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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Impedance Calculation
In an R-L circuit, the impedance is a crucial concept that represents the total opposition to the flow of alternating current. Both resistance and inductance contribute to this opposition. For a series R-L circuit, the impedance \(Z\) is calculated using the formula: \[Z = \sqrt{R^2 + (\omega L)^2}\] Here, \
  • \(R\) stands for the resistance of the resistor.
  • \(L\) is the inductance of the inductor.
  • \(\omega\) represents the angular frequency, which is related to the frequency of the source.
The value of impedance shows how much the circuit resists the alternating current due to both its resistive and inductive parts. A larger impedance means less current flows through the circuit for a given voltage. Understanding impedance helps in determining how different components affect the circuit's performance.
RMS Voltage and Current
When working with alternating current (AC) circuits like an R-L circuit, RMS (Root Mean Square) values are essential. RMS voltage \(V_{RMS}\) and RMS current \(I_{RMS}\) represent effective or equivalent DC values for power calculations. The RMS values help us to directly use AC voltages and currents in power formulas to evaluate how much power a load consumes or dissipates.Power in an R-L circuit is calculated using these RMS values. The formula is: \[P = V_{RMS}I_{RMS} \cos \theta\]
  • \(V_{RMS}\) is the effective AC voltage across the circuit.
  • \(I_{RMS}\) is the effective AC current through the circuit.
  • \(\cos \theta\) is the power factor, discussing later, representing the efficiency of the circuit in using the energy.
RMS values help in practical sizing and designing of circuit components for safe and efficient energy use in AC systems.
Phase Angle and Power Factor
The phase angle \(\theta\) in an R-L circuit measures the difference in phase between the voltage and current waveforms. It’s a significant aspect of AC analysis. You can calculate this angle using the formula: \[\theta = \tan^{-1} \left(\frac{\omega L}{R}\right)\] In this formula:
  • \(\omega L\) is the reactive part due to inductance.
  • \(R\) is the resistive part.
The power factor \(\cos \theta\) is crucial because it tells you how effectively a circuit uses power. It’s: \[\cos \theta = \frac{R}{Z}\] Power factor, ranging from 0 to 1, indicates efficiency:
  • A power factor close to 1 means most of the power is effectively used, reducing energy losses.
  • A lower power factor implies more energy loss as heat.
Understanding phase angle and power factor helps improve circuit efficiency, guiding in the selection of components and adjustments to reduce wasted energy.

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Most popular questions from this chapter

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.

In a LCR circuit capacitance is changed from \(\mathrm{C}\) to \(2 \mathrm{C}\). For the resonant frequency to remain unchanged, the inductance should be change from \(L\) to (a) \(4 \mathrm{~L}\) (b) \(2 \mathrm{~L}\) (c) \(L / 2\) (d) \(\mathrm{L} / 4\)

\(\mathrm{A}\) coil of inductance \(\mathrm{L}\) has an inductive reactance of \(\mathrm{X}_{\mathrm{L}}\) in an AC circuit in which the effective current is \(\mathrm{I}\). The coil is made from a super-conducting material and has no resistance. The rate at which power is dissipated in the coil is. (a) 0 (b) \(\mathrm{IX}_{\mathrm{L}}\) (c) \(I^{2} \mathrm{X}_{\mathrm{L}}\) (d) \(\mathrm{IX}^{2} \mathrm{~L}\)

The diagram shows a capacitor \(\mathrm{C}\) and resistor \(\mathrm{R}\) connected in series to an ac source. \(\mathrm{V}_{1}\) and \(\mathrm{V}_{2}\) are voltmeters and \(\mathrm{A}\) is an ammeter, consider the following statements.(a) Readings in \(\mathrm{A}\) and \(\mathrm{V}_{2}\) are always in phase. (b) Reading in \(\mathrm{V}_{1}\) is ahead in phase with reading in \(\mathrm{V}_{2}\). (c) Reading in \(\mathrm{A}\) and \(\mathrm{V}_{1}\) are always in phase. (d) Which of these statements are is correct (a) 1) only (b) 2) only (c) 1 ) and 2) only (d) 2 ) and 3) only

The instantaneous voltage through a device of impedance \(20 \Omega\) is \(\varepsilon=80 \sin 100 \pi t\). The effective value of the current is, (a) \(3 \mathrm{~A}\) (b) \(2.828 \mathrm{~A}\) (c) \(1.732 \mathrm{~A}\) (d) \(4 \mathrm{~A}\)

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