Solving Circuit Equations with Mesh and Nodal Analysis

[chrizzilla]

Homework Statement

The circuit shown has two inputs, V_{s} and i_{s} and one output V_{o}. The output is related to the input by the equation V_{o}=ai_{s}+bV_{s} where a and b are constants to be determined. determine the values a and b by (a) writing and solving mesh equations and (b) writing and solving node equations.

Homework Equations

Kirchoff's Voltage Law (Sum of voltages in a closed loop = 0)
Kirchoff's Current Law (Sum of currents at a node = 0)
Ohm's Law (V=IR)

The Attempt at a Solution

Loop 1 (i_{1})
-V+96(i_{1}-i_{2})+120(i_{1}-i_{3})=0

Loop 2 (i_{2})
96(i_{2}-i_{1})+32i_{2}-V_{s}=0

Loop 3 (i_{3})
120(i_{3}-i_{2})+V_{s}+30i_{3}=0

Constraint
i_{1} = i_{s}

How am I to solve for anything? I have 5 equations and eight unknowns (V, V_{s}, i_{s}, i_{1}, i_{2}, i_{3}, a, b). I know these equations are right because my professor helped the class set up all of the homework problems on Monday. I just don't know how to find V. Please help!

 

[berkeman]

Homework Statement

The circuit shown has two inputs, V_{s} and i_{s} and one output V_{o}. The output is related to the input by the equation V_{o}=ai_{s}+bV_{s} where a and b are constants to be determined. determine the values a and b by (a) writing and solving mesh equations and (b) writing and solving node equations.

Homework Equations

Kirchoff's Voltage Law (Sum of voltages in a closed loop = 0)
Kirchoff's Current Law (Sum of currents at a node = 0)
Ohm's Law (V=IR)

The Attempt at a Solution

Loop 1 (i_{1})
-V+96(i_{1}-i_{2})+120(i_{1}-i_{3})=0

Loop 2 (i_{2})
96(i_{2}-i_{1})+32i_{2}-V_{s}=0

Loop 3 (i_{3})
120(i_{3}-i_{2})+V_{s}+30i_{3}=0

Constraint
i_{1} = i_{s}

How am I to solve for anything? I have 5 equations and eight unknowns (V, V_{s}, i_{s}, i_{1}, i_{2}, i_{3}, a, b). I know these equations are right because my professor helped the class set up all of the homework problems on Monday. I just don't know how to find V. Please help!

What do you get if you use KCL equations instead? ( it works either way )

 

[lorenb]

there's no way you'll get a real number from your equations. considering the only values given are resistances. The is probably why we are trying to get a solution that has some constants.(a and b). not really sure how to solve for a and b though

 

[HenzNett]

a hint for V: when you have 2+ independent sources, consider superposition.

also, you don't have 5 equations with 8 unknowns. Is, Vs are given. a and b is what eventually you are looking for, in terms of the other variables.

try it again, and you should have just the right amount of equations to help you solve for a and b.

 

[chrizzilla]

a hint for V: when you have 2+ independent sources, consider superposition.

also, you don't have 5 equations with 8 unknowns. Is, Vs are given. a and b is what eventually you are looking for, in terms of the other variables.

try it again, and you should have just the right amount of equations to help you solve for a and b.

Hmm, I am confused. I know that V0=30*I3 and Is=I1, but I don't know how to get V, so I can't actually get Vs.

 

[The Electrician]

Hmm, I am confused. I know that V0=30*I3 and Is=I1, but I don't know how to get V, so I can't actually get Vs.

You don't need to "get" Vs; Vs is a given. You need to "get" Vo, and it will be in terms of Vs and Is.

Get rid of that Loop 1 equation you have; this one:

-V+96(i_{1}-i_{2})+120(i_{1}-i_{3})=0

Your loop1 equation should be just the constraint equation, I1 = Is.

Take that together with your loop2 and loop3 equations and solve.

Your unknowns are I1, I2 and I3. Vs and Is are not unknowns; they are given. Just leave them in symbolic form and solve your 3 simultaneous equations.

You will get expressions for I1, I2 and I3. V will be given by: V = 32*I2 + 30*I3; Vo will be just Vo = 30 * I3

As a hint, and to help you have confidence in your work, the expression for I3 is:

I3 = \frac{120*Is -Vs}{150}

 

[chrizzilla]

You don't need to "get" Vs; Vs is a given. You need to "get" Vo, and it will be in terms of Vs and Is.

Get rid of that Loop 1 equation you have; this one:

-V+96(i_{1}-i_{2})+120(i_{1}-i_{3})=0

Your loop1 equation should be just the constraint equation, I1 = Is.

Take that together with your loop2 and loop3 equations and solve.

Your unknowns are I1, I2 and I3. Vs and Is are not unknowns; they are given. Just leave them in symbolic form and solve your 3 simultaneous equations.

You will get expressions for I1, I2 and I3. V will be given by: V = 32*I2 + 30*I3; Vo will be just Vo = 30 * I3

As a hint, and to help you have confidence in your work, the expression for I3 is:

I3 = \frac{120*Is -Vs}{150}

hmm, well, that is not what I have gotten. solving for I2 using the second equation, I get i_{2} = \frac{96*i_{s}+V_{s}}{128} and then using that in the third equation, I get i_{3} = \frac{960*i_{s}-V_{s}}{1920}, which is nowhere near what you got.

Also, everyone keeps saying Is and Vs are given. How do you figure? I don't see how that is possible!

 

[The Electrician]

hmm, well, that is not what I have gotten. solving for I2 using the second equation, I get i_{2} = \frac{96*i_{s}+V_{s}}{128} and then using that in the third equation, I get i_{3} = \frac{960*i_{s}-V_{s}}{1920}, which is nowhere near what you got.

You have for the third equation:

120(i_{3}-i_{2})+V_{s}+30i_{3}=0

It should be:

120(i_{3}-i_{1})+V_{s}+30i_{3}=0

Try solving with this corrected equation 3.

Also, everyone keeps saying Is and Vs are given. How do you figure? I don't see how that is possible!

I don't understand what the problem is. You've already treated Is and Vs as givens when you derived:

i_{2} = \frac{96*i_{s}+V_{s}}{128}

The unknowns in this problem are I1, I2 and I3; Is and Vs are assumed to be known. After you derive the expressions involving Is and Vs, you can substitute numeric values for them later if that should be needed.

When you learned algebra, you should have learned how to work with literal variables.

If you want to solve the general quadratic equation:

a x^2 + b x + c = 0

you understand that the unknown is x and the variables a, b and c are given, even though they're letters (literals) rather than numeric quantities.

This circuit problem is the same.

 

[chrizzilla]

You have for the third equation:

120(i_{3}-i_{2})+V_{s}+30i_{3}=0

It should be:

120(i_{3}-i_{1})+V_{s}+30i_{3}=0

Try solving with this corrected equation 3.

Oops, you're right, that's my bad.

I don't understand what the problem is. You've already treated Is and Vs as givens when you derived:

i_{2} = \frac{96*i_{s}+V_{s}}{128}

The unknowns in this problem are I1, I2 and I3; Is and Vs are assumed to be known. After you derive the expressions involving Is and Vs, you can substitute numeric values for them later if that should be needed.

When you learned algebra, you should have learned how to work with literal variables.

If you want to solve the general quadratic equation:

a x^2 + b x + c = 0

you understand that the unknown is x and the variables a, b and c are given, even though they're letters (literals) rather than numeric quantities.

This circuit problem is the same.

You will have to forgive me, I misunderstood what you meant by "given." I assumed that you were seeing that these were given as numbers somewhere.

using V0=30I3 and substituting for I3 I get
30*((120*Is -Vs)/(150))=a*i_{s}+b*V_{s}

or

(120i_{s}-V_{s})/5=a*i_{s}+b*V_{s}

At this point I still need to find the values of a and b, obviously, and I need a value for Is and Vs in order to do that. I can't see how this is accomplished with the information that I have.

 

[The Electrician]

a = 120/5 and b = -1/5

a is simply the coefficient of Is in your expression and b is the coefficient of Vs.

 

[chrizzilla]

a = 120/5 and b = -1/5

a is simply the coefficient of Is in your expression and b is the coefficient of Vs.

wow... that's correct. why did I not see that? I feel like an idiot now... Thank you so much good sir.