RMS Current through RLC combination ciruit

[KokoDaSilvaback]

Homework Statement

Referencing the attached picture:

a) Show that the RMS current in the 1 kΩ resistor is 6.5mA.

b) If the AC voltage source was replaced by a battery, what would the current in the resistor be?

Homework Equations

V = IR
Z = √(R^2 + X^2)

The Attempt at a Solution

To begin, I calculated the impedance of the inductor and the capacitor, as well as the inductor and resistor. From these I found Z_L&C = 1528.72Ω, and Z_L&R = 843.56Ω.

Next (and this is where I think I may have gone wrong), I added the values of these impedances together to find Z_Total = 2372.28Ω.

Rearranging Ohm's Law, I find that I_RMS = 4.2 mA, which I assume can't be correct since I'm supposed to show that the current through the resistor is 6.5 mA. I assume I've made an error in calculating my impedance values, but after verifying with other online sources, I can't seem to find where my mistake is coming from. Any insight would be greatly appreciated.

 

[NascentOxygen]

I calculated the impedance of the inductor and the capacitor, as well as the inductor and resistor. From these I found Z_L&C = 1528.72Ω, and Z_L&R = 843.56Ω.

You need to preserve the real and imaginery components of these complex impedances before the two can be summed.

 

[KokoDaSilvaback]

You need to preserve the real and imaginery components of these complex impedances before the two can be summed.

So in that case would I treat the reactance (X_L and X_C) as entirely imaginary and the resistance as real? Once those are found, then calculate the impedance of the circuit?

 

[NascentOxygen]

So in that case would I treat the reactance (X_L and X_C) as entirely imaginary and the resistance as real? Once those are found, then calculate the impedance of the circuit?

Yes. Not only are the reactances imaginery, but X_C has a negative sign.

So what is the complex impedance of the R || L combination?

 

[KokoDaSilvaback]

Yes. Not only are the reactances imaginery, but X_C has a negative sign.

So what is the complex impedance of the R || L combination?

For a parallel circuit I have: $$Z = \frac {1} {\sqrt{(1/R)^2 + (1/jwl)^2}},$$ but when I perform that calculation, the inductance is no longer imaginary, as I have to use the complex conjugate in order to properly calculate the impedance of the parallel branch. The value I got from this calculation was 843.56

 

[NascentOxygen]

It's true that L appears in both the real and imaginary expressions of your Z. You have to keep Z as a complex number, to be able to add further series impedances to it.

 

[KokoDaSilvaback]

So rather than take the magnitude of the parallel impedance, simply add 1/R and 1/X_L, then add those values to the rest of the reactance values, leaving the impedance calculation for the final step?

 

[NascentOxygen]

Using the complex conjugate, express Z as the sum of a real component and an imaginary component.

 

[KokoDaSilvaback]

Ok, after rerunning the number and keeping my real portion separate from my imaginary portion, I end up with an impedance of 1528.72Ω, which when used to divide the 10V_RMS, gives me I_RMS = 6.54mA. That value closely matches the value sought, but do I now need to calculate the voltage drop across each component? Or is the resistor the only component that is considered since the inductor and capacitor behave differently due to the oscillating voltage?

 

[NascentOxygen]

I end up with an impedance of 1528.72Ω, which when used to divide the 10VRMS, gives me IRMS = 6.54mA. That value closely matches the value sought

That reads as though you have found the total current. The question asks for that current in the resistor, and not including that in its parallel inductance.

 

[KokoDaSilvaback]

The question asks for that current in the resistor, and not including that in its parallel inductance.

So I'm back to square one essentially? Since the current will split between the parallel branches, there is no way 6.5mA will go to the resistor.

 

[NascentOxygen]

So I'm back to square one essentially? Since the current will split between the parallel branches, there is no way 6.5mA will go to the resistor.

I can't say without knowing the relative impedances of R and L. So test this result before concluding you have gone wrong somewhere—because you might actually be correct.

 

[KokoDaSilvaback]

So, the impedance of the parallel portion is $(R^{-1}+ (jwl)^{-1})^{-1}$, correct? That preserves the real and imaginary components, which can then be added in series with the other impedance values.

 

[NascentOxygen]

So, the impedance of the parallel portion is $(R^{-1}+ (jwl)^{-1})^{-1}$, correct? That preserves the real and imaginary components, which can then be added in series with the other impedance values.

Yes, so long as you keep the $j$ in your expression for $Z$. But it won't save you any work whether you do the calculation in one step or 2 smaller steps because you need a complex conjugate in either case.