How to Calculate Impedance in an AC Circuit with Series Components?

In summary, the problem involves a series circuit with a 25-Ω resistance, a 30-mH inductance, and a 12-µF capacitor connected to a 90-V (rms) ac source with a frequency of 500 Hz. To solve the circuit, the impedance for each element must be calculated using the equations Z_L=j\omega L and Z_C=\frac{1}{j\omega C}. The angular frequency, \omega, can be found by setting it equal to 2\pi f. Once the impedances are calculated, the circuit can be solved to find the current, voltage across each element, phase angle, and power dissipated.
  • #1
9.8m*s^-2
1
0

Homework Statement



A 25-Ω resistance is connected in series with a 30-mH inductance and a 12-µF capacitor and are connected to a 90-V (rms) ac. If the frequency of the ac source is 500 Hz, calculate

(i) the current in the circuit.
(ii) the voltage across each element
(iii) the phase angle
(iv) the power dissipated in the circuit



Homework Equations





The Attempt at a Solution



I believe I have to begin this problem by calculating the impedance but I do not understand how to begin doing that.

Thank you.
 
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  • #2
[tex]X_L=\omega L[/tex]
[tex]X_C=\frac{1}{\omega C}[/tex]
[tex]Z= R + i(X_L - X_C)[/tex]
 
  • #3
Well Impedence for each element is given by the equations that xcvxcvvc gave except with one little adjustment:

[tex]Z_L=j\omega L[/tex]
[tex]Z_C=\frac{1}{j\omega C}[/tex]
(The ones he gave were reactance)

[tex]\omega[/tex] is the angular frequency which can be calculated by setting it equal to [tex]2\pi f[/tex]

From there you can solve the circuit.
 
  • #4
Also, just in case you don't see it:
[tex] j = \sqrt{-1}[/tex]
[tex]\frac{1}{j}= \frac{j}{j}\frac{1}{j} =\frac{j}{-1}=-j[/tex]

which is why that capacitive reactance is subtracted when forming the expression for impedance.
 

Related to How to Calculate Impedance in an AC Circuit with Series Components?

What is impedance calculation?

Impedance calculation is the process of determining the total opposition to an alternating current (AC) circuit. It takes into account the resistance, inductance, and capacitance of the circuit components and is represented by the symbol Z.

Why is impedance calculation important?

Impedance calculation is important because it helps engineers and scientists understand the behavior of AC circuits. It allows for the accurate prediction of currents and voltages in the circuit, which is necessary for designing and troubleshooting electrical systems.

How is impedance calculated?

Impedance is calculated using Ohm's law, where Z equals the square root of the sum of the squares of the resistance (R), inductance (L), and capacitance (C). The formula is Z = √(R² + (ωL)² + (1/ωC)²), where ω is the angular frequency of the AC current.

What are some common applications of impedance calculation?

Impedance calculation is commonly used in the design of electrical circuits and systems, such as power distribution networks, audio and radio frequency circuits, and telecommunications systems. It is also used in medical devices, such as electrocardiograms and impedance tomography, to measure the electrical properties of biological tissues.

What are some factors that can affect impedance calculation?

Some factors that can affect impedance calculation include the frequency of the AC current, the type of circuit components used, and the temperature of the circuit. Other factors, such as the length and thickness of conductors, can also impact impedance and must be taken into consideration during the calculation process.

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