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Q104PE

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 \(1.00\;k\Omega \)** **resistor, a \(150\;\mu H\)** **inductor, and a \(25.0\;nF\)** **capacitor. (a) Find the power factor at \(f = 7.50\;Hz\). (b) What is the phase angle at this frequency? (c) What is the average power at this frequency? (d) Find the average power at the circuit's resonant frequency.**

(a.)The power factor is** \(0.0011781\). **

(b.) The phase angle at this frequency is** \({89.9325^\circ }\)**.

(c.) The average power at this frequency is \(0.0011781{P_{\max }}\).

(d.) The average power at the circuit's resonant frequency is \(0.0011781{P_{\max }}\).

**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.**

**The electrical energy is absorbed by the resistor in the process of acting as a barrier to the flow of electricity by lowering the voltage, and it is released as heat.**

- The resistance value: \(1.00\;k\Omega \)
- The inductance value: \(150\;\mu H\)
- The capacitance value: \(25.0\;nF\)
- The frequency value: \(f = 7.5Hz\)

(a) We are aware that the power factor is expressed as:

\(\cos \phi = \frac{R}{Z}\)

This means we'll have to figure out the circuit's reactance, \(Z\). We already know that the reactance is expressed as

\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)

This means we'll need to figure out the capacitor and inductor's active resistances, which are provided as

\(\begin{aligned} {X_C} &= \frac{1}{{2\pi fC}}\\{X_L} &= 2\pi fL\end{aligned}\)

The frequency in our case is \(f = 7.50\;Hz\) which means we will have,

\(\begin{aligned} {X_C} &= \frac{1}{{2\pi \times 7.5{\rm{ }}Hz \times 80.0\;nF \times {{10}^{ - 6}}\;F}}\\ &= \frac{1}{{2\pi \times 120{\rm{ }}Hz \times 8.00 \times {{10}^{ - 6}}\;F}}\\ &= 848826\;\Omega \end{aligned}\)

\(\begin{aligned} {X_L} &= 2\pi \times 7.5\;Hz \times 100\;\mu H\left( {\frac{{{{10}^{ - 6}}\;H}}{{1\;\mu H}}} \right)\\ &= \;0.00471\;\Omega \end{aligned}\)

As a result, the reactance will be

\(\begin{aligned} Z &= \sqrt {{{1000}^2} + {{\left( {0.00471\;\Omega - 848826\;\Omega } \right)}^2}} \\ &= 848827\;\Omega \end{aligned}\)

As a result, the power factor will be

\(\begin{aligned} \cos \phi &= \frac{R}{Z}\\ &= \frac{{1000}}{{848827}}\\ &= 0.0011781\end{aligned}\)

Hence, the result is obtained as \(0.0011781\).

** **

(b)

We know the value of the power factor - let's call it \(k\) - and that it is the cosine of the desired angle. As a result, we can identify the latter as

\(\begin{aligned} k &= \cos \phi \\ \Rightarrow \phi &= \arccos {\rm{ }}k\end{aligned}\)

As a result, our angle will be in degrees.

\(\begin{aligned} \phi &= \arccos 0.0011781\\ &= {89.9325^\circ }.\end{aligned}\)

Therefore, the phase angle at this frequency is** \({89.9325^\circ }\)**.

c)

For this, we can expect the power to be

\(P = {U_{rms}}{I_{rms}}\)

However, we have no idea what the current or voltage values are. To conclude, one of these would be required. We can only predict that we will succeed.

\(\begin{aligned} {P_{rms{\rm{ }}}} &= {P_{\max }}\cos \phi \\ &= 0.0011781{P_{\max }}\end{aligned}\)

Therefore, the average power is \(0.0011781{P_{\max }}\).

d)

For this, we can expect the power to be

\(P = {U_{rms}}{I_{rms}}\)

However, we have no idea what the current or voltage values are. To conclude, one of these would be required. We can only predict that we will succeed.

\(\begin{aligned} {P_{rms{\rm{ }}}} &= {P_{\max }}\cos \phi \\ &= 0.0011781{P_{\max }}\end{aligned}\)

Hence, the average power is \(0.0011781{P_{\max }}\).

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