Rms current in circuit with capacitor resistor and rms output

In summary: So, when I see the symbol j, I know to use the same mathematical operation as when I see the symbol i.
  • #1
a_ferret
3
0

Homework Statement



A 39.5 uF capacitor is connected to a 47.0 ohm resistor and a generator whose rms output is 25.2 V at 60.0 Hz. Calculate the rms current in the circuit.
(also asks for voltage drop across resistor, capacitor and the phase angle for the circuit, but I mostly want to get the first one first)

Homework Equations



I=current, V=voltage, R=resistance, f=frequency

Irms=1/sqrt(2)*Imax
Irms=delta Vc, rms/Xc
Xc=1/(2pi*f*c)
delta Vc,rms=Irms*Xc

and then I start going in circles

Other formulas that might be appropriate:
Vmax=Imax*Z
Z=sqrt(R^2+(Xl-Xc)^2)
Xl=2pi*f*L
L=?? not given in this problem
Power average=Irms^2*R

The Attempt at a Solution


combining lots of these formulas trying to come up with the correct Vrms has proven unsuccessful. I came up with 17.819 A, .37515 A, .21739 A, none of which were right. None of them were successful and I don't remember or care to explain how I got them precisely because I wasn't confident in those anyway.
Overall I still haven't been able to link my given information directly to the answer I need
 
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  • #2
Formulate the expression for Z, then you know that I = V/Z.
 
  • #3
but what do I use for L while setting it up for Z? I don't have it directly in the given information and I don't think I can derive it. Should I just use a default value like 1 or 0?

ps thanks for a response :D
 
  • #4
Calculate the impedance Z of the RC network.

[tex] Z = R + \frac{1}{j \omega C} [/tex]

(true for this particular problem)

[tex] Z = R +j \left( \frac{-1}{\omega C} \right)[/tex]

For this problem, the resistance is R, and the reactance is -1/(ωC).

In general,

[tex] Z = \Re \{ Z \} + j \Im \{ Z \} [/tex]

Use the Pythagorean Theorem to find the magnitude of Z.

[tex] |Z| = \sqrt{\left( \Re \{ Z \} \right)^2 + \left( \Im \{ Z \} \right)^2} [/tex]

And finally [the complex version of] Ohms law to find the current.

[tex] i_{RMS} = \frac{v_{RMS}}{|Z|} [/tex]
 
  • #5
What is j for those equations?
 
  • #6
a_ferret said:
What is j for those equations?
Here I used the symbol j to represent [itex] \sqrt{-1}[/itex]. This is common in electrical engineering courses, since the symbol i is already taken, representing current.
 

What is the definition of RMS current?

RMS current, also known as root mean square current, is a method of measuring the average current in an alternating current (AC) circuit. It takes into account both the magnitude and direction of the current, and is calculated by taking the square root of the mean of the squared values of the current over one period of the AC signal.

How does a capacitor affect the RMS current in a circuit?

A capacitor is an electrical component that stores and releases electrical energy. In a circuit with a capacitor and resistor, the capacitor will initially draw more current as it charges up, but as it reaches its full charge, the current will decrease. This results in a lower RMS current compared to a circuit without a capacitor, as the capacitor is able to store and release energy, reducing the overall current flow.

Why do we use RMS output rather than peak output?

RMS output is a more accurate representation of the power delivered by an AC signal compared to peak output. This is because RMS takes into account the varying values of the signal over time, whereas peak output only looks at the highest value of the signal. RMS is also the standard unit for measuring power in AC circuits.

What factors can affect the RMS current in a circuit?

The RMS current in a circuit can be affected by various factors such as the resistance and capacitance values, the frequency of the AC signal, and the overall impedance of the circuit. Additionally, any changes in the voltage or load in the circuit can also affect the RMS current.

How is RMS current different from average current?

Average current is simply the arithmetic mean of the values of the current over one period of an AC signal. This does not take into account the direction of the current and may not accurately represent the overall power in the circuit. RMS current, on the other hand, is a more precise measurement that considers the direction and magnitude of the current, providing a more accurate representation of the power in an AC circuit.

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