Hi
The question :
Find R3 and C3 which cause a Wien bridge with supply frequency of 2.5Hz to null
Assume R1=3.1kΩ, R2=25kΩ, R4=100kΩ and C1=5.2µF

Z2=25k ,Z4=100k
Z1=1/(jωc1)+3100
=3100-j12243

Z3=R3//C3
= R3/jωc3
(R3+1/jωc3)

Z3=(Z1*Z4)/Z2

R3/jωc3 =12400-j48972
(R3+1/jωc3)

The answers are 1.22 µF, 205 kΩ
I substitute the answer to the equation and the equation is work
But here are two unknowns in this equation. How can I find the R3 and C3 ?

Thanks

 

You can do complex math. But you can not apply Node-Voltage Method or Mesh-Current Method? Really?

 

Hi
The question :
Find R3 and C3 which cause a Wien bridge with supply frequency of 2.5Hz to null
Assume R1=3.1kΩ, R2=25kΩ, R4=100kΩ and C1=5.2µF

Z2=25k ,Z4=100k
Z1=1/(jωc1)+3100
=3100-j12243

Z3=R3//C3
= R3/jωc3
(R3+1/jωc3)

Z3=(Z1*Z4)/Z2

R3/jωc3 =12400-j48972
(R3+1/jωc3)

The answers are 1.22 µF, 205 kΩ
I substitute the answer to the equation and the equation is work
But here are two unknowns in this equation. How can I find the R3 and C3 ?

Thanks

Hi,

As mentioned already, you can use complex math.

One way is to calculate the transfer function for the output of the bridge, then set that equal to zero because at null the output will be zero. You can then try to separate R3 and C3 so that they both do not appear in any terms together, which may involve multiplying or dividing by one or the other and maybe some factoring. You then get the two equations by separating the real and imaginary parts and since they both must equal zero both parts get set equal to zero. You then have two equations in two unknowns so you can solve for R3 and C3 explicitly.

You can also try this by calculating the phase and amplitude and forming two equations that way, but they may be harder to solve because of the trig functions involved, or maybe even easier for that same reason.

 

Is complex math means this method?

1/(jωc+1/R3) =12400-j48972
arg [jωc+1/R3 ] =arg [1/(12400-j48972)]
tan^-1 (jωcR3 )= 75.79
cR3 = 2.481
I get this expression but it seems not work

You can do complex math. But you can not apply Node-Voltage Method or Mesh-Current Method? Really?

I don't know the voltage (E) so I apply Z3=(Z1*Z4)/Z2 to solve it

Hi,

As mentioned already, you can use complex math.

One way is to calculate the transfer function for the output of the bridge, then set that equal to zero because at null the output will be zero. You can then try to separate R3 and C3 so that they both do not appear in any terms together, which may involve multiplying or dividing by one or the other and maybe some factoring. You then get the two equations by separating the real and imaginary parts and since they both must equal zero both parts get set equal to zero. You then have two equations in two unknowns so you can solve for R3 and C3 explicitly.

You can also try this by calculating the phase and amplitude and forming two equations that way, but they may be harder to solve because of the trig functions involved, or maybe even easier for that same reason.

Do you mean Laplace transform ?
I think it may be more complicated and I haven't learned transfer function

 

Let's see how much you do know and at what point your comprehension needs to be improved. So let's take it one step at a time.

The stated goal is to null the bridge at that particular frequency.

Q1) What does it mean for the bridge to be nulled?

Q2) What will be the indication that this has been achieved?

 

Let's see how much you do know and at what point your comprehension needs to be improved. So let's take it one step at a time.

The stated goal is to null the bridge at that particular frequency.

Q1) What does it mean for the bridge to be nulled?

Q2) What will be the indication that this has been achieved?

1) change the frequency to 0 ?
2) the reactance of capacitor become zero ?

 

1) change the frequency to 0 ?

The first line of the question states, "...with supply frequency of 2.5Hz..." so how could the frequency be zero?

What is it that the detector measures? At the null point, what does the detector show?

 

Is complex math means this method?

1/(jωc+1/R3) =12400-j48972
arg [jωc+1/R3 ] =arg [1/(12400-j48972)]
tan^-1 (jωcR3 )= 75.79
cR3 = 2.481
I get this expression but it seems not work



I don't know the voltage (E) so I apply Z3=(Z1*Z4)/Z2 to solve it


Do you mean Laplace transform ?
I think it may be more complicated and I haven't learned transfer function

Hi,

Well you might have the right idea, but let me explain a little bit more.

First, you can just calculate the left and right side node voltages using the voltage divider formula. You can then subtract one from the other and set the result equal to zero because that's what happens when the bridge is balanced or 'nulled'. As WBahn suggested, do you understand what it means to null a bridge?

Assuming you get that far, you have two expressions, one for each side, and you subtract them:
a+j*b
c+j*d

then subtracting and set equal to zero:
(a+j*b)-(c+j*d)=0

Doing the subtraction we get:
(a-c)+j*(b-d)=0

Here we have real part a-c and imaginary part b-d and because the real part can not influence the imaginary part and the imaginary part can not influence the real part, it must mean that BOTH real and imaginary parts must be BOTH equal to zero, so we separate them and form two equations:
a-c=0
b-d=0

And there we would be able to solve for two variables because we have two equations, as long as they are linearly independent.
What happens is we often end up with a term that is not linearly independent in an equation like:
j*R3*C3+1=0

and that's not linearly independent, so we divide by say R3 and we get:
j*C3+1/R3=0

and now we are ready to use simultaneous equations to solve for R3 and C3, although we might instead try to solve for 1/R3 and then later compute the reciprocal.

The way you seem to have approached it was to use the phase angle and amplitude. When you do it that way you have to use the equation for the phase angle and the equation of the amplitude to solve for the two unknowns. We can try it that way too, but i think the complex number version works easier.

First though you probably answer the questions brought up by WBahn so we can find out where you are in your studies.
BTW no you dont really need Laplace Transforms, you can just use the expressions you seem familiar with like j*w*R*C or the like.

 

Here is the strategy I used solving for R3 & C3. At the null point, the voltage on either side of the bridge must be equal (so the detector reads zero). These voltages, VL & VR, can be found with the voltage divider rule, which then leads to the result that (at the null point) the ratio of the complex impedance equals the ratio of the resistance. So that gives one equation and two unknowns (R3 & C3) since everything else is a constant. The hard part (but not too hard if you have a symbolic algebra engine that does complex algebra) is to separate the complex impedance ratio into its real, F(R3,C3), and imaginary, G(R3,C3), parts. However, since the resistance ratio R2/R4 has no imaginary part, then G(R3,C3)=0 and F(R3,C3)=R2/R4. That is two equations & two variables, solve for R3 & C3.

 

1) change the frequency to 0 ?
2) the reactance of capacitor become zero ?

Okay, it's good we started where we did because your misconceptions go all the way back to here. No point even attempting to go forward until you understand what the problem is even about.

You can't arbitrarily choose the frequency because the4 problem explicitly states what the frequency is.

To null a bridge means to make adjustments to the circuit until the voltage across the bridging element is exactly zero (or, in practice, as close to zero as we can make it). So with that in mind, try answer both questions again and we'll see if we are ready to move to the next step.

 

Okay, it's good we started where we did because your misconceptions go all the way back to here. No point even attempting to go forward until you understand what the problem is even about.

You can't arbitrarily choose the frequency because the4 problem explicitly states what the frequency is.

To null a bridge means to make adjustments to the circuit until the voltage across the bridging element is exactly zero (or, in practice, as close to zero as we can make it). So with that in mind, try answer both questions again and we'll see if we are ready to move to the next step.

WBahn, I have a procedural question.
Can we assume that Detector is some kind of very high impedance device and replace it with open circuit?

 

WBahn, I have a procedural question.
Can we assume that Detector is some kind of very high impedance device and replace it with open circuit?

It actually doesn't matter what it is because if the bridge is nulled, then there is no voltage across it and, hence no current through it (unless it has zero impedance, which is unreasonable to assume). That's the beauty of null-voltage methods -- the finite resistance of the measuring device has no impact on the measurement. Historically, pretty low-resistance galvanometers were used because they tend to me much more sensitive to slight voltage differences across them and so you get a much more sensitive measurement (though it is much more twitchy, as well).

 

It actually doesn't matter what it is because if the bridge is nulled, then there is no voltage across it and, hence no current through it (unless it has zero impedance, which is unreasonable to assume). That's the beauty of null-voltage methods -- the finite resistance of the measuring device has no impact on the measurement. Historically, pretty low-resistance galvanometers were used because they tend to me much more sensitive to slight voltage differences across them and so you get a much more sensitive measurement (though it is much more twitchy, as well).

Ah. Ok. I asked my question before I read RBR's explanation of nulification condition to the exercise. Thank you for explaining and the historical example.

 

thanks for all the reply
I know where I went wrong now

 

This thread has been dead for ten days with no sign that the OP will ever post the solution. So in order to close the loop, here is a copy of a Maple worksheet that implements the given strategy for finding the null point values.

 

Why do we need Maple to solve this? Most people do not have Maple (or other similar package).

We know that the ratio of the two impedances in the bridge must match, which means:

\(
\frac{R_2}{R_4} \; = \; \frac{Z_1}{Z_3}
\)

where

\(
Z_1 \; = \; R_1 \, + \, \frac{1}{j \omega C_1}
Z_3 \; = \; \frac{R_3 \frac{1}{j \omega C_3}}{R_3 \, + \, \frac{1}{j \omega C_3}}
\)

In about four lines of simple algebra this reduces to

\(
j \omega C_1 \frac{R_2 R_3}{R_4} \; = \; 1 \; + \; j \omega \(R_1 C_1 \; + \; R_3 C_3 \) \; - \; \omega^2 R_1 C_1 R_3 C_3
\)

EDIT: correct the error pointed out by TheEngineer in which I had the difference of the two RC terms in the first equation above. If I had asked if the final result made sense, I would have caught this.

Now you just set real and imaginary terms equal to each other and you have:

\(
C_1 \frac{R_2 R_3}{R_4} \; = \; R_1 C_1 \; + \; R_3 C_3
1 \; = \; \omega^2 R_1 C_1 R_3 C_3
\)

We know everything except R3 and C3. From the second equation we have

\(
R_3 C_3 \; = \; \frac{1}{\omega^2 R_1 C_1}
\)

Plug that into the first equation and solve for R3

\(
C_1 \frac{R_2 R_3}{R_4} \; = \; R_1 C_1 \; + \; \frac{1}{\omega^2 R_1 C_1}
R_3 \; = \; \frac{R_4}{R_2 C_1} \( R_1 C_1 \; + \; \frac{1}{\omega^2 R_1 C_1} \)
R_3 \; = \; R_4 \( \frac{R_1}{R_2} \; + \; \frac{1}{\omega^2 R_1 R_2 C_1^2} \)
\)

Now you just substitute this in to the prior expression for R3·C3 and solve for C3.

 

When electronic calculators first became widely available in the 70's, instructors everywhere were aghast! Now students will be able to solve their problems without working or learning anything. I saw the answer to that non-problem--disallow calculators for a while during learning of basics, then give harder problems.

We don't need Maple (or similar packages) for the given problem, but this is the 21st century. We can use Maple and our powerful ubiquitous computers to solve much harder problems; problems that would be intractable otherwise.

To learn Maple, start out with simple problems and then move on to the really difficult ones. At first a person could solve simple problems by hand and compare to the result obtained from Maple. This way one becomes confident in their ability to use Maple.

I see LTSpice used a lot for even simple circuits. This gives a person lots of practice which begets proficiency with the tool.

Beginning students should be able to solve this problem by hand, but they should also be able to use modern software tools, because that's what they will be using on the job.

 

When electronic calculators first became widely available in the 70's, instructors everywhere were aghast! Now students will be able to solve their problems without working or learning anything. I saw the answer to that non-problem--disallow calculators for a while during learning of basics, then give harder problems.

I agree -- the problem is that this is seldom how it works in practice. Usually students are effectively ONLY shown how to let a tool do their thinking for them.

In this particular case, I think the use of something like Maple is extremely out of place. The TS clearly does NOT have anything approaching either the understanding of the concepts involved nor the mathematical literacy to tackle them on their own. So showing them how to use Maple to solve it just does exactly what should be avoided -- tries to sidestep a lack of understanding and skill and opt for letting a tool do the thinking for them.

I was fortunate and, to my thinking, went to high school in a narrow window in which this balance was well-struck -- not by planning, but by coincidence. Scientific calculators had just gotten cheap enough (~$35) for chemistry and physics courses to require them while the math classes were still entrenched in the no-calculator curriculum. The result is that we were firmly educated and skilled in the fundamentals while also comfortable using the available technology as a tool (and not a thinking-replacement). The class before mine was the last class to use slide rules at my high school (I always wished I had been in that class, but I covered that skill on my own a few years later). My class and the two that followed did everything by hand, using log and trig names where needed. The class after that had used four-function calculators starting in first grade and the math curriculum changed in lock step with them as they progressed through school.

Now many, many students understand logarithms literally only to the degree that they are shown which button on a calculator to press when they see something like ln(x). I ran across a trigonometry text about a decade ago that introduced the sine of an angle by showing which button to press on a TI-83 calculator when you wanted to evaluate sin(something).

 

Why do we need Maple to solve this? Most people do not have Maple (or other similar package).

We know that the ratio of the two impedances in the bridge must match, which means:

\(
\frac{R_2}{R_4} \; = \; \frac{Z_1}{Z_3}
\)

where

\(
Z_1 \; = \; R_1 \, + \, \frac{1}{j \omega C_1}
Z_3 \; = \; \frac{R_3 \frac{1}{j \omega C_3}}{R_3 \, + \, \frac{1}{j \omega C_3}}
\)

In about four lines of simple algebra this reduces to

\(
j \omega C_1 \frac{R_2 R_3}{R_4} \; = \; 1 \; + \; j \omega \(R_1 C_1 \; - \; R_3 C_3 \) \; - \; \omega^2 R_1 C_1 R_3 C_3
\)

By the way, using Maple (or similar package) would have avoided the error you made in your four lines of simple algebra (I think it's just a sign error).

That's one of the advantages to using what I sometimes see called a "math assistant". The silly algebra mistakes are avoided. Real world problems are usually bigger than this one and I very much appreciate the handing over of the drudgery of much algebra to my computer.

 

In this particular case, I think the use of something like Maple is extremely out of place. The TS clearly does NOT have anything approaching either the understanding of the concepts involved nor the mathematical literacy to tackle them on their own. So showing them how to use Maple to solve it just does exactly what should be avoided -- tries to sidestep a lack of understanding and skill and opt for letting a tool do the thinking for them.

I am fully with you in the respect that the powerful math tool shouldn't be used to substitute for understanding how to do the problem at all. Fortunately, when someone tries to use one of these modern math packages without understanding what they're doing, it's just GIGO.

You are correct about the TS's lack of understanding, and I think we did help him in that regard, so the Maple example was only for doing the algebra. The equating of real and imaginary parts still needed to be understood and carried out, either by hand or with a math package.

I sometimes show a solution to problems on this forum using Mathematica to show the student what can be done with modern tools, taking them to be an assistant, not a substitute for understanding, but hoping to inspire them to learn the modern tools.