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