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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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Most popular questions from this chapter

A solenoid of \(1.5 \mathrm{~m}\) long with inner diameter of \(4 \mathrm{~cm}\) has three layers of windings of 1000 turns each \& carries a current of $2 \mathrm{~A}$. The magnetic flux through a cross section of the solenoid is nearly..... (a) \(2.5 \times 10^{-7} \mathrm{wb}\) (b) \(6.31 \times 10^{-6} \mathrm{wb}\) (c) \(2.1 \times 10^{-6} \mathrm{wb}\) (d) \(4.1 \times 10^{-5} \mathrm{wb}\)

The power factor of an ac circuit having resistance \((\mathrm{R})\) and inductance (L) connected in series and an angular velocity w is, (a) \((\mathrm{R} / \mathrm{cL})\) (b) $\left[\mathrm{R} /\left\\{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}\right\\}^{(1 / 2)}\right]$ (c) \((\omega L / R)\) (d) $\left[\mathrm{R} /\left\\{\mathrm{R}^{2}-\omega^{2} \mathrm{~L}^{2}\right\\}^{(1 / 2)}\right]$

When 100 volt dc is applied across a coil, a current of \(1 \mathrm{~A}\) flows through it. When 100 volt ac at 50 cycle \(\mathrm{s}^{-1}\) is applied to the same coil, only \(0.5\) A current flows. The impedance of the coil is, (a) \(100 \Omega\) (b) \(200 \Omega\) (c) \(300 \Omega\) (d) \(400 \Omega\)

A transformer of efficiency \(90 \%\) draws an input power of \(4 \mathrm{~kW}\). An electrical appliance connected across the secondary draws a current of $6 \mathrm{~A}$. The impedance of device is ....... (a) \(60 \Omega\) (b) \(50 \Omega\) (c) \(80 \Omega\) (d) \(100 \Omega\)

In general in an alternating current circuit. (a) The average value of current is zero. (b) The average value of square of current is zero. (c) Average power dissipation is zero. (d) The phase difference between voltage and current is zero.
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