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Q102PE

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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 $1.00k\Omega$ resistor, a $150\mu H$ inductor, and a $25.0nF$ capacitor. (a) Find the circuit's impedance at $500Hz$ . (b) Find the circuit's impedance at $7.50kHz$ . (c) If the voltage source has ${V}_{rms}=408V$ , what is ${I}_{rms}$ at each frequency? (d) What is the resonant frequency of the circuit? (e) What is ${I}_{rms}$ at resonance?

1. The circuit's impedance at $500Hz$is $12.74k\Omega$ .
2. The circuit's impedance at $5kHz$ is $1307.1\Omega$ .
3. The ${I}_{rms}$ at each frequency is $312mA$
4. The resonant frequency of the circuit is $82187Hz$ .
5. The ${I}_{rms}$ at resonance is$408mA$ .
See the step by step solution

Step 1: Definition of inductor and capacitor

A passive electrical component called an inductor stores energy in the form of a magnetic field. An inductor is a wire loop or coil in its most basic form. The coil's inductance is proportional to the number of turns it has.

A capacitor (also called a condenser) is a two-terminal passive electrical component that stores energy electrostatically in an electric field. Practical capacitors come in a variety of shapes and sizes, but they all have at least two electrical conductors (plates) separated by a dielectric (i.e., insulator).

Step 2: Given data

The resistance of the RLC circuit is $R=1.00k\Omega \left(\frac{{10}^{3}\Omega }{1k\Omega }\right)=1.00×{10}^{3}\Omega$

The capacitance of the RLC circuit is $C=25.0nF\left(\frac{{10}^{-9}F}{1nF}\right)=2.50×{10}^{-8}F$

The inductance of the RLC circuit is $L=150\mu \left(\frac{{10}^{-6}H}{1\mu }\right)=1.50×{10}^{-4}H$

Step 3: 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)

${X}_{L}=2\pi fL$ …………………..(3)

Step 4: Calculate the circuit's impedance(a)

We have${f}_{1}=500Hz$ , therefore substituting the given data in equations (2) and (3), we get,

role="math" localid="1654512552476" ${X}_{C}=\frac{1}{2\pi ×500Hz×2.50×{10}^{-8}F}\phantom{\rule{0ex}{0ex}}=12.74×{10}^{3}\Omega \left(\frac{1k\Omega }{{10}^{3}\Omega }\right)\phantom{\rule{0ex}{0ex}}=12.74k\Omega \phantom{\rule{0ex}{0ex}}{X}_{L}=2\pi ×500Hz×1.50×{10}^{-4}H\phantom{\rule{0ex}{0ex}}=0.314\Omega$

Now, we use the above-obtained values of XC and XL in equation (3), and we get

$Z=\sqrt{{\left(1.00×{10}^{3}\Omega \right)}^{2}+{\left(0.314\Omega -12.74×{10}^{3}\Omega \right)}^{2}}\phantom{\rule{0ex}{0ex}}=12.74×{10}^{3}\Omega \left(\frac{1k\Omega }{{10}^{3}\Omega }\right)\phantom{\rule{0ex}{0ex}}=12.74k\Omega$

Therefore, the circuit's impedance at $500Hz$ is $12.74k\Omega$.

Step 5: Find the circuit's impedance(b)

We have ${f}_{2}=7.50kHz\left(\frac{{10}^{3}Hz}{1kHz}\right)=7.50×{10}^{3}Hz$, therefore substituting the given data in equations (2) and (3), we get,

${X}_{C}=\frac{1}{2\pi ×7.50×{10}^{3}Hz×2.50×{10}^{-8}F}\phantom{\rule{0ex}{0ex}}=848.8\Omega \phantom{\rule{0ex}{0ex}}{X}_{L}=2\pi ×7.50×{10}^{3}Hz×1.50×{10}^{-4}H\phantom{\rule{0ex}{0ex}}=7.07\Omega$

Now, we use the above-obtained values of XC and XL in equation (3), and we get

$Z=\sqrt{{\left(1.00×{10}^{3}\Omega \right)}^{2}+{\left(7.07\Omega -848.8\Omega \right)}^{2}}\phantom{\rule{0ex}{0ex}}=1307.1\Omega$

Therefore, the circuit's impedance at $7.50kHz$ is $1307.1\Omega$.

Step 6: Find   at each frequency(c)

According to Ohm's law, the root-mean-square intensity may be calculated using the reactance and the rms potential difference such that,

${I}_{rms}=\frac{{V}_{rms}}{Z}$ …………….(4)

For the frequency 500 Hz , the value of impedance is $Z=12.74k\Omega$, therefore

${I}_{rms}=\frac{408V}{12.74k\Omega }\phantom{\rule{0ex}{0ex}}=32.1mA$

For the frequency 7.50 kHz , the value of impedance is $Z=1307.1\Omega$, therefore

${I}_{rms}=\frac{408V}{1307.1\Omega }\phantom{\rule{0ex}{0ex}}=312.2mA$

Therefore, the ${I}_{rms}$at 500 Hz is 32.1 mA and at 7.50 kHz is 312.2 mA .

Step 7: Find the resonant frequency of the circuit(d)

The resonant frequency can be calculated using the expression,

${f}_{0}=\frac{1}{2\pi \sqrt{LC}}$……………(5)

That indicates that in our case, the end outcome will be

${f}_{0}=\frac{1}{2\pi \sqrt{\left(1.50×{10}^{-4}H\right)×\left(2.50×{10}^{-8}F\right)}}\phantom{\rule{0ex}{0ex}}=82187Hz$

Therefore, the resonant frequency of the circuit is 82187 Hz .

Step 8: Find Irms at resonance (e)

Remember that the resonant frequency occurs when the capacitor and inductor reactances are equal. If we look at the total reactance formula, this suggests that our resistance will be the same as the resistor. To put it another way, the rms current will be

${I}_{rms}=\frac{408V}{1000\Omega }\phantom{\rule{0ex}{0ex}}=408mA$

Therefore, the ${I}_{rms}$ at resonance is 408 mA.