RLC Circuit: Find Applied Source Voltage

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Homework Help Overview

The discussion revolves around finding the applied source voltage in an RLC circuit. Participants are exploring the implications of Kirchhoff's Voltage Law and the relationships between voltage, resistance, and reactance in the context of alternating current circuits.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to apply Kirchhoff's Voltage Law but are questioning whether they can simply add voltages across components due to phase differences. There are discussions about treating voltages as vectors and considering their magnitudes. Some participants express confusion about how to calculate voltages without given values for resistance and reactance.

Discussion Status

The discussion is active, with participants raising questions about the validity of their approaches and the lack of necessary values for calculations. Some guidance has been offered regarding the use of voltage dividers and vector representation, but there is no consensus on the correct method or outcome.

Contextual Notes

Participants note that specific values for resistance, reactance, and other circuit parameters are not provided, which complicates their ability to reach a definitive solution.

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Homework Statement



[PLAIN]http://img255.imageshack.us/img255/3048/56893213.jpg
Find the voltage of the applied source.

Homework Equations



I=V/R
Kirchoff's Voltage Law

The Attempt at a Solution



Applying KVL.
V= 40 + 50 +20
=110

Is this question that simple?
Or am I missing something?
 
Last edited by a moderator:
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The voltages across the capacitor and the inductor don't have the same phase, so you can't add them.
 
But the values of resistance and reactance aren't given for capacitor and inductor, so how should I calculate their voltages?
Or I just can't add them anyway? So how to reach the answer?
 
What if I took them as vectors?
VR = 40
VC = -20i
VL = 50i

V=40+30i
The magnitude of this is sqrt (402 + 302) = sqrt (1600 + 900) = sqrt(2500) = 50 volts
Is this correct?
 
You don't know what the direction of those vectors is.

Use the equation for a voltage divider

V_R = \frac {R} {X}

V_L = \frac {i \omega L } { X}

V_C = ...

where X is the total reactance of the circuit, a complex number that is a
function of \omega R, C and L

now compute the magnitude of V_L/V_R and V_C/V_R

this will give a relation between R and \omega L and \omega C

which you can substitute in the equation for V_R above.
 
But the values for R and \omega L and \omega C are not given.
 

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