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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.

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.

## 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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