Help with AC Circuit containing a motor

In summary, the current through the motor is phasor-like and depends on the voltage and the impedance of the load. When a capacitor is added in parallel to the impedance, the current is minimized to 12 A.
  • #1
docsxp
10
0
Hello, this is not a homework question, but is a past year exam question which I'm stumped at. Any help will be appreciated.

1. Homework Statement

We're designing a current reduction system for an AC motor which is modeled as an inductor ##L##in series with a resistor ##R##. This motor is connected to 230##V_{rms}## at ##50hz##. An RMS meter is used to measure the current the motor takes. We need to find:
(1) Motor Impedance
(2) RMS current is measured as 15A. Voltage is ##v_a cos(\omega t)## and current is ##i_a cos(\omega t + \theta)##. Write down ##\omega##, ##v_a## and ##i_a##.
(3) Voltage is now ##v_a e^{j\omega t}##, write down current and hence obtain an equation that relates ##R## and ##L##.
(4) A variable capacitor is added in parallel that reduces current to a minimum of 12A when C = 0.1F. Write down the total impedance and find L.

Homework Equations


##V=IZ##
##Z_t = R + X_l##

The Attempt at a Solution


(1) ##Z = R + j\omega L##
(2)
##i_{peak} = \sqrt 2 \times 15 = 21.21 A##
##v_{peak} = \sqrt 2 \times 230 = 325.269 V##
##\omega = 2\times \pi \times 50hz##
(3) This is the part that I don't really get
Using ##V=IZ## we have
##i(t) = \frac{v_a}{\sqrt{R^2 + (j\omega L)^2}} cos(\omega t + \theta)## which I think can also be written as ##\frac{v_a}{\sqrt{R^2 + (j\omega L)^2}} e^{j\omega t + \theta}##
where ##\theta## would be ##tan^{-1} \frac{R}{\omega L}##

Is this the relationship they are looking for?

(4) Capacitor is added in parallel to existing impedance ##R+j \omega L##. Hence ##\frac{1}{Z} = \frac{1}{R + j\omega L} + j\omega C##

This is as far as I've gotten. Could someone guide me along please? Thank you.
 
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  • #2
Hi docsxp, Welcome to Physics Forums.

docsxp said:
Hello, this is not a homework question, but is a past year exam question which I'm stumped at. Any help will be appreciated.

1. Homework Statement

We're designing a current reduction system for an AC motor which is modeled as an inductor ##L##in series with a resistor ##R##. This motor is connected to 230##V_{rms}## at ##50hz##. An RMS meter is used to measure the current the motor takes. We need to find:
(1) Motor Impedance
(2) RMS current is measured as 15A. Voltage is ##v_a cos(\omega t)## and current is ##i_a cos(\omega t + \theta)##. Write down ##\omega##, ##v_a## and ##i_a##.
(3) Voltage is now ##v_a e^{j\omega t}##, write down current and hence obtain an equation that relates ##R## and ##L##.
(4) A variable capacitor is added in parallel that reduces current to a minimum of 12A when C = 0.1F. Write down the total impedance and find L.
Can you verify the value of that capacitance? 0.1F seems awfully large to place across a 230V mains. I'd expect it to pass a current of several thousand amperes! I suspect that the units are not Farads but perhaps milli-Farads (mF).

Homework Equations


##V=IZ##
##Z_t = R + X_l##

The Attempt at a Solution


(1) ##Z = R + j\omega L##
(2)
##i_{peak} = \sqrt 2 \times 15 = 21.21 A##
##v_{peak} = \sqrt 2 \times 230 = 325.269 V##
##\omega = 2\times \pi \times 50hz##
(3) This is the part that I don't really get
Using ##V=IZ## we have
##i(t) = \frac{v_a}{\sqrt{R^2 + (j\omega L)^2}} cos(\omega t + \theta)## which I think can also be written as ##\frac{v_a}{\sqrt{R^2 + (j\omega L)^2}} e^{j\omega t + \theta}##
where ##\theta## would be ##tan^{-1} \frac{R}{\omega L}##

Is this the relationship they are looking for?
I think that by expressing the voltage as ##v_a e^{j\omega t}## they mean to treat the voltage as a phasor (although I don't know why they chose to use the peak value rather than the RMS value).

In that case you can write ##I = \frac{v_a}{Z} = \frac{v_a}{R + j \omega L}##. Then taking magnitudes,

$$I_a = \frac{v_a}{\sqrt{R^2 + \omega^2 L^2}}$$

You can define a quantity ##X_o = \frac{v_a}{I_a}## (it'll come in handy later) and then write an expression relating R and L to it. Note that when you take the magnitude of the impedance you don't include the "j" in the imaginary term.
(4) Capacitor is added in parallel to existing impedance ##R+j \omega L##. Hence ##\frac{1}{Z} = \frac{1}{R + j\omega L} + j\omega C##

This is as far as I've gotten. Could someone guide me along please? Thank you.
You can proceed as before and write an expression for the magnitude of the current phasor with the capacitor in play. Then rearrange as before to find the magnitude of ##E/I## (you have both for the second case: 230 V and 12 A RMS. Use peak if you wish). Define ##X_1 = \frac{v_a}{I_1}## where ##I_1## is your "new" current of 12 A. You should find that you can make use of the ##X_o## expression defined above to simplify things a bit and eliminate the R variable.
 
  • #3
gneill said:
although I don't know why they chose to use the peak value rather than the RMS value

When you use ##|Z| e^{j\Phi}## notation do you not use ##V_{peak}##? Do you still use ##V_{rms}##?
 
  • #4
gneill said:
You can define a quantity Xo=vaIaX_o = \frac{v_a}{I_a} (it'll come in handy later) and then write an expression relating R and L to it

So I attempted the question again and I'm still confused. When you say that I have to use ##v_a## do you mean the magnitude or do I have to treat it as a phasor? If we're focusing on magnitude then,

##i_a = \frac{v_a}{\sqrt{R^2 + (\omega L)^2}}##
##21.21 = \frac{325.269}{\sqrt{R^2 + (\omega L)^2}}##

And for the capacitor added, the overall impedance of the circuit, in terms of magnitude becomes ##|Z| = \sqrt{R^2 + (\frac{1}{\omega C} - \omega L)^2}##

Am I correct? Because the phasors for capacitors and inductors point in the opposite directions.

When we define ##X_0## do we use phasor notation?
 
  • #5
docsxp said:
When you use ##|Z| e^{j\Phi}## notation do you not use ##V_{peak}##? Do you still use ##V_{rms}##?
When working with AC phasors the usual convention is to use the RMS values for AC voltage and current.

When RMS values are used, calculations of power (I*V) are straight forward.
 
  • #6
docsxp said:
So I attempted the question again and I'm still confused. When you say that I have to use ##v_a## do you mean the magnitude or do I have to treat it as a phasor? If we're focusing on magnitude then,

##i_a = \frac{v_a}{\sqrt{R^2 + (\omega L)^2}}##
##21.21 = \frac{325.269}{\sqrt{R^2 + (\omega L)^2}}##

And for the capacitor added, the overall impedance of the circuit, in terms of magnitude becomes ##|Z| = \sqrt{R^2 + (\frac{1}{\omega C} - \omega L)^2}##

Am I correct? Because the phasors for capacitors and inductors point in the opposite directions.

When we define ##X_0## do we use phasor notation?

When dealing with phasor values for calculation you use the magnitudes of the phasors along with their phase angles. Here you have measured values of current using an RMS current meter, so only the magnitudes are available to you.

Your formula for the overall impedance magnitude is not correct. Your formula would apply if the capacitor was in series with the R and L, but the capacitor in this case is connected in parallel with the RL branch. How do impedances in parallel combine?

The given values of voltage and current are phasor magnitudes. So ##X_o = 230 V / 15 A## would yield the magnitude of the impedance, that is, ##X_o = \sqrt{R^2 + \omega^2 L^2}##.
 
  • #7
gneill said:
How do impedances in parallel combine?
Like resistors in parallel. ##Z = \frac{1}{\frac{1}{j\omega L} + j\omega C}## But this will give rise to a very complex impedance.

Unless we leave it in ##\frac{1}{Z}## form.

I will try again with the corrected formula.
 
  • #8
gneill said:
When dealing with phasor values for calculation you use the magnitudes of the phasors along with their phase angles

gneill, thanks for sticking with me for this long. Anyway I did the question again.

Using the combined impedance ##Z = \frac{1}{\frac{1}{j\omega L} + j\omega C}## we can actually calculate after simplification. However this system will be most efficient when there is no complex reactance, that is to say, in the phasor diagram I want to achieve a phasor lying entirely in real. The resistor here already is in the real, but the inductor is introducing complex reactance.

Hence I did ##\frac{1}{j\omega C} = -j\omega L## Because I want Im(Z) = 0

##\frac{1}{\omega C} = \omega L##
##L = 0.000101H##

Is this the correct answer?
 
  • #9
No that doesn't work. You'll find that the real part of the impedance is influenced by the complex parts of the separate impedances and vice versa. Your impedance is given by:

$$ Z = \frac{1}{\frac{1}{R + j\omega L} + j\omega C} $$

Simplify it and take the magnitude. Note that the magnitude of this impedance will be equal to V/I for the 12 Amp case. Yes, it's a bit of a math grind...

You should see an opportunity to substitute in the Xo that was defined earlier, and R can be replaced using the same expression.
 

Related to Help with AC Circuit containing a motor

1. What is an AC circuit containing a motor?

An AC circuit containing a motor is an electrical circuit that uses alternating current (AC) to power a motor. The motor converts electrical energy into mechanical energy, causing it to move and perform a specific task.

2. How do I troubleshoot an AC circuit containing a motor?

To troubleshoot an AC circuit containing a motor, you should first check the power supply to ensure it is working properly. Then, check the connections and wiring to make sure they are secure and not damaged. You can also use a multimeter to test for continuity and voltage. If these steps do not identify the issue, it may be a problem with the motor itself and it may need to be replaced.

3. What are some common problems with AC circuits containing motors?

Some common problems with AC circuits containing motors include loose or damaged connections, faulty wiring, and motor overload. These issues can result in the motor not functioning properly or even burning out.

4. How can I prevent motor overload in an AC circuit?

To prevent motor overload in an AC circuit, you can use a motor controller or soft starter to regulate the amount of current going to the motor. It is also important to properly size the motor for the task it will be performing and to avoid running the motor for extended periods of time without breaks.

5. Can I use a DC motor in an AC circuit?

No, you cannot use a DC motor in an AC circuit. DC motors are designed to operate with direct current, while AC circuits use alternating current. Attempting to use a DC motor in an AC circuit can result in damage to the motor and can be dangerous.

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