Voltage amplitudes across inductor and resistor

AI Thread Summary
The discussion focuses on calculating the voltage amplitudes across a 22.0 mH inductor and a 145.0 ohm resistor in a series circuit connected to an AC generator with a peak voltage of 1.2 kV at a frequency of 1250 Hz. The user initially struggled to find the current but eventually calculated the impedance and determined the RMS current to be approximately 3.77 A. Using this current, the voltage across the inductor was found to be 648.4 V and across the resistor 546.7 V, totaling approximately 1195.1 V, which is close to the source voltage of 1200 V. The discussion clarified that while the sum of the voltages does not equal the RMS voltage, the sum of their squares does match the square of the RMS voltage. Overall, the calculations and relationships between voltages in the circuit were explored and verified.
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Homework Statement


A series combination of a 22.0 mH inductor and a 145.0 ohm resistor are connected across the output terminal of an ac generator with peak voltage 1.2 kV. (a) At f= 1250 Hz, what are the voltage amplitudes across the inductor and across the resistor? (b) Do the voltage amplitudes add to give the source voltage? Explain. (c) Draw a phasor diagram to show the addition of the voltages.


Homework Equations


V_L = I X_L
V_R = IR
X_L = wL
Vrms = Vmax / (sqrt 2)


The Attempt at a Solution


Originally I attempted to solve for the voltages using V_L and V_R equations. I found X_L = 172 ohms. R was given. At this point I was stuck as I could not solve for current?
 
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How is the impedance, Z, related to R & XL ?
 
well Z= sqrt ( R^2 + X_L^2)
also Z= V/I
 
Get Z, then get IMAX and/or IRMS.

It's a series circuit, so what do you know about the instantaneous current?
 
z= sqrt( (145ohm^2)=(172 ohm)^2 = 225 ohm
given Vrms = Vmax / (sqrt 2) = 1200/ (sqrt2) = 849 Vrms
i'm going to assume Irms= Vrms/Z
so 849 Vrms/ 225 ohm = 3.77 A

back to the original equations gives
V_L= IX_L
= (3.11 A)(172 ohm) = 534 V
V_R=IR
= (3.11A)(145 ohm) = 451 V

?
 
sorry * that was supposed to be a plus not equals in line 1
 
Now, check. Does VL2 + VR2 = (VRMS)2 ?

You used 3.11A for I, rather than 3.77 A.
 
oh my math was actually wrong on that
V_L= 648.4 V
V_R = 546.7 V
V_L + V_R = 1200 V as expected but using your equation and squaring each gives 848 V which matches the Vrms. Which are they referring to when they ask if it matches the "source voltage" the Vrms of the peak V?
 
648.4 + 546.7 = 1195.1 ≠ 1200, although it's close. This is just a coincidence b/c VL ≈ VR.

Definitely, VL + VR ≠ VRMS --- these are all RMS voltages.But see if the sum of the squares equals the square of the RMS .
 
  • #10
yes the sum of the squares = 848V
I had calculated Vrms to be 849 V

also thank you so much for your help and explanations, its greatly appreciated!
 
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