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College Physics (Urone)
Found in: Page 864

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Short Answer

An RLC series circuit has a 200Ω resistor and a 25.0 mH inductor. At 8000 Hz, the phase angle is 45.0 . (a) What is the impedance? (b) Find the circuit's capacitance. (c) If Vrms=408 V is applied, what is the average power supplied?

  1. The impedance is 282.9Ω.
  2. The circuit's capacitance two possible solutions are 18.8 nF or 13.6 nF.
  3. The average power supplied is 416.2 W.
See the step by step solution

Step by Step Solution

Step 1: Concept introduction

In an electronic circuit, a resistor is an electrical component that controls or regulates the passage of electrical current. Resistors can also be used to supply a set voltage to an active device like a transistor.

Step 2: Given data

The resistance of the RLC circuit is R=200 Ω

The capacitance of the RLC circuit is C=25.0 mF10-3F1 mF=2.50×10-2F

The frequency of the circuit is f=8000 Hz

The phase angle is ϕ=45

Step 2: Write the formula for the capacitor

The reactance of an RLC circuit can be expressed as,

Z=R2+(XL-XC)2 ………………(1)

This means we'll need to figure out the capacitor and inductor's active resistances, which can be given as,

XC=12πfC ………………….(2)

XL=2πL …………………..(3)

The phase angle can be expressed as,

cosϕ=RZ ………………….(4)

Step 2: Find the impedance(a)

Using the given data and rearranging equation (4), we get,

Z=Rcosϕ =200Ω0.707 =282.9Ω

Therefore, the impedance is 282.9 Ω

Step 3: Calculate the circuit's capacitance(b)

We may rearrange the concept of impedance to determine the capacitive reactance from equation (1), we get


This means that the difference between inductive and capacitive reactances will have an absolute value of

XL-XC=Z2-R2 …………………(5)

The inductive reactance can be found using equation (3), such that

XL=2π×8000×2.50×10-2F =1256.6Ω

Therefore using the value of XLin equation (5), we obtain that the value of capacitive reactance can be XC1=1057Ωor XC2=1457 Ωis the capacitive reactance.

In these situations, we can look for capacity as follows using equation (2), such that,


Using the given data in the above expression we get,

For , XC1=1057Ω,C1=12π×8000×1057Ω =18.8×10-9F1nF10-9F =18.8 nF

And for XC2=1457Ω, we have

C2=12π×8000×1457Ω =13.6×10-9F1nF10-9F =13.6 nF

Therefore, regarding the circuit's capacitance, two possible solutions are 18.8 nF 13.6 nF.

Step 4: Find the average power supplied(d)

The average power will be calculated by using the formula:

Pavg=VrmsIrms =Vrms2Z

Substituting the given data in the above equation, we get,

Pavg=408 V2282.9Ω =416.2 W

Therefore, the average power supplied is 416.2 W.


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