# Voltage amplitudes across inductor and resistor

## 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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SammyS
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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

SammyS
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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

SammyS
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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?

SammyS
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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 .

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!