What is the peak current for this circuit?

In summary, the RLC circuit has a sinusoidal voltage supplied to it at 632 kHz with a peak voltage of 748 kV; a 25 kΩ resistance; a 16 µF capacitance; and a 30 H inductance. The peak current for this circuit can be found by dividing the peak voltage by the impedance, which results in a peak current of 26 μA. However, there may be inconsistencies in the units used in the question, so it is best to try various permutations of units to arrive at the correct answer.
  • #1
jacksonwiley
17
0

Homework Statement



An RLC circuit has a sinusoidal voltage supplied to it at 632 kHz with a peak voltage of 748 kΩ; a 25 kΩ resistance; a 16 µF capacitance; and a 30 H inductance. What is the peak current for this circuit?
A) 30 μA
B) 26 μA
C) 6.3 μA
D) 11 μA



Homework Equations



Xc= 1/(2∏ƒC)
Z = √R^2 + (Xl-Xc)^2
Xl= 2∏ƒL


The Attempt at a Solution



Irms = Vrms / √R^2 + [(2∏ƒL) - (1/(2∏ƒC))]^2

when i plug everything in I am getting 4.7 x 10^-8 which is nowhere correct. I'm not sure if I'm doing something wrong in relation to the theory aspect when making the variable equation to plug everything into; but it seems right so I'm a little confused
 
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  • #2
We can't check your working when you don't show it.

But I note that the question asks for peak current, and you show a formula with RMS. This may not necessarily be the problem, though.

You assumed the given elements are all to be connected in series. Was this stated?
 
  • #3
NascentOxygen said:
We can't check your working when you don't show it.

But I note that the question asks for peak current, and you show a formula with RMS. This may not necessarily be the problem, though.

You assumed the given elements are all to be connected in series. Was this stated?

it doesn't say if its in series or parallel.
so i was just using the generalized RLC circuit equations.

(748,000)/ √(25000)^2 + [{2∏*(632,000)} - (1/(2∏*632,000* 16x10^-6))^2]
that's the work, its a lot to take in.
that gives the Irms .

i tried taking the voltage and using ohm's law but that also doesn't work so I'm really confused.
 
  • #4
jacksonwiley said:

Homework Statement



An RLC circuit has a sinusoidal voltage supplied to it at 632 kHz with a peak voltage of 748 kΩ; a 25 kΩ resistance; a 16 µF capacitance; and a 30 H inductance. What is the peak current for this circuit?
A) 30 μA
B) 26 μA
C) 6.3 μA
D) 11 μA



Homework Equations



Xc= 1/(2∏ƒC)
Z = √R^2 + (Xl-Xc)^2
Xl= 2∏ƒL


The Attempt at a Solution



Irms = Vrms / √R^2 + [(2∏ƒL) - (1/(2∏ƒC))]^2

when i plug everything in I am getting 4.7 x 10^-8 which is nowhere correct. I'm not sure if I'm doing something wrong in relation to the theory aspect when making the variable equation to plug everything into; but it seems right so I'm a little confused

Check the problem wording: The peak voltage can not be given in kΩ. Also, the inductance looks too large.

If you divide the peak voltage with the impedance, you get peak current. Dividing rms voltage with the impedance results in rms current.

ehild
 
  • #5
ehild said:
Check the problem wording: The peak voltage can not be given in kΩ. Also, the inductance looks too large.

If you divide the peak voltage with the impedance, you get peak current. Dividing rms voltage with the impedance results in rms current.

ehild

yeah i think that's just a mistake in the problem that I was given, because i double checked and it has voltage in ohms which is weird.

but the peak voltage/impedance works i think. thanks for the help!
 
  • #6
jacksonwiley said:
yeah i think that's just a mistake in the problem that I was given, because i double checked and it has voltage in ohms which is weird.

but the peak voltage/impedance works i think. thanks for the help!

aw man okay, i did that and got 26 but its not right. i don't know what I'm missing.
 
  • #7
OK, I missed the voltage being expressed in kΩ. In that case, you need a whole different set of obtuse formulae to solve it. :smile:

In other words, it's anyone's guess what the question should be. Do they mean V, or kV? And the same holds for the other data.

You could try working backwards from each answer, since there is consistency there in the units, and see what you end up with. Maybe try various permutations of units, e.g., Hz or kHz, H or mH, etc.

Good luck!
 

Related to What is the peak current for this circuit?

1. What is peak current?

Peak current refers to the maximum amount of electrical current that flows through a circuit at any given time. It is typically measured in units of amperes (A).

2. How is peak current different from average current?

Peak current is the maximum current that flows through a circuit, while average current is the average amount of current over a given period of time. Peak current is useful for determining the maximum load that a circuit can handle, while average current is more relevant for determining the overall power consumption of a circuit.

3. How is peak current calculated?

Peak current can be calculated by dividing the peak voltage by the total resistance of the circuit. This is known as Ohm's law, which states that current (I) is equal to voltage (V) divided by resistance (R), or I = V/R. However, in more complex circuits, peak current may need to be calculated using more advanced equations and techniques.

4. Why is it important to know the peak current of a circuit?

Knowing the peak current of a circuit is important for several reasons. It can help determine the maximum load that a circuit can handle without overheating or causing damage. It can also be used to select the right components and materials for a circuit, ensuring that they are able to withstand the maximum current flow. Additionally, understanding peak current can help with troubleshooting and identifying potential issues in a circuit.

5. How can peak current be measured?

Peak current can be measured using a specialized instrument called an ammeter, which is designed to measure electrical current. It is important to use the correct type of ammeter for the specific circuit being measured, as different ammeters have different ranges and capabilities. Additionally, peak current can also be calculated using mathematical equations and simulations.

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