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Found in: Page 864

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

An RLC series circuit has a $200\Omega$ resistor and a $25.0mH$ inductor. At $8000Hz$, the phase angle is $45.{0}^{\circ }$ . (a) What is the impedance? (b) Find the circuit's capacitance. (c) If ${V}_{rms}=408V$ is applied, what is the average power supplied?

1. The impedance is $282.9\Omega$.
2. The circuit's capacitance two possible solutions are $18.8nF$ or $13.6nF$.
3. The average power supplied is $416.2W$.
See the 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 $R=200\Omega$

The capacitance of the RLC circuit is $C=25.0mF\left(\frac{{10}^{-3}F}{1mF}\right)=2.50×{10}^{-2}F$

The frequency of the circuit is $f=8000Hz$

The phase angle is $\varphi ={45}^{\circ }$

Step 2: Write the formula for the capacitor

The reactance of an RLC circuit can be expressed as,

$Z=\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}$ ………………(1)

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

${X}_{C}=\frac{1}{2\pi fC}$ ………………….(2)

${X}_{L}=2\pi L$ …………………..(3)

The phase angle can be expressed as,

$\mathrm{cos}\varphi =\frac{R}{Z}$ ………………….(4)

Step 2: Find the impedance(a)

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

$Z=\frac{R}{\mathrm{cos}\varphi }\phantom{\rule{0ex}{0ex}}=\frac{200\Omega }{0.707}\phantom{\rule{0ex}{0ex}}=282.9\Omega$

Therefore, the impedance is $282.9\Omega$

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

${\left({X}_{L}-{X}_{C}\right)}^{2}={Z}^{2}-{R}^{2}$

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

${X}_{L}-{X}_{C}=\sqrt{{Z}^{2}-{R}^{2}}$ …………………(5)

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

${X}_{L}=2\pi ×8000×\left(2.50×{10}^{-2}F\right)\phantom{\rule{0ex}{0ex}}=1256.6\Omega$

Therefore using the value of ${X}_{L}$in equation (5), we obtain that the value of capacitive reactance can be ${X}_{C1}=1057\Omega$or ${X}_{C2}=1457\Omega$is the capacitive reactance.

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

$C=\frac{1}{2\pi f{X}_{C}}$

Using the given data in the above expression we get,

For , ${X}_{C1}=1057\Omega ,\phantom{\rule{0ex}{0ex}}{C}_{1}=\frac{1}{2\pi ×8000×1057\Omega }\phantom{\rule{0ex}{0ex}}=18.8×{10}^{-9}F\left(\frac{1nF}{{10}^{-9}F}\right)\phantom{\rule{0ex}{0ex}}=18.8nF$

And for ${X}_{C2}=1457\Omega$, we have

${C}_{2}=\frac{1}{2\pi ×8000×1457\Omega }\phantom{\rule{0ex}{0ex}}=13.6×{10}^{-9}F\left(\frac{1nF}{{10}^{-9}F}\right)\phantom{\rule{0ex}{0ex}}=13.6nF$

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:

${P}_{avg}={V}_{rms}{I}_{rms}\phantom{\rule{0ex}{0ex}}=\frac{{V}_{rms}^{2}}{Z}$

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

${P}_{avg}=\frac{{\left(408V\right)}^{2}}{282.9\Omega }\phantom{\rule{0ex}{0ex}}=416.2W$

Therefore, the average power supplied is 416.2 W.

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