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A tricky one...by smunger81
Tags: tricky 
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#1
Oct405, 12:13 PM

P: 12

Here's a tricky one...
A resistor is in series with an inductor, with the current (I) lagging behind the voltage (V) by 60degrees. I am given the impedance (Z) of the circuit, which is 350ohms. How the heck can I find the resistance (R) and the inductive reactance (XL) when I know only the impedance and the phase angle. XL=2pifL Z=sq.root(R^2+(XLXC)^2 What equation would I use...since I don't know f, L, or C!?! Thanks! 


#2
Oct405, 12:45 PM

P: 254

Well you don't have a capacitor in your circuit to start off with so you don't actually need to know it, nor does it appear in the impedance expression. It usually helps to construct phasor diagrams in this situation. If you construct a phasor diagram you can see that the phase angle [tex]\phi[/tex] between the current and voltage is related by the expression:
tan([tex]\phi[/tex]) = [tex]X_L/R[/tex] From this you can express [tex]X_L[/tex] in terms of R or R in terms of [tex]X_L[/tex] and then solve the original impedance expression of: [tex]Z^2[/tex] = [tex]R^2[/tex] +[tex]X_L^2[/tex] 


#3
Oct405, 12:49 PM

P: 12

I knew that about the capacitor...I was unaware you could manipulate the impedance expression like that but it makes sense since there is no capacitor! Thank you so much for your help...so much.



#4
Oct405, 01:05 PM

P: 254

A tricky one...
Yeah the only reason it is [tex]X_LX_C[/tex] for the impedance in an RLC series circuit is because the voltage across the inductor leads the current by 90 degrees while the voltage across the capacitor lags the current by 90 degrees.
But phasors sometimes help to make these questions a lot easier. 


#5
Oct405, 02:05 PM

P: 12

Do you mean drawing a phasor diagram makes it easier? Could you explain that last statement a little more if you get time...thanks.



#6
Oct505, 09:41 AM

P: 254

Sorry yeah, what I meant was that for problems like that one it's easier to see how to answer a question if you use phasors. I've attached what I mean.



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