the wire resistance is assumed to be zero, the points are all electrically equivalent (zero voltage drop). You have to calculate current from an region with a defined, finite voltage drop, which is the resistors. Now, the three paths through the resistors are all equivalent (assuming the resistors are equal), so the current travels through all three of them equally.

Also, note that "shorted out" means that an alternative path with zero resistance exists around a component. This doesn't occur here.

R1 = R2 = 10k and V = 30V. I'm supposed to find output voltage (V across R2) when (a) R_load = 0 and (b) R_load is 10k. I get 15V for (a) and 10V for (b). I can get the answer mathematically, which really doesn't mean that I understood what the problem was asking. From Ohm's law, it's clear that the function V=IR is linear. Given a constant I, V should decrease when R decreases. In this given example, net I is not the same in (a) and (b). In fact, change in I is less than the change in R (net values). Why? I assumed that since a function is linear, the drops must be the same too. Maybe I don't even understand the meaning of a linear function! :surprised
Check these answers again; one of them is wrong. That might clear up the confusion.

C
I came to two conclusions about this one. The textbook claims that they're equal, and even after thinking about it, I just don't get it.

Oh snaps, I forgot to upload the last image. Sorry =).

My calculations for

V_in = 30V
Total R = 20k
Net I = 1.5mA
Voltage drop across R2 = I*R = 1.5mA*10k = 15V
This answer actually does make sense, since the R's are equal and the setup is a voltage divider, the voltage gets divided up equally.

Then R_Load and R2 are in parallel,
Total R = 15k
Total V = 30V
Net I = 2mA
Voltage drop across the second half = I*R = 2mA*5k = 10V

Oh, I think I know what I did wrong - I assumed the bottom half to be parallel, when it is not. So recalculating, I should first find the drop across the first half, which is 20V, and the rest of the 10V gets split up equally into 5V across both R2 and R_load. So I guess V_out is 5V, instead of 10V? But I still don't get constant slope for a voltage-current "plot." Is it because I modified the circuit?

Mapes
Homework Helper
Gold Member
R_2 and R_load are in parallel. If R_load is 0 (that is, a short circuit), wouldn't the voltage across R_2 be 0 also?

Oh, hmm. I thought that when R_load is 0, they meant that they took away the resistor. I never thought about it that way. The exact wording of the book was "Find the output voltage with no load attached (the open circuit voltage)" and part (b) says "find the output voltage with a 10k load (treat as voltage divider, with R2 and R_load combined into a single resistor)."

Hope that cleared up some confusion.

Mapes
Homework Helper
Gold Member
OK, that would correspond to $R_\mathrm{load}=\infty$. With that condition, I'll agree with 15V and 10V. Ohm's Law is linear because the current through $R_2$ is 1.5mA and 1.0mA, respectively. In the second case an additional 1.0mA flows through $R_\mathrm{load}$. Another way to look at it is that 2.0mA flows through the effective resistance of $R_2$ and $R_\mathrm{load}$ in parallel, which is 5K.

Ah. Thanks Mapes, that helped. Do you mind helping with the third one too? :D.

Mapes