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Two alternating voltage generators produce e.m.f.s of the same amplitude \(E_{0}\) but with a phase difference of \(\pi / 3\). The resultant emf is a. \(E_{0} \sin (\omega t+\pi / 3)\) b. \(E_{0} \sin (\omega t+\pi / 6)\) c. \(\sqrt{3} E_{0} \sin (\omega t+\pi / 6)\) d. \(\sqrt{3} E_{0} \sin (\omega t+\pi / 2)\)

Short Answer

Expert verified
The correct answer is c: \(\sqrt{3} E_{0} \sin(\omega t+\pi/6)\).

Step by step solution

01

Express both EMFs Mathematically

The first EMF can be written as \(E_1 = E_0 \sin(\omega t)\) and the second EMF, considering a phase shift of \(\pi/3\), is \(E_2 = E_0 \sin(\omega t + \pi/3)\).
02

Apply the Sine Addition Formula

Using the sine addition formula: \(\sin(a + b) = \sin a \cos b + \cos a \sin b\), we find \(E_2 = E_0 (\sin \omega t \cdot \cos \pi/3 + \cos \omega t \cdot \sin \pi/3)\).
03

Substitute Numerical Values for Trigonometric Functions

Given \(\cos \pi/3 = 1/2\) and \(\sin \pi/3 = \sqrt{3}/2\), we substitute these into the formula: \(E_2 = E_0 (\sin \omega t \cdot 1/2 + \cos \omega t \cdot \sqrt{3}/2)\).
04

Combine the Two EMFs

The resultant EMF, \(E\), is the sum of \(E_1\) and \(E_2\): \(E = E_0 \sin(\omega t) + E_0/2 \sin(\omega t) + E_0 \sqrt{3}/2 \cos(\omega t)\).
05

Factor and Simplify the Expression

Factor out \(E_0\) from the expression: \(E = E_0(3/2 \sin \omega t + \sqrt{3}/2 \cos \omega t)\). This can be simplified using the form \(R \sin(\omega t + \alpha)\) where \(R = \sqrt{(3/2)^2 + (\sqrt{3}/2)^2} = \sqrt{3}\).
06

Determine Phase Angle

The phase angle \(\alpha\) is determined using \(\tan \alpha = \frac{\sqrt{3}/2}{3/2} = \frac{1}{\sqrt{3}}\), which corresponds to \(\alpha = \pi/6\). Thus, the expression simplifies to \(E = \sqrt{3}E_0 \sin(\omega t + \pi/6)\).

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Key Concepts

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

Alternating Voltage Generators
Alternating voltage generators are essential components in electrical engineering. They create alternating current (AC) by converting mechanical energy into electrical energy, using the principle of electromagnetic induction. In simple terms, as the generator's rotor spins within a magnetic field, it produces an alternating voltage or e.m.f (electromotive force). These devices are crucial in supplying power across various applications ranging from industrial machinery to household electronics.

This alternating nature means the voltage periodically reverses direction, following a sine wave pattern. Understanding this pattern is important because it helps in analyzing how these currents interact when more than one source is connected together, like in the case of the problem we are examining.

Multiple generators can be connected in parallel, and when they are, their outputs can add together creating a single resultant voltage. The phase of output from each generator is crucial and can lead to situations where they either aid or oppose one another, owing much to the concept of phase difference.
Phase Difference
Phase difference refers to the difference in phase angle between the voltages produced by alternating voltage generators. It is a significant concept in AC circuits since it affects how two waveforms align in time when combined.

Phase is measured in degrees or radians, and one complete cycle of a waveform corresponds to 360 degrees or 2π radians. A phase difference arises when there is a horizontal shift in the waveforms. For instance, in our exercise, two generators have a phase difference of π/3 radians (or 60 degrees).

The phase difference determines the constructive or destructive interference of the waves. Constructive interference occurs when the waves are in phase, leading to a higher resultant amplitude. In contrast, destructive interference takes place when the waves are out of phase, which can reduce or cancel the resultant amplitude.

Understanding phase difference is critical because it informs how the outputs from different power sources can be efficiently managed and optimized.
Sine Addition Formula
The sine addition formula is a trigonometric identity used to simplify calculations involving the sine of the sum of two angles. It’s expressed as: \[\sin(a + b) = \sin a \cos b + \cos a \sin b\] This identity is particularly useful in problems involving alternating voltages.

For the problem at hand, we used this formula to manage the phase shift in the second generator’s output relative to the first. By applying the sine addition formula, the expression for the second e.m.f can be expanded to every individual trigonometric component, making calculations simpler.

This step is essential as it allows us to break complex expressions into simpler components that are much easier to manipulate mathematically.

Once simplified, these components can be added accurately, letting us find the resultant voltage when the two e.m.f.s are combined, considering all phase differences and amplitudes. This shows how the sine addition formula is fundamental in solving AC circuit problems efficiently.

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Most popular questions from this chapter

An \(L C R\) circuit contains resistance of \(100 \Omega\) and a supple of \(200 \mathrm{~V}\) at 300 radian angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by \(60^{\circ}\). If on the other hand, only inductor is taken out the current leads by \(60^{\circ}\) with the applied voltage. The current flowing in the circuit is a. \(1 \mathrm{~A}\) b. \(1.5 \mathrm{~A}\) c. \(2 \mathrm{~A}\) d. \(2.5 \mathrm{~A}\)

A coil has an inductance of \(0.7 \mathrm{H}\) and is joined in series with a resistance of \(220 \Omega\). When an alternating e.m.f. of \(220 \mathrm{~V}\) at 50 cps is applied to it, then the wattless component of the current in the circuit is a. \(5 \mathrm{~A}\) b. \(0.5 \mathrm{~A}\) c. \(0.7 \mathrm{~A}\) d. \(7 \mathrm{~A}\)

An nc voltagc is represented by \(E=220 \sqrt{2} \cos (50 \pi) t\) How many times will the current becone zero in 1 s? a. 50 times b. 100 times c. 30 times d. 25 times

A capacitor of \(10 \mathrm{mF}\) and an inductor of \(1 \mathrm{H}\) are joined in series. An ac of \(50 \mathrm{~Hz}\) is applied to this combination. What is the impedance of the combination? a. \(28 / \pi \Omega\) b \(14 / \pi \Omega\) C. \(10 / \pi \Omega\) d. \(20 / \pi \Omega\)

When an AC source of e.m.f. \(e=E_{0} \sin (100 t)\) is connected across a circuit, the phase diffcrence between the c.m.f. \(e\) and the current \(i\) in the circuit is observed to be \(\pi / 4\) as shown in Fig. 10.54. If the circuit consists possibly only of \(R-C\) or \(R-C\) or \(L-C\) series, find the relationship between the two elements. a. \(R=1 \mathrm{k} \Omega, C=10 \mu \mathrm{F}\) b \(R=1 \mathrm{k} \Omega, C=1 \mu \mathrm{F}\) c \(R=1 \mathrm{k} \Omega, L=10 \mathrm{H}\) d \(R=1 k \Omega, L=1 \mathrm{H}\)

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