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Q105PE

Expert-verifiedFound in: Page 864

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?**

- The impedance is $282.9\Omega $.
- The circuit's capacitance two possible solutions are $18.8nF$ or $13.6nF$.
- The average power supplied is $416.2W$.

**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.**

The resistance of the RLC circuit is $R=200\Omega $

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

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

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

The reactance of an RLC circuit can be expressed as,

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

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 $

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

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

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 \times 8000\times \left(2.50\times {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 , ${X}_{C1}=1057\Omega ,\phantom{\rule{0ex}{0ex}}{C}_{1}=\frac{1}{2\pi \times 8000\times 1057\Omega}\phantom{\rule{0ex}{0ex}}=18.8\times {10}^{-9}F\left(\frac{1nF}{{10}^{-9}F}\right)\phantom{\rule{0ex}{0ex}}=18.8nF$

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

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

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

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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