Circuit Lab B/C

mjcox2000
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Re: Circuit Lab B/C

Postby mjcox2000 » February 21st, 2019, 7:53 pm

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?
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Re: Circuit Lab B/C

Postby Cathy-TJ » February 21st, 2019, 8:39 pm

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
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mjcox2000
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Re: Circuit Lab B/C

Postby mjcox2000 » February 22nd, 2019, 12:38 pm

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!
MIT ‘23
TJHSST ‘19
Longfellow MS
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Circuit Lab: 1st
Thermodynamics: 2nd
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2018:
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Cathy-TJ
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Re: Circuit Lab B/C

Postby Cathy-TJ » February 22nd, 2019, 5:19 pm

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.
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mjcox2000
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Re: Circuit Lab B/C

Postby mjcox2000 » February 22nd, 2019, 6:19 pm

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]
MIT ‘23
TJHSST ‘19
Longfellow MS
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:
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Cathy-TJ
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Re: Circuit Lab B/C

Postby Cathy-TJ » February 22nd, 2019, 7:12 pm

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]
That was my answer! You got the next question!
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Chemistry Lab, Circuit Lab, Disease Detectives

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Re: Circuit Lab B/C

Postby mjcox2000 » February 23rd, 2019, 8:17 pm

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.)
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Re: Circuit Lab B/C

Postby UTF-8 U+6211 U+662F » February 23rd, 2019, 9:47 pm

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.

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Re: Circuit Lab B/C

Postby mjcox2000 » February 24th, 2019, 6:10 am

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.
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Mission Possible: 6th

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Re: Circuit Lab B/C

Postby UTF-8 U+6211 U+662F » February 24th, 2019, 9:02 am

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.
Agh! Well that sucks.
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


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