# RMS Current through RLC combination ciruit

• KokoDaSilvaback
In summary: 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.
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?

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 ZL&C = 1528.72Ω, and ZL&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 ZTotal = 2372.28Ω.

Rearranging Ohm's Law, I find that IRMS = 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.

I calculated the impedance of the inductor and the capacitor, as well as the inductor and resistor. From these I found ZL&C = 1528.72Ω, and ZL&R = 843.56Ω.
You need to preserve the real and imaginery components of these complex impedances before the two can be summed.

NascentOxygen said:
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 (XL and XC) as entirely imaginary and the resistance as real? Once those are found, then calculate the impedance of the circuit?

KokoDaSilvaback said:
So in that case would I treat the reactance (XL and XC) 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 XC has a negative sign.

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

NascentOxygen said:
Yes. Not only are the reactances imaginery, but XC 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

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.

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

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

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 10VRMS, gives me IRMS = 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?

KokoDaSilvaback said:
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.

NascentOxygen said:
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.

KokoDaSilvaback said:
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.

Last edited:
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.

KokoDaSilvaback said:
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.

## 1. What is RMS current and how is it different from regular current?

RMS (Root Mean Square) current is a measure of the average current in an AC circuit. Unlike regular current, which is measured as the peak value of the current, RMS current takes into account the fluctuation of the current over time.

## 2. How do you calculate RMS current through an RLC combination circuit?

To calculate RMS current through an RLC combination circuit, you can use the formula IRMS = VRMS/Z, where VRMS is the RMS voltage and Z is the total impedance of the circuit. Z can be calculated using the formula Z = √(R2 + (XL - XC)2), where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.

## 3. How does the RLC combination circuit affect the RMS current?

The RLC combination circuit can affect the RMS current in a few ways. The presence of resistance in the circuit will cause the RMS current to decrease, while the presence of inductance or capacitance can cause the RMS current to increase or decrease depending on the frequency of the AC signal. Additionally, the phase difference between the voltage and current in the circuit can also affect the RMS current.

## 4. Can the RMS current be higher than the peak current in an RLC combination circuit?

Yes, it is possible for the RMS current to be higher than the peak current in an RLC combination circuit. This can occur when the voltage and current are out of phase, meaning they do not peak at the same time. In this case, the RMS current will be higher than the peak current and can be calculated using the formula IRMS = Ipeak/√2.

## 5. How can the RMS current be used in practical applications?

The RMS current is an important measure in AC circuits, as it represents the average power being dissipated in the circuit. This makes it useful in designing and selecting components for a circuit, as well as for calculating the power consumption. It is also important in ensuring the safety of electrical systems, as it is used to determine the maximum current that a circuit can handle without overheating or causing damage.

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