Voltage amplitudes across inductor and resistor

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Homework Help Overview

The discussion revolves around a problem involving a series circuit consisting of a 22.0 mH inductor and a 145.0 ohm resistor connected to an AC generator with a peak voltage of 1.2 kV. Participants are exploring the voltage amplitudes across the inductor and resistor at a frequency of 1250 Hz, as well as the relationship between these voltages and the source voltage.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of voltage amplitudes using the equations for inductive and resistive voltages. There are attempts to determine the impedance and its relationship to resistance and inductive reactance. Questions arise regarding the addition of voltages and their relation to the source voltage.

Discussion Status

Participants are actively engaging with the problem, attempting calculations and checking their work. Some have provided guidance on finding impedance and current, while others are verifying their results and questioning the relationship between the calculated voltages and the source voltage.

Contextual Notes

There are indications of confusion regarding the definitions of source voltage and RMS voltage, as well as discrepancies in calculations that are being addressed throughout the discussion.

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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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