Finding Vout/Vin of an RL filter

[stude]

I need to find V_out/V_in for an RL filter. The filter in question is set up as low pass filter with a resistor of R=9 ohms, a inductance of 76mH and angular frequency of 1.6ω_c.

To my knowledge V_in should be the vector sum of the voltages of each component so the equation would be V_in=√((IR)²+(IωL)²).

When trying to solve this problem I have set V_out as both the voltage of the inductor and the voltage over the resistor but neither of these has brought the correct answer

 

[NascentOxygen]

Hi stude. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Your formula for the magnitude of V_in is right. V_out is the voltage across R, i.e., between one end of the resistor and ground.

How could you recognize your answer as probably being correct, even if you weren't told the "correct" answer? (Remember, the variable here is ω.)

 

[stude]

I'm not sure what you're asking with the last line, it's online homework site so it tells me I'm wrong, I just don't know what to do to be right.

 

[NascentOxygen]

Maybe you aren't correctly expressing Vout/Vin? What are you writing? There won't be an "I" involved.

Or maybe they want the answer in dB? What exactly is the question the computer asks?

Quite possibly they are expecting you to specify the phase shift as well as magnitude change?

 

[stude]

It asks "what is the ratio between the output and input voltage amplitudes" and then "Vout/Vin=" followed by a blank to answer in

 

[ehild]

Show your work in detail. We can find your mistake if we see what you have done. Note that the frequency is 1.6 times the cut-off frequency. What does it mean? ehild

 

[stude]

V_out/V_in=V_r/√(V_r²+V_l²)

V_out/V_in=(IR)/√((IR)²+(IωL)²)

V_out/V_in=(I*9)/√((I*9)²+(I*1.6*ω_c*0.076)²)

ω_c=1/RC, I made the assumption that I could use 1/RL because my book doesn't actually address this kind of problem.

V_out/V_in=(I*9)/√((I*9)²+(I*1.6*1/(9*0.076)*0.076)²)

V_out/V_in=0.999

 

[ehild]

V_out/V_in=V_r/√(V_r²+V_l²)

V_out/V_in=(IR)/√((IR)²+(IωL)²)

V_out/V_in=(I*9)/√((I*9)²+(I*1.6*ω_c*0.076)²)

ω_c=1/RC, I made the assumption that I could use 1/RL because my book doesn't actually address this kind of problem.

That is wrong. At the critical frequency, the reactance is equal of the resistance. If it is an RC circuit, X_C=1/(ωC), and from X_C=R ω_c=1/(RC) follows.

If it is an RL circuit, X_L=ωL, from X_L=R, ω_c=R/L follows.

ehild

 

[stude]

That was the trick, thanks a lot

 

[ehild]

That was the trick, thanks a lot

You are welcome.


ehild