Circuit Lab B/C

[mjcox2000]

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?

[Cathy-TJ]

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?

Answer (Click to Show Hidden Text)

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

[mjcox2000]

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?

Answer (Click to Show Hidden Text)

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!

[Cathy-TJ]

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.

[mjcox2000]

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.

Answer (Click to Show Hidden Text)

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]

[Cathy-TJ]

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.

Answer (Click to Show Hidden Text)

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!

[mjcox2000]

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.)

[UTF-8 U+6211 U+662F]

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.)

Not really sure if I'm on the right track but (Click to Show Hidden Text)

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.

[mjcox2000]

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.)

Not really sure if I'm on the right track but (Click to Show Hidden Text)

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.

You're on the right track, but (Click to Show Hidden Text)

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.

[UTF-8 U+6211 U+662F]

You're on the right track, but (Click to Show Hidden Text)

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.

Take Two (Click to Show Hidden Text)

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