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Chapter 14: Problem 2100

An inductor of inductance \(\mathrm{L}\) and resistor of resistance \(\mathrm{R}\) are joined in series and connected by a source of frequency 0 power dissipated in the circuit is, (a) $\left[\left\\{\mathrm{R}^{2}+\mathrm{c}^{2} \mathrm{~L}^{2}\right\\} / \mathrm{V}\right]$ (b) $\left[\left\\{\mathrm{V}^{2} \mathrm{R}\right\\} /\left\\{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}\right\\}\right]$ (c) $\left[\mathrm{V} /\left\\{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}\right\\}\right]$ (d) $\left[\sqrt{ \left.\left\\{R^{2}+\omega^{2} L^{2}\right\\} / V^{2}\right]}\right.$

Short Answer

Expert verified
The short answer is: \(P = \frac{V^2 R}{R^2 + ω^2 L^2}\)

Step by step solution

01

Determining the Impedance (Z) of the Circuit

For a series R-L circuit, the impedance can be calculated using the formula: \[Z = \sqrt{R^2 + (ωL)^2}\] Where, Z = impedance of the circuit, R = resistance of the resistor, ω = angular frequency, L = inductance of the inductor. Now we can plug in the given values for R, ω, and L to find the impedance Z of this circuit.
02

Finding the RMS Voltage (V_RMS) and Current (I_RMS)

The power dissipated in the circuit is given by the product of the RMS voltage and RMS current times the power factor (cos θ): \(P = V_{RMS}I_{RMS} \cos \theta\) Where, P = power dissipated, V_RMS = RMS voltage across the circuit, I_RMS = RMS current through the circuit, θ = phase angle between voltage and current, given by: \(\theta = \tan^{-1} \frac{ωL}{R}\), and the power factor is given by: \(\cos \theta = \frac{R}{Z}\)
03

Calculating the Power Dissipated in Terms of RMS Voltage and Current

To find the power dissipated in the circuit (P), we can rewrite the given equation: \(P = V_{RMS}I_{RMS}\frac{R}{Z}\) Since \(I_{RMS} = \frac{V_{RMS}}{Z}\) We can substitute the current in the equation: \(P = V_{RMS}\left(\frac{V_{RMS}}{Z}\right)\frac{R}{Z}\) Which simplifies to: \(P = \frac{V_{RMS}^2 R}{Z^2}\) Finally, we substitute the impedance formula from step 1: \(P = \frac{V_{RMS}^2 R}{(R^2 + (ωL)^2)}\) Considering the given options: (a) \(\frac{R^2 + ω^2 L^2}{V}\) (b) \(\frac{V^2 R}{R^2 + ω^2 L^2}\) (c) \(\frac{V}{R^2 + ω^2 L^2}\) (d) \(\sqrt{ \frac{R^2 + ω^2 L^2}{V^2}}\) Comparing the final formula with the given options, the correct option is (b): \(P = \frac{V^2 R}{R^2 + ω^2 L^2}\)

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