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Chapter 28: Problem 14

Much of Europe uses AC power at \(230 \mathrm{V}\) rms and \(50 \mathrm{Hz}\). Express this AC voltage in the form of Equation \(28.3,\) taking \(\phi_{V}=0\)

Short Answer

Expert verified
The AC voltage of \(230V\) rms and \(50Hz\) in Europe is expressed in the form of Equation 28.3 as \(V(t) = 325 \cos(100蟺t)\)

Step by step solution

01

Find the peak voltage \(V_m\)

From \(V_{rms} = V_m/\sqrt{2}\), we can solve for \(V_m\) as: \(V_m = V_{rms} \times \sqrt{2}\). Substituting \(V_{rms}\) as \(230 V\), this gives us \(V_m = 230 V \times \sqrt{2}\).
02

Calculate the actual value

Doing the multiplication, we find that \(V_m 鈮� 325 V\). This is the peak voltage.
03

Substitute into Equation 28.3

Equation 28.3 is \(V(t) = V_mcos(蠅t + 蠁)\) where \(V(t)\) is the instantaneous voltage, \(V_m\) is the peak voltage, \(蠅\) is the angular frequency, \(t\) is the time, and \(蠁\) is the phase angle. We are given that \(\phi_{V} = 0\) and \(蠅 = 2蟺f\) where \(f\) is the frequency. Substituting \(V_m 鈮� 325V\), \(蠅 = 2蟺 \times 50Hz\) and \(蠁 = 0\) into Equation 28.3, we get the final equation \(V(t) 鈮� 325cos(100蟺t)\).

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Most popular questions from this chapter

An RLC circuit includes a \(1.5-\mathrm{H}\) inductor and a \(250-\mu \mathrm{F}\) capacitor rated at 400 V. The circuit is connected across a sine-wave generator with \(V_{\mathrm{p}}=32 \mathrm{V} .\) What minimum resistance will ensure that the capacitor voltage does not exceed its rated value when the circuit is at resonance?

One-eighth of a cycle after the capacitor in an \(L C\) circuit is fully charged, what are the following as fractions of their peak values: (a) capacitor charge, (b) energy in the capacitor, (c) inductor current, (d) energy in the inductor?

The voltage across two components in series is zero. Is it possible that the voltages across the individual components aren't zero? Give an example.

Why is Equation 28.5 not a full description of the relation between voltage and current in a capacitor?

For \(R L C\) circuits in which the resistance isn't too high, the \(Q\) factor may be defined as the ratio of the resonant frequency to the difference between the two frequencies where the power dissipated in the circuit is half the power dissipated at resonance. Using suitable approximations, show that this definition leads to \(Q=\omega_{0} L / R,\) with \(\omega_{0}\) the resonant frequency.
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