Find the Capacitance based on 3 different switch positions

• Jaccobtw
In summary,You can find the impedance for each of the three positions with V/I. However I'm not sure how to find the inductive and capacitive reactance. Thanks for your help.
Jaccobtw
Homework Statement
An AC circuit contains a capacitor, an inductor, and two resistors of equal resistance as shown in the figure. The source provides 20V rms at 60Hz. When the switch is open as shown, the rms current is 183mA. When the switch is closed in position a, the rms current is 298mA. When the switch is closed in position b, the rms current is 137mA. What is the value of the capacitor in F?
Relevant Equations
$$X_C = \frac{1}{\omega C}$$
$$X_L = \omega L$$
$$Z = \sqrt{R^{2} + (X_L - X_C)^{2}}$$
$$I_0 = \frac{V_0}{Z}$$

You can find the impedance for each of the three positions with V/I. However I'm not sure how to find the inductive and capacitive reactance. Thanks for your help.

Yes, one step is as you say to take the quotient ##{\frac{V_{rms}}{I_{rms}}}## for each of the three cases, but then for each of the three cases you must calculate the impedance in terms of R,L and C. Equate the three expressions containing R,L,C with the three respective quotients and you ll have a system of 3 equations with 3 unknowns the R, L,C.
I can tell you that the impedance ##Z_S## when the switch is as shown is given by the 3rd equation of your OP, where ##X_L,X_C## as in first and second equations of your OP.
What is the impedance(always in terms of R,L,C) ##Z_a## when the switch is on a and what is the impedance ##Z_b## when the switch is on b? Answer these two questions first please.

Jaccobtw
Delta2 said:
What is the impedance(always in terms of R,L,C) ##Z_a## when the switch is on a and what is the impedance ##Z_b## when the switch is on b? Answer these two questions first please.
On a:
$$Z = \sqrt{(R/2)^{2} + (\omega L - \frac{1}{2C \omega})^{2}}$$

On b:

$$Z = \sqrt{(R)^{2} + (\omega L - \frac{1}{C \omega})^{2}}$$

For b the inductance is shorted. There is not potential drop on it.

Jaccobtw
nasu said:
For b the inductance is shorted. There is not potential drop on it.
So its zero?

The potential drop on it is zero. This is what "no potential drop" means. Or you mean to ask about something else?

Jaccobtw
nasu said:
The potential drop on it is zero. This is what "no potential drop" means. Or you mean to ask about something else?
The inductive reactance on B is zero because the potential across it is zero?

It's like the coil is not connected. It does not contribute to the total impedance. But this does not mean that it's reactance is zero. You cannot define a reactance for a coil not part of a circuit.

Jaccobtw
Jaccobtw said:
Homework Statement:: An AC circuit contains a capacitor, an inductor, and two resistors of equal resistance as shown in the figure. The source provides 20V rms at 60Hz. When the switch is open as shown, the rms current is 183mA. When the switch is closed in position a, the rms current is 298mA. When the switch is closed in position b, the rms current is 137mA. What is the value of the capacitor in F?
Relevant Equations:: $$X_C = \frac{1}{\omega C}$$
$$X_L = \omega L$$
$$Z = \sqrt{R^{2} + (X_L - X_C)^{2}}$$
$$I_0 = \frac{V_0}{Z}$$
View attachment 302005
You can find the impedance for each of the three positions with V/I. However I'm not sure how to find the inductive and capacitive reactance. Thanks for your help.
Frankly, I would try to find the value of ##R## first.

What is the difference in the circuit for the open position versus the switch being in position ##a## ?

alan123hk
Jaccobtw said:
On a:
$$Z = \sqrt{(R/2)^{2} + (\omega L - \frac{1}{2C \omega})^{2}}$$

On b:

$$Z = \sqrt{(R)^{2} + (\omega L - \frac{1}{C \omega})^{2}}$$
For a, why you say the reactance of capacitor is equal to ##\frac{1}{2C\omega}##? Does the capacitor change value to 2C?
For b what happens is that the coil is like is not present in the circuit, it is "shorted out" as we say in EE argo. Yes you are kind of right in :
Jaccobtw said:
The inductive reactance on B is zero because the potential across it is zero?
Because the coil is shorted out, the potential drop across it is zero (the same would happen if we had a capacitor there instead of a coil or even a resistor), so yes its like the inductive reactance of the circuit configuration in b is zero.

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Jaccobtw
SammyS said:
Frankly, I would try to find the value of R first.
Very clever way.

Jaccobtw said:
You can find the impedance for each of the three positions with V/I. However I'm not sure how to find the inductive and capacitive reactance. Thanks for your help.
You can try first writing the equations for the three different impedances.

## {\left(\frac{V}{I_0}\right)}^2 = {Z_0}^2##,
##{\left(\frac{V}{I_a}\right)}^2 = {Z_a}^2##,
## {\left(\frac{V}{I_b}\right)}^2 = {Z_b}^2##

At this point you will find it easy to first solve for ##R##, then ##X_c## and ##X_L##.

Delta2
SammyS said:
Frankly, I would try to find the value of R first.
Yes that might simplify the process of solving the system of equations, given that the system contains a term ##X_LX_C## which makes it non linear, otherwise it would be linear w.r.t ##R^2,L^2,\frac{1}{C^2} ##.

Jaccobtw
Delta2 said:
For a, why you say the reactance of capacitor is equal to ##\frac{1}{2C\omega}##? Does the capacitor change value to 2C?
I thought since the resistance is half the capacitance would double. My thoughts got mixed up

Delta2
Jaccobtw said:
I thought since the resistance is half the capacitance would double. My thoughts got mixed up
Well no the capacitance doesn't double, neither changes anything regarding the capacitor connection to the rest of the circuit.

Jaccobtw
Delta2 said:
Well no the capacitance doesn't double, neither changes anything regarding the capacitor connection to the rest of the circuit.
OK, so you know that the resistance in the circuit with the switch in position ##a## is ##\dfrac R 2 ## . And now, hopefully, you know that the capacitance and inductance are the same as for the switch being in the original position, which was Open.

When working with the equation: ##\quad\quad \displaystyle Z = \sqrt{R^{2} + (X_L - X_C)^{2}~}~,\quad## I strongly urge you to use it in the following form.

Use ##Z_0## and ##Z_a##, as suggested by the notation of @alan123hk in Post #11 . You should get a surprisingly simple result for ##\quad\displaystyle {Z_0}^2 - {Z_a}^2~,\quad##which you can then solve for ##R##.

Also, I suggest that you carry your solution all the way through solving for ##R## before plugging in any of the given values, although this may sound tedious.

When you reach the appropriate point, I strongly suggest solving for the reactance(s) ##X_C## and/or ##X_L## , waiting until after that to find the actual capacitance and inductance values.

Last edited:
alan123hk, Delta2 and Jaccobtw
SammyS said:
Use ##Z_0## and ##Z_a##, as suggested by the notation of @alan123hk in Post #11 . You should get a surprisingly simple result for ##\quad\displaystyle {Z_0}^2 - {Z_a}^2~,\quad##which you can then solve for ##R##.
Wait, why would you subtract the impedances?

Jaccobtw said:
Wait, why would you subtract the impedances?
Really?

Try it for yourself. We can't do it all for you.

The post you replied to says why.

SammyS said:
Really?

Try it for yourself. We can't do it all for you.

The post you replied to says why.
Thanks for your help. That took me an especially long time to figure out. It just wasn't intuitive for me to subtract the impedances so I equated them. That didn't work so I tried what you said.

Jaccobtw said:
Thanks for your help. That took me an especially long time to figure out. It just wasn't intuitive for me to subtract the impedances so I equated them. That didn't work so I tried what you said.
What value did you get for the resistance, R ?

SammyS said:
What value did you get for the resistance, R ?
Yeah and the capacitance

Jaccobtw said:
Yeah and the capacitance
Yeah ?

What were the values you got for those?

Delta2
SammyS said:
Yeah ?

What were the values you got for those?
99.6 ohms for resistance

alan123hk and Delta2
Jaccobtw said:
99.6 ohms for resistance
I got very nearly those same numbers.

Jaccobtw

1. What is capacitance and how is it related to switch positions?

Capacitance is the ability of a system to store an electrical charge. It is related to switch positions because the capacitance value of a system can change depending on the position of the switch, which can alter the electrical charge stored in the system.

2. How do you calculate capacitance based on 3 different switch positions?

To calculate capacitance based on 3 different switch positions, you will need to measure the electrical charge stored in the system for each position. Then, you can use the formula C = Q/V, where C is the capacitance, Q is the charge, and V is the voltage, to determine the capacitance for each switch position.

3. What are the units of capacitance?

Capacitance is measured in units of Farads (F). However, in most cases, the values of capacitance are expressed in smaller units such as microfarads (μF) or picofarads (pF).

4. How does capacitance affect the performance of a circuit?

The capacitance of a circuit can affect its performance in several ways. For example, a higher capacitance can cause a circuit to take longer to charge and discharge, which can impact its overall speed. It can also affect the stability and frequency response of the circuit.

5. Can capacitance be changed by adjusting switch positions?

Yes, the capacitance of a system can be changed by adjusting switch positions. This is because the position of the switch can alter the distance between the conductors, which can affect the capacitance value. Additionally, some switches may have built-in capacitors that can be activated by changing their positions.

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