2019: Circuit Lab: 1st Thermodynamics: 2nd Sounds of Music: 5th Mission Possible: 6th 2018: Code Busters (trial): 1st 2015: Simple Machines: 2nd SumoBots (trial): 2nd 2014: WIDI: 2nd Simple Machines: 3rd
To revive this thread:
Diagram: https://drive.google.com/file/d/1-IqHS7 ... sp=sharing
Resistors R1, R2, and R3, voltage source V1, and current source I1 are arranged as in the diagram. These circuit elements have the following values:
R1: 5Ω
R2: 3Ω
R3: 10Ω
V1: 8V
I1: 6A
What is the voltage across and current through each of the three resistors?
Voltages: R1: 65/9 V (Right is positive) R2: 41/3 V (Bottom is positive) R3: 130/9 V (Right is positive) Currents: R1: 13/9 A (Right to Left) R2: 41/9 A (Bottom to Top) R3: 13/9A (Bottom to Top) Edited because oops
That looks right! (Although, if this were a test, you might want to evaluate those fractions and truncate them to 1 sig fig.) Your turn!To revive this thread:
Diagram: https://drive.google.com/file/d/1-IqHS7 ... sp=sharing
Resistors R1, R2, and R3, voltage source V1, and current source I1 are arranged as in the diagram. These circuit elements have the following values:
R1: 5Ω
R2: 3Ω
R3: 10Ω
V1: 8V
I1: 6A
What is the voltage across and current through each of the three resistors?Voltages: R1: 65/9 V (Right is positive) R2: 41/3 V (Bottom is positive) R3: 130/9 V (Right is positive) Currents: R1: 13/9 A (Right to Left) R2: 41/9 A (Bottom to Top) R3: 13/9A (Bottom to Top) Edited because oops
2019: Circuit Lab: 1st Thermodynamics: 2nd Sounds of Music: 5th Mission Possible: 6th 2018: Code Busters (trial): 1st 2015: Simple Machines: 2nd SumoBots (trial): 2nd 2014: WIDI: 2nd Simple Machines: 3rd
Hi!
Write and simplify a boolean algebra expression for the following diagram:
(A link longer than my attention span)
http://www.101computing.net/logic-gates ... itle=Title
For fun, make a truth table.
A' OR B' OR C Truth table: [code]A|B|C|Result 0|0|0|1 0|0|1|1 0|1|0|1 0|1|1|1 1|0|0|1 1|0|1|1 1|1|0|0 1|1|1|1[/code]
2019: Circuit Lab: 1st Thermodynamics: 2nd Sounds of Music: 5th Mission Possible: 6th 2018: Code Busters (trial): 1st 2015: Simple Machines: 2nd SumoBots (trial): 2nd 2014: WIDI: 2nd Simple Machines: 3rd
That was my answer! You got the next question!Hi!
Write and simplify a boolean algebra expression for the following diagram:
(A link longer than my attention span)
http://www.101computing.net/logic-gates ... itle=Title
For fun, make a truth table.A' OR B' OR C Truth table: [code]A|B|C|Result 0|0|0|1 0|0|1|1 0|1|0|1 0|1|1|1 1|0|0|1 1|0|1|1 1|1|0|0 1|1|1|1[/code]
2019: Circuit Lab: 1st Thermodynamics: 2nd Sounds of Music: 5th Mission Possible: 6th 2018: Code Busters (trial): 1st 2015: Simple Machines: 2nd SumoBots (trial): 2nd 2014: WIDI: 2nd Simple Machines: 3rd
Diagram: https://drive.google.com/file/d/1pqa8dq ... sp=sharing
Note: this problem requires calculus. (It's possible there's a particularly clever solution that bypasses the calculus, but I can't think of one.)
You are an electrical engineer tasked with charging a capacitor , which is connected in series with resistor and in parallel with resistor . You want to charge the capacitor to a final voltage , and you have at your disposal a current source . (See the diagram for the circuit configuration, and note that at time , the capacitor is fully discharged. Also note that is constant; i.e. the current cannot vary with time.)
In the aim of efficiency, you want to dissipate as little power as possible in the resistors as you charge the capacitor to voltage . However, your colleague, who designed the circuit with , , and and chose the value of , won't let you change any of those values. The only value you can play around with is the charging current .
a) Find an equation, in terms of , , , and , and , for the time at which the capacitor is fully charged to voltage .
b) Find an equation, in terms of , , , and , and , for the amount of power dissipated in the resistors in the course of charging the capacitor to voltage .
c) Find an equation, in terms of , , , and , for the value of that offers minimum power dissipation. (I'm not convinced this part has a closed-form solution -- if not, do what you can.)
a) [math]Q=CV[/math] [math]I_C=C \cdot \frac{dV}{dt}[/math] [math]I-VR_p = C \cdot \frac{dV}{dt}[/math] [math]dt = C \cdot \frac{dV}{I-VR_p}[/math] [math]t + C_{int} = C \cdot -\frac{\ln(I-VR_p)}{R_p}[/math] (the constant of integration also being a C makes this slightly confusing) [math]t = \frac{-C\ln(I-VR_p)}{R_p} + k[/math] (so I changed it to a k) [math]t = \frac{-C\ln(I-VR_p) + C\ln I}{R_p}[/math] (but plugging in (0,0) gives us a value for k) [math]\boxed{t_f = \frac{-C\ln(I-V_fR_p) + C\ln I}{R_p}}[/math] b) I'm going to assume you mean energy for the next problem? [math]P = I^2R_s + \frac{V^2}{R_p}[/math] [math]E = I^2R_s \frac{-C\ln(I-V_fR_p) + C\ln I}{R_p} + \int\frac{V^2}{R_p}dt[/math] Noting that [math]\frac{dV}{dt} = \frac{I-VR_p}{C}[/math], so [math]dt = \frac{CdV}{I-VR_p}[/math] [math]\boxed{E = I^2R_s \frac{-C\ln(I-V_fR_p) + C\ln I}{R_p} + \int^{\frac{-C\ln(I-V_fR_p) + C\ln I}{R_p}}_0\frac{CV^2}{IR_p-VR^2_p}dV}[/math] This technically satisfies what you asked for, but it's really ugly, so I'm guessing I either did it wrong or missed something. c) I feel like you're supposed to take the derivative of part B with respect to I for this but since I got something weird for part B, I'm going to call it a night.
Diagram: https://drive.google.com/file/d/1pqa8dq ... sp=sharing
Note: this problem requires calculus. (It's possible there's a particularly clever solution that bypasses the calculus, but I can't think of one.)
You are an electrical engineer tasked with charging a capacitor , which is connected in series with resistor and in parallel with resistor . You want to charge the capacitor to a final voltage , and you have at your disposal a current source . (See the diagram for the circuit configuration, and note that at time , the capacitor is fully discharged. Also note that is constant; i.e. the current cannot vary with time.)
In the aim of efficiency, you want to dissipate as little power as possible in the resistors as you charge the capacitor to voltage . However, your colleague, who designed the circuit with , , and and chose the value of , won't let you change any of those values. The only value you can play around with is the charging current .
a) Find an equation, in terms of , , , and , and , for the time at which the capacitor is fully charged to voltage .
b) Find an equation, in terms of , , , and , and , for the amount of power dissipated in the resistors in the course of charging the capacitor to voltage .
c) Find an equation, in terms of , , , and , for the value of that offers minimum power dissipation. (I'm not convinced this part has a closed-form solution -- if not, do what you can.)a) [math]Q=CV[/math] [math]I_C=C \cdot \frac{dV}{dt}[/math] [math]I-VR_p = C \cdot \frac{dV}{dt}[/math] [math]dt = C \cdot \frac{dV}{I-VR_p}[/math] [math]t + C_{int} = C \cdot -\frac{\ln(I-VR_p)}{R_p}[/math] (the constant of integration also being a C makes this slightly confusing) [math]t = \frac{-C\ln(I-VR_p)}{R_p} + k[/math] (so I changed it to a k) [math]t = \frac{-C\ln(I-VR_p) + C\ln I}{R_p}[/math] (but plugging in (0,0) gives us a value for k) [math]\boxed{t_f = \frac{-C\ln(I-V_fR_p) + C\ln I}{R_p}}[/math] b) I'm going to assume you mean energy for the next problem? [math]P = I^2R_s + \frac{V^2}{R_p}[/math] [math]E = I^2R_s \frac{-C\ln(I-V_fR_p) + C\ln I}{R_p} + \int\frac{V^2}{R_p}dt[/math] Noting that [math]\frac{dV}{dt} = \frac{I-VR_p}{C}[/math], so [math]dt = \frac{CdV}{I-VR_p}[/math] [math]\boxed{E = I^2R_s \frac{-C\ln(I-V_fR_p) + C\ln I}{R_p} + \int^{\frac{-C\ln(I-V_fR_p) + C\ln I}{R_p}}_0\frac{CV^2}{IR_p-VR^2_p}dV}[/math] This technically satisfies what you asked for, but it's really ugly, so I'm guessing I either did it wrong or missed something. c) I feel like you're supposed to take the derivative of part B with respect to I for this but since I got something weird for part B, I'm going to call it a night.
your third equation should have [math]I-\frac{V}{R_p}[/math], not [math]I-VR_p[/math], and that error messed up almost everything else you have. With that fix, you would have gotten part a right, and I haven't evaluated part b, but the way you approach it seems reasonable. Wolfram Alpha or Mathematica might help with the more involved integrals.
2019: Circuit Lab: 1st Thermodynamics: 2nd Sounds of Music: 5th Mission Possible: 6th 2018: Code Busters (trial): 1st 2015: Simple Machines: 2nd SumoBots (trial): 2nd 2014: WIDI: 2nd Simple Machines: 3rd
Agh! Well that sucks.your third equation should have [math]I-\frac{V}{R_p}[/math], not [math]I-VR_p[/math], and that error messed up almost everything else you have. With that fix, you would have gotten part a right, and I haven't evaluated part b, but the way you approach it seems reasonable. Wolfram Alpha or Mathematica might help with the more involved integrals.
a) [math]Q=CV[/math] [math]I_C=C \cdot \frac{dV}{dt}[/math] [math]I-\frac{V}{R_p} = C \cdot \frac{dV}{dt}[/math] [math]dt = C \cdot \frac{dV}{I-\frac{V}{R_p}}[/math] [math]t + C_{int} = -CR_p\ln(I-\frac{V}{R_p})[/math] (the constant of integration also being a C makes this slightly confusing) [math]t = -CR_p\ln(I-\frac{V}{R_p}) + k[/math] (so I changed it to a k) [math]t = -CR_p\ln(I-\frac{V}{R_p}) + CR_p\ln I[/math] (but plugging in (0,0) gives us a value for k) [math]\boxed{t_f = -CR_p\ln(I-\frac{V_f}{R_p}) + CR_p\ln I}[/math] b) I'm going to assume you mean energy for the next problem? [math]P = I^2R_s + \frac{V^2}{R_p}[/math] [math]E = I^2R_s \cdot (-CR_p\ln(I-\frac{V_f}{R_p}) + CR_p\ln I) + \int^{t_f}_0\frac{V^2}{R_p}dt[/math] Now, let's solve for V in terms of t... [math]t = -CR_p\ln(I-\frac{V}{R_p}) + CR_p\ln I[/math] [math]\frac{CR_p\ln I - t}{CR_p} = \ln(I-\frac{V}{R_p})[/math] [math]\ln I - \frac{t}{CR_p} = \ln(I-\frac{V}{R_p})[/math] [math]Ie^{-\frac{t}{CR_p}} = I - \frac{V}{R_p}[/math] [math]V = IR_p(1-e^{-\frac{t}{CR_p}})[/math] [math]E = CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + \int^{t_f}_0\frac{(IR_p(1-e^{-\frac{t}{CR_p}}))^2}{R_p}dt[/math] [math]E = CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + I^2R_p\int^{t_f}_0(1-e^{-\frac{t}{CR_p}})^2dt[/math] [math]E = CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + I^2R_p\left[t+\frac{e^{-2\frac{t}{CR_p}}-4e^{{-\frac{t}{CR_p}}}}{-\frac{2}{CR_p}}\right]^{t_f}_0[/math] [math]\boxed{E = CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + I^2R_p\left(-CR_p\ln(I-\frac{V_f}{R_p}) + CR_p\ln I+\frac{\frac{(I-\frac{V_f}{R_p})^2}{I^2}-4\frac{I-\frac{V_f}{R_p}}{I}}{-\frac{2}{CR_p}}-\frac{3CR_p}{2}\right)}[/math] For part c: [math]\boxed{0 = \frac{d}{dV}\left[CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + I^2R_p\left(-CR_p\ln(I-\frac{V_f}{R_p}) + CR_p\ln I+\frac{\frac{(I-\frac{V_f}{R_p})^2}{I^2}-4\frac{I-\frac{V_f}{R_p}}{I}}{-\frac{2}{CR_p}}-\frac{3CR_p}{2}\right)\right]}[/math] Although you'd need to test all of the roots and see which one yields the lowest value
Return to “2019 Question Marathons”
Users browsing this forum: No registered users and 1 guest