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.
The drift velocity is given by
[math]v_D = \frac{j}{nq}[/math]
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:
[math]n = \frac{8.94 g/cm^3}{64 g/mol} \times 1 electron/atom \times N_A atoms/mol \times 10^{-3} cm^3/mm^3 = 8.4\times 10^{19} electrons/mm^3[/math]
Putting everything together, we have
[math]v_D = \frac{18.8 A/mm^2}{8.4 \times 10^{19} electrons/mm^3 \times 1.6\times 10^{-19} C/electron} = 1.4 mm/s = 1400 \mu m/s[/math]
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.
mdv2o5 wrote:Determine the output voltage as a function of the source voltage, R1, and R2.
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.
Labeling the voltage of the node to the bottom left [math]v^-[/math] and renaming [math]v_s = v^+[/math] for convenience:
[math]A(v^+ - v^-) = v_o[/math] (How op-amps work.)
[math](v^- - v_o)R_2 + (v^-)R_1 = 0[/math] (KCL)
[math](v^-)(R_1 + R_2) = (v_o)R_2[/math]
[math]v^- = \frac{v_oR_2}{R_1 + R_2}[/math] (Looking back, I could've just used the voltage divider rule and obtained this directly.)
[math]A\left(v^+ - A\frac{v_oR_2}{R_1 + R_2}\right) = v_o[/math] (Substitution. Now we just solve for [math]v_o[/math].)
[math]Av^+ - A\frac{v_oR_2}{R_1 + R_2} = v_o[/math]
[math]Av^+ = v_o\left(1 + \frac{AR_2}{R_1 + R_2}\right)[/math]
[math]v_o = \frac{Av_s}{1 + \frac{AR_2}{R_1 + R_2}} \approx \frac{v_s(R_1 + R_2)}{R_2}[/math] (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 [math]v_s[/math] is the same thing as [math]v^+[/math]) Close! Check the KCL expression again (and remember that I = V/R)UTF-8 U+6211 U+662F wrote:mdv2o5 wrote:Determine the output voltage as a function of the source voltage, R1, and R2.
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.Labeling the voltage of the node to the bottom left [math]v^-[/math] and renaming [math]v_s = v^+[/math] for convenience: [math]A(v^+ - v^-) = v_o[/math] (How op-amps work.) [math](v^- - v_o)R_2 + (v^-)R_1 = 0[/math] (KCL) [math](v^-)(R_1 + R_2) = (v_o)R_2[/math] [math]v^- = \frac{v_oR_2}{R_1 + R_2}[/math] (Looking back, I could've just used the voltage divider rule and obtained this directly.) [math]A\left(v^+ - A\frac{v_oR_2}{R_1 + R_2}\right) = v_o[/math] (Substitution. Now we just solve for [math]v_o[/math].) [math]Av^+ - A\frac{v_oR_2}{R_1 + R_2} = v_o[/math] [math]Av^+ = v_o\left(1 + \frac{AR_2}{R_1 + R_2}\right)[/math] [math]v_o = \frac{Av_s}{1 + \frac{AR_2}{R_1 + R_2}} \approx \frac{v_s(R_1 + R_2)}{R_2}[/math] (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 [math]v_s[/math] is the same thing as [math]v^+[/math])
Whoops. I always mess up Ohm's law for some reason (I should really make sure to double check).mdv2o5 wrote:Close! Check the KCL expression again (and remember that I = V/R)
Labeling the voltage of the node to the bottom left [math]v^-[/math] and renaming [math]v_s = v^+[/math] for convenience:
[math]A(v^+ - v^-) = v_o[/math] (How op-amps work.)
[math]\frac{v^- - v_o}{R_2} + \frac{v^-}{R_1} = 0[/math] (KCL)
[math]v^-(\frac1{R_1} + \frac1{R_2}) = \frac{v_o}{R_2}[/math]
[math]v^- = \frac{v_o}{R_2\left(\frac1{R_1} + \frac1{R_2}\right)}[/math]
[math]v^- = \frac{v_oR_1}{R_1 + R_2}[/math](Looking back, I could've just used the voltage divider rule and obtained this directly.)
[math]A\left(v^+ - \frac{v_oR_1}{R_1 + R_2}\right) = v_o[/math] (Substitution. Now we just solve for [math]v_o[/math].)
[math]Av^+ - A\frac{v_oR_1}{R_1 + R_2} = v_o[/math]
[math]Av^+ = v_o\left(1 + \frac{AR_1}{R_1 + R_2}\right)[/math]
[math]v_o = \frac{Av_s}{1 + \frac{AR_1}{R_1 + R_2}} \approx \frac{v_s(R_1 + R_2)}{R_1}[/math] (The gain [math]A[/math] of the op-amp is very large and so the 1 in the equation can be approximated away with little concern. Also, recall that [math]v_s[/math] is the same thing as [math]v^+[/math])
I again hid some of the algebraic steps, but it should all (hopefully) be correct this time. Looks good!UTF-8 U+6211 U+662F wrote:Whoops. I always mess up Ohm's law for some reason (I should really make sure to double check).mdv2o5 wrote:Close! Check the KCL expression again (and remember that I = V/R)Labeling the voltage of the node to the bottom left [math]v^-[/math] and renaming [math]v_s = v^+[/math] for convenience: [math]A(v^+ - v^-) = v_o[/math] (How op-amps work.) [math]\frac{v^- - v_o}{R_2} + \frac{v^-}{R_1} = 0[/math] (KCL) [math]v^-(\frac1{R_1} + \frac1{R_2}) = \frac{v_o}{R_2}[/math] [math]v^- = \frac{v_o}{R_2\left(\frac1{R_1} + \frac1{R_2}\right)}[/math] [math]v^- = \frac{v_oR_1}{R_1 + R_2}[/math](Looking back, I could've just used the voltage divider rule and obtained this directly.) [math]A\left(v^+ - \frac{v_oR_1}{R_1 + R_2}\right) = v_o[/math] (Substitution. Now we just solve for [math]v_o[/math].) [math]Av^+ - A\frac{v_oR_1}{R_1 + R_2} = v_o[/math] [math]Av^+ = v_o\left(1 + \frac{AR_1}{R_1 + R_2}\right)[/math] [math]v_o = \frac{Av_s}{1 + \frac{AR_1}{R_1 + R_2}} \approx \frac{v_s(R_1 + R_2)}{R_1}[/math] (The gain [math]A[/math] of the op-amp is very large and so the 1 in the equation can be approximated away with little concern. Also, recall that [math]v_s[/math] is the same thing as [math]v^+[/math]) I again hid some of the algebraic steps, but it should all (hopefully) be correct this time.
...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
[math]v^- = v_S = \frac{R_1}{R_1 + R_2}v_O[/math].
This solves directly to give us
[math]v_O = \frac{R_1 + R_2}{R_1}v_S[/math]
However, your method above is definitely a good way to formally show the relationship and get some practice with op amps!All right, thanks!mdv2o5 wrote:Your turn!...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 [math]v^- = v_S = \frac{R_1}{R_1 + R_2}v_O[/math]. This solves directly to give us [math]v_O = \frac{R_1 + R_2}{R_1}v_S[/math] However, your method above is definitely a good way to formally show the relationship and get some practice with op amps!
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)?
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 [math]\Omega[/math] 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: [url]https://www.physics.ohio-state.edu/~p616/safety/fatal_current.html[/url]
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.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)?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 [math]\Omega[/math] 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: [url]https://www.physics.ohio-state.edu/~p616/safety/fatal_current.html[/url]
