Circuit Lab B/C

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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mjcox2000
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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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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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Re: Circuit Lab B/C

Postby mjcox2000 » February 24th, 2019, 5:44 pm

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
That's right!
Instead of integrating for [math]V_{capacitor}(t)[/math], you can eliminate [math]R_s[/math] from the circuit since it impacts just the power loss and not the charging time. Then, [math]I[/math] is in parallel with [math]R_p[/math], and you can use Norton-Thevenin equivalence to see that the capacitor is simply charged by a voltage source [math]IR_p[/math] through resistance [math]R_p[/math], so you can apply a stock formula rather than integrating.

Your part b formula is right, but it can be algebraically simplified to [math]-\frac12 c \bigg[V_f (2 I R_p + V_f) + 2 I^2 R_p (R_p + R_s) ln\bigg(1 - \frac{V_f}{iR_p}\bigg)\bigg].[/math]

I haven't proved it, but intuition tells me there can only be one point at which the derivative of part b is 0. If current is under [math]V_f/R_p[/math], infinite energy is wasted because the target voltage is never reached. With current slightly above that threshold, it takes a very long time to charge, so a lot of energy is wasted in the series resistor. As current increases, time decreases, which should monotonically decrease the energy wasted in the series resistor. However, when current is very high, the parallel resistor wastes a lot of energy because it shorts around the capacitor. If I'm not mistaken, the limiting energy waste for small and large currents mean that there should be just one local minimum. (Plots I've done seem to confirm that.)
Your turn!
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Re: Circuit Lab B/C

Postby UTF-8 U+6211 U+662F » February 25th, 2019, 7:34 pm

Explain how you could solve for the resistance between two ends of a bridge circuit using Kirchoff's rules.

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Re: Circuit Lab B/C

Postby wec01 » February 26th, 2019, 3:07 pm

Explain how you could solve for the resistance between two ends of a bridge circuit using Kirchoff's rules.
I believe you could just use Kirchoff's loop rule to set up three equations involving different currents then use algebra to determine the total current. From there you could just use Ohm's law to solve for the total resistance.
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Re: Circuit Lab B/C

Postby UTF-8 U+6211 U+662F » February 26th, 2019, 4:11 pm

Explain how you could solve for the resistance between two ends of a bridge circuit using Kirchoff's rules.
I believe you could just use Kirchoff's loop rule to set up three equations involving different currents then use algebra to determine the total current. From there you could just use Ohm's law to solve for the total resistance.
Yep, your turn. Also mention source injection, as applying Kirchoff's laws is rather difficult without it.

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Re: Circuit Lab B/C

Postby wec01 » February 26th, 2019, 6:05 pm

Image

Calculate the values of I1, I2, and I3 given:

Image
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