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Chapter 21: Problem 2

A European outlet supplies \(220 \mathrm{V}\) (rms) at \(50 \mathrm{Hz}\). How many times per second is the magnitude of the voltage equal to $220 \mathrm{V} ?$

Short Answer

Expert verified
Answer: The magnitude of the voltage reaches 220V 100 times per second.

Step by step solution

01

Express the voltage as a function of time

We know that the voltage of a sinusoidal wave can be represented as: $$v(t) = V_m \cdot \sin(\omega t)$$ Where \(V_m\) is the peak voltage, \(\omega\) is the angular frequency and \(t\) is the time. To find the peak voltage, we can use the formula: $$V_\text{rms} = \frac{V_m}{\sqrt{2}}$$ Given \(V_\text{rms} = 220V\), we can find the peak voltage as follows: $$V_m = V_\text{rms} \cdot \sqrt{2} = 220V \cdot \sqrt{2}$$ And the angular frequency can be found using: $$\omega = 2\pi f$$ Where \(f\) is the frequency. For this problem, \(f=50Hz\). This gives us: $$\omega = 2\pi \cdot 50Hz$$ Now we can write our sinusoidal voltage function: $$v(t) = 220\sqrt{2} \cdot \sin(100\pi t)$$
02

Solve the equation for when the magnitude is 220V

We need to find the points in time when the magnitude of \(v(t)\) equals the given voltage magnitude of \(220V\). We can set up the equation as follows: $$|220\sqrt{2} \cdot \sin(100\pi t)| = 220V$$ To solve the equation, we can divide both sides by \(220V\): $$|\sin(100\pi t)| = \frac{1}{\sqrt{2}}$$ Since sine has a maximum value of 1, it can achieve a value of \(\frac{1}{\sqrt{2}}\) twice in each period. Therefore, the value of \(220V\) is reached twice in each cycle.
03

Find the total number of times \(220V\) is reached per second

We know that the frequency is \(50Hz\), meaning there are 50 cycles of the signal in a second. Since the voltage reaches \(220V\) twice in one cycle, it will do so \(50 \times 2 = 100\) times in a second. So the magnitude of the voltage is equal to \(220V\) 100 times per second.

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Most popular questions from this chapter

Show that over one complete cycle, the average value of a sine function squared is \(\frac{1}{2} .\) [Hint: Use the following trigonometric identities: \(\sin ^{2} a+\cos ^{2} a=1 ; \cos 2 a=\) \(\left.\cos ^{2} a-\sin ^{2} a .\right]\)
Two ideal inductors \((0.10 \mathrm{H}, 0.50 \mathrm{H})\) are connected in series to an ac voltage source with amplitude \(5.0 \mathrm{V}\) and frequency \(126 \mathrm{Hz}\). (a) What are the peak voltages across each inductor? (b) What is the peak current that flows in the circuit?
An RLC series circuit is driven by a sinusoidal emf at the circuit's resonant frequency. (a) What is the phase difference between the voltages across the capacitor and inductor? [Hint: since they are in series, the same current \(i(t) \text { flows through them. }]\) (b) At resonance, the rms current in the circuit is \(120 \mathrm{mA}\). The resistance in the circuit is \(20 \Omega .\) What is the rms value of the applied emf? (c) If the frequency of the emf is changed without changing its rms value, what happens to the rms current? (W) tutorial: resonance)
A capacitor is rated at \(0.025 \mu \mathrm{F}\). How much rms current flows when the capacitor is connected to a \(110-\mathrm{V}\) rms, \(60.0-\mathrm{Hz}\) line?
A capacitor and a resistor are connected in parallel across an ac source. The reactance of the capacitor is equal to the resistance of the resistor. Assuming that \(i_{\mathrm{C}}(t)=I \sin \omega t,\) sketch graphs of \(i_{\mathrm{C}}(t)\) and \(i_{\mathrm{R}}(t)\) on the same axes.
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