Circuit Lab B/C

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Circuit Lab B/C

Postby UTF-8 U+6211 U+662F » September 3rd, 2018, 7:21 pm

Reposting my question from General Chat :)

Copper has one valence electron, with a density of 8.94 grams per cubic centimeter and an atomic weight of approximately 64 grams per mole. Suppose a copper wire has a current density of 18.8 amperes per square millimeter. Find the drift velocity inside the wire.

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Re: Circuit Lab B/C

Postby mdv2o5 » September 5th, 2018, 7:10 am

UTF-8 U+6211 U+662F wrote:Reposting my question from General Chat :)

Copper has one valence electron, with a density of 8.94 grams per cubic centimeter and an atomic weight of approximately 64 grams per mole. Suppose a copper wire has a current density of 18.8 amperes per square millimeter. Find the drift velocity inside the wire.


Answer
The drift velocity is given by

where j is the current density, n is the number density of the charge carrier (the electron), and q is the charge on the charge carrier. j is given and q can be looked up from a reference table as the charge of an electron. We just need to calculate n:

Putting everything together, we have

This is pretty ridiculously fast for a drift velocity, but given a current density of 18.8 A/mm^2, I suppose it's reasonable.

If the answer above is right, then here's the next question:
(this one is for Div C according to the new rules)
Determine the output voltage as a function of the source voltage, R1, and R2.
Image

A note on studying for op amps and also a bit of a hint:
Ideally, inverting and non-inverting op amp configurations should be memorized/included on your notes, but it's always good to be able to derive these formulas using basic circuit laws and the ideal op amp assumptions since it's really easy to create an op amp circuit that does not fall neatly into a standard configuration.

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Re: Circuit Lab B/C

Postby UTF-8 U+6211 U+662F » September 5th, 2018, 4:42 pm

mdv2o5 wrote:Determine the output voltage as a function of the source voltage, R1, and R2.
Image

A note on studying for op amps and also a bit of a hint:
Ideally, inverting and non-inverting op amp configurations should be memorized/included on your notes, but it's always good to be able to derive these formulas using basic circuit laws and the ideal op amp assumptions since it's really easy to create an op amp circuit that does not fall neatly into a standard configuration.

Interesting introduction problem to op-amps for me!
Labeling the voltage of the node to the bottom left and renaming for convenience:

(How op-amps work.)

(KCL)



(Looking back, I could've just used the voltage divider rule and obtained this directly.)

(Substitution. Now we just solve for .)





(The gain of the op-amp is very large and so the 1 in the equation can be approximated away with little concern. Also, recall that is the same thing as )


Edit: Small error. See next post for corrected work and answer.
Last edited by UTF-8 U+6211 U+662F on September 5th, 2018, 6:05 pm, edited 1 time in total.

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Re: Circuit Lab B/C

Postby mdv2o5 » September 5th, 2018, 5:47 pm

UTF-8 U+6211 U+662F wrote:
mdv2o5 wrote:Determine the output voltage as a function of the source voltage, R1, and R2.
Image

A note on studying for op amps and also a bit of a hint:
Ideally, inverting and non-inverting op amp configurations should be memorized/included on your notes, but it's always good to be able to derive these formulas using basic circuit laws and the ideal op amp assumptions since it's really easy to create an op amp circuit that does not fall neatly into a standard configuration.

Interesting introduction problem to op-amps for me!
Labeling the voltage of the node to the bottom left and renaming for convenience:

(How op-amps work.)

(KCL)



(Looking back, I could've just used the voltage divider rule and obtained this directly.)

(Substitution. Now we just solve for .)





(The gain of the op-amp is very large and so the 1 in the equation can be approximated away with little concern. Also, recall that is the same thing as )

Close! Check the KCL expression again (and remember that I = V/R)

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Re: Circuit Lab B/C

Postby UTF-8 U+6211 U+662F » September 5th, 2018, 6:04 pm

mdv2o5 wrote:Close! Check the KCL expression again (and remember that I = V/R)

Whoops. I always mess up Ohm's law for some reason (I should really make sure to double check).
Take two
Labeling the voltage of the node to the bottom left and renaming for convenience:

(How op-amps work.)

(KCL)





(Looking back, I could've just used the voltage divider rule and obtained this directly.)

(Substitution. Now we just solve for .)





(The gain of the op-amp is very large and so the 1 in the equation can be approximated away with little concern. Also, recall that is the same thing as )

I again hid some of the algebraic steps, but it should all (hopefully) be correct this time.

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Re: Circuit Lab B/C

Postby mdv2o5 » September 5th, 2018, 6:29 pm

UTF-8 U+6211 U+662F wrote:
mdv2o5 wrote:Close! Check the KCL expression again (and remember that I = V/R)

Whoops. I always mess up Ohm's law for some reason (I should really make sure to double check).
Take two
Labeling the voltage of the node to the bottom left and renaming for convenience:

(How op-amps work.)

(KCL)





(Looking back, I could've just used the voltage divider rule and obtained this directly.)

(Substitution. Now we just solve for .)





(The gain of the op-amp is very large and so the 1 in the equation can be approximated away with little concern. Also, recall that is the same thing as )

I again hid some of the algebraic steps, but it should all (hopefully) be correct this time.


Looks good!

A neat little trick...
...that's useful for solving op amp problems to avoid a lot of algebra is to make the ideal op amp assumption earlier. In an ideal op amp, no current flows into or out of the +/- terminals and v+ = v-. This means that the circuit from v0 to R2 to R1 to ground forms a perfect voltage divider (i.e. no current flows into the negative terminal). Thus we see that
.
This solves directly to give us

However, your method above is definitely a good way to formally show the relationship and get some practice with op amps!


Your turn!

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Re: Circuit Lab B/C

Postby UTF-8 U+6211 U+662F » September 5th, 2018, 6:50 pm

mdv2o5 wrote:
A neat little trick...
...that's useful for solving op amp problems to avoid a lot of algebra is to make the ideal op amp assumption earlier. In an ideal op amp, no current flows into or out of the +/- terminals and v+ = v-. This means that the circuit from v0 to R2 to R1 to ground forms a perfect voltage divider (i.e. no current flows into the negative terminal). Thus we see that
.
This solves directly to give us

However, your method above is definitely a good way to formally show the relationship and get some practice with op amps!


Your turn!

All right, thanks!

Onto basic circuit safety:
Touching two nodes of a DC voltage of above approximately what voltage can be lethal? What kind of factors affect whether a shock might be lethal? What current range is usually lethal? What are some harmful effects of having a current flow through you that is below the lethal limit (i.e. not enough to get you killed)?

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Re: Circuit Lab B/C

Postby Jacobi » September 17th, 2018, 8:07 am

UTF-8 U+6211 U+662F wrote:All right, thanks!

Onto basic circuit safety:
Touching two nodes of a DC voltage of above approximately what voltage can be lethal? What kind of factors affect whether a shock might be lethal? What current range is usually lethal? What are some harmful effects of having a current flow through you that is below the lethal limit (i.e. not enough to get you killed)?

Answer
Actually, it is current, not voltage that kills. 100-200 milliamps of live current can be lethal. However, high-voltage shocks respond more easily to artificial respiration. Assuming 1000 body resistance, 100-200 volts is the lethal range. Based on how much resistive body material is between the contact points, there may be more or less dangerous current.
Even if you aren't killed, it's still dangerous to be shocked. Shock, upset breathing, and paralysis can result from mid-level shocks.

Source: https://www.physics.ohio-state.edu/~p616/safety/fatal_current.html

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Re: Circuit Lab B/C

Postby UTF-8 U+6211 U+662F » September 17th, 2018, 11:52 am

Jacobi wrote:
UTF-8 U+6211 U+662F wrote:All right, thanks!

Onto basic circuit safety:
Touching two nodes of a DC voltage of above approximately what voltage can be lethal? What kind of factors affect whether a shock might be lethal? What current range is usually lethal? What are some harmful effects of having a current flow through you that is below the lethal limit (i.e. not enough to get you killed)?

Answer
Actually, it is current, not voltage that kills. 100-200 milliamps of live current can be lethal. However, high-voltage shocks respond more easily to artificial respiration. Assuming 1000 body resistance, 100-200 volts is the lethal range. Based on how much resistive body material is between the contact points, there may be more or less dangerous current.
Even if you aren't killed, it's still dangerous to be shocked. Shock, upset breathing, and paralysis can result from mid-level shocks.

Source: https://www.physics.ohio-state.edu/~p616/safety/fatal_current.html

Correct!
Although, it's important to note that for current to happen, you do need a voltage. That's why you see signs like "High Voltage" on fences. The rule that current not voltage kills isn't entirely true because of this and is only a guideline. Be very wary of voltage, as it's impossible to know the resistance of your body at any given moment.

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Re: Circuit Lab B/C

Postby Jacobi » September 18th, 2018, 2:08 pm

Write the voltage drop equations for a capacitor and a resistor.

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Re: Circuit Lab B/C

Postby UTF-8 U+6211 U+662F » September 18th, 2018, 5:57 pm

Jacobi wrote:Write the voltage drop equations for a capacitor and a resistor.

Answer



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Re: Circuit Lab B/C

Postby Jacobi » September 19th, 2018, 5:03 am

UTF-8 U+6211 U+662F wrote:
Jacobi wrote:Write the voltage drop equations for a capacitor and a resistor.

Answer



Awesome! You next.

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Re: Circuit Lab B/C

Postby Jacobi » September 19th, 2018, 5:04 am

UTF-8 U+6211 U+662F wrote:
Jacobi wrote:Write the voltage drop equations for a capacitor and a resistor.

Answer



Awesome! You next.

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Re: Circuit Lab B/C

Postby UTF-8 U+6211 U+662F » September 19th, 2018, 6:56 am

Jacobi wrote:
UTF-8 U+6211 U+662F wrote:
Jacobi wrote:Write the voltage drop equations for a capacitor and a resistor.

Answer



Awesome! You next.

All right!

Describe the behavior of an RC circuit which is a) charging and b) discharging.

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Re: Circuit Lab B/C

Postby Jacobi » September 23rd, 2018, 9:02 am

UTF-8 U+6211 U+662F wrote:All right!

Describe the behavior of an RC circuit which is a) charging and b) discharging.


Answer
a) The voltage through the resistor starts high and decays exponentially as the capacitor becomes saturated. The charged capacitor has an extremely high voltage drop.
b) The voltage through the resistor starts high and decays exponentially as the capacitor discharges. The capacitor acts as a voltage source in the absence of other sources.


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