## Circuit Lab B/C

mjcox2000
Member Posts: 120
Joined: May 9th, 2014, 3:34 am
State: VA

### Re: Circuit Lab B/C

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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Cathy-TJ
Member Posts: 28
Joined: January 20th, 2019, 7:13 pm
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State: VA

### Re: Circuit Lab B/C

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
Member Posts: 120
Joined: May 9th, 2014, 3:34 am
State: VA

### Re: Circuit Lab B/C

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
2019:
Circuit Lab: 1st
Thermodynamics: 2nd
Sounds of Music: 5th
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Cathy-TJ
Member Posts: 28
Joined: January 20th, 2019, 7:13 pm
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### Re: Circuit Lab B/C

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
Member Posts: 120
Joined: May 9th, 2014, 3:34 am
State: VA

### Re: Circuit Lab B/C

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:
WIDI: 2nd
Simple Machines: 3rd

Cathy-TJ
Member Posts: 28
Joined: January 20th, 2019, 7:13 pm
Division: C
State: VA

### Re: Circuit Lab B/C

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!
Nationals Placings:
2019 Circuit Lab - 1
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Chemistry Lab, Circuit Lab, Disease Detectives

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mjcox2000
Member Posts: 120
Joined: May 9th, 2014, 3:34 am
State: VA

### Re: Circuit Lab B/C

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

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)

$Q=CV$

$I_C=C \cdot \frac{dV}{dt}$

$I-VR_p = C \cdot \frac{dV}{dt}$

$dt = C \cdot \frac{dV}{I-VR_p}$

$t + C_{int} = C \cdot -\frac{\ln(I-VR_p)}{R_p}$ (the constant of integration also being a C makes this slightly confusing)

$t = \frac{-C\ln(I-VR_p)}{R_p} + k$ (so I changed it to a k)

$t = \frac{-C\ln(I-VR_p) + C\ln I}{R_p}$ (but plugging in (0,0) gives us a value for k)

$\boxed{t_f = \frac{-C\ln(I-V_fR_p) + C\ln I}{R_p}}$

b) I'm going to assume you mean energy for the next problem?

$P = I^2R_s + \frac{V^2}{R_p}$

$E = I^2R_s \frac{-C\ln(I-V_fR_p) + C\ln I}{R_p} + \int\frac{V^2}{R_p}dt$

Noting that $\frac{dV}{dt} = \frac{I-VR_p}{C}$, so $dt = \frac{CdV}{I-VR_p}$

$\boxed{E = I^2R_s \frac{-C\ln(I-V_fR_p) + C\ln I}{R_p} + \int^{\frac{-C\ln(I-V_fR_p) + C\ln I}{R_p}}_0\frac{CV^2}{IR_p-VR^2_p}dV}$

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.

mjcox2000
Member Posts: 120
Joined: May 9th, 2014, 3:34 am
State: VA

### Re: Circuit Lab B/C

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)

$Q=CV$

$I_C=C \cdot \frac{dV}{dt}$

$I-VR_p = C \cdot \frac{dV}{dt}$

$dt = C \cdot \frac{dV}{I-VR_p}$

$t + C_{int} = C \cdot -\frac{\ln(I-VR_p)}{R_p}$ (the constant of integration also being a C makes this slightly confusing)

$t = \frac{-C\ln(I-VR_p)}{R_p} + k$ (so I changed it to a k)

$t = \frac{-C\ln(I-VR_p) + C\ln I}{R_p}$ (but plugging in (0,0) gives us a value for k)

$\boxed{t_f = \frac{-C\ln(I-V_fR_p) + C\ln I}{R_p}}$

b) I'm going to assume you mean energy for the next problem?

$P = I^2R_s + \frac{V^2}{R_p}$

$E = I^2R_s \frac{-C\ln(I-V_fR_p) + C\ln I}{R_p} + \int\frac{V^2}{R_p}dt$

Noting that $\frac{dV}{dt} = \frac{I-VR_p}{C}$, so $dt = \frac{CdV}{I-VR_p}$

$\boxed{E = I^2R_s \frac{-C\ln(I-V_fR_p) + C\ln I}{R_p} + \int^{\frac{-C\ln(I-V_fR_p) + C\ln I}{R_p}}_0\frac{CV^2}{IR_p-VR^2_p}dV}$

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 $I-\frac{V}{R_p}$, not $I-VR_p$, 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.
MIT ‘23
TJHSST ‘19
Longfellow MS
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### Re: Circuit Lab B/C

your third equation should have $I-\frac{V}{R_p}$, not $I-VR_p$, 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)

$Q=CV$

$I_C=C \cdot \frac{dV}{dt}$

$I-\frac{V}{R_p} = C \cdot \frac{dV}{dt}$

$dt = C \cdot \frac{dV}{I-\frac{V}{R_p}}$

$t + C_{int} = -CR_p\ln(I-\frac{V}{R_p})$ (the constant of integration also being a C makes this slightly confusing)

$t = -CR_p\ln(I-\frac{V}{R_p}) + k$ (so I changed it to a k)

$t = -CR_p\ln(I-\frac{V}{R_p}) + CR_p\ln I$ (but plugging in (0,0) gives us a value for k)

$\boxed{t_f = -CR_p\ln(I-\frac{V_f}{R_p}) + CR_p\ln I}$

b) I'm going to assume you mean energy for the next problem?

$P = I^2R_s + \frac{V^2}{R_p}$

$E = I^2R_s \cdot (-CR_p\ln(I-\frac{V_f}{R_p}) + CR_p\ln I) + \int^{t_f}_0\frac{V^2}{R_p}dt$

Now, let's solve for V in terms of t...

$t = -CR_p\ln(I-\frac{V}{R_p}) + CR_p\ln I$

$\frac{CR_p\ln I - t}{CR_p} = \ln(I-\frac{V}{R_p})$

$\ln I - \frac{t}{CR_p} = \ln(I-\frac{V}{R_p})$

$Ie^{-\frac{t}{CR_p}} = I - \frac{V}{R_p}$

$V = IR_p(1-e^{-\frac{t}{CR_p}})$

$E = CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + \int^{t_f}_0\frac{(IR_p(1-e^{-\frac{t}{CR_p}}))^2}{R_p}dt$

$E = CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + I^2R_p\int^{t_f}_0(1-e^{-\frac{t}{CR_p}})^2dt$

$E = CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + I^2R_p\left[t+\frac{e^{-2\frac{t}{CR_p}}-4e^{{-\frac{t}{CR_p}}}}{-\frac{2}{CR_p}}\right]^{t_f}_0$

$\boxed{E = CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + I^2R_p\left(-CR_p\ln(I-\frac{V_f}{R_p}) + CR_p\ln I+\frac{\frac{(I-\frac{V_f}{R_p})^2}{I^2}-4\frac{I-\frac{V_f}{R_p}}{I}}{-\frac{2}{CR_p}}-\frac{3CR_p}{2}\right)}$

For part c:

$\boxed{0 = \frac{d}{dV}\left[CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + I^2R_p\left(-CR_p\ln(I-\frac{V_f}{R_p}) + CR_p\ln I+\frac{\frac{(I-\frac{V_f}{R_p})^2}{I^2}-4\frac{I-\frac{V_f}{R_p}}{I}}{-\frac{2}{CR_p}}-\frac{3CR_p}{2}\right)\right]}$

Although you'd need to test all of the roots and see which one yields the lowest value