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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

Make a figure analogous to Fig. 21.4 for an ideal inductor in an ac circuit. Start by assuming that the voltage across an ideal inductor is \(v_{\mathrm{L}}(t)=V_{\mathrm{L}}\) sin \(\omega t .\) Make a graph showing one cycle of \(v_{\mathrm{L}}(t)\) and \(i(t)\) on the same axes. Then, at each of the times \(t=0, \frac{1}{8} T, \frac{2}{8} T, \ldots, T,\) indicate the direction of the current (or that it is zero), whether the current is increasing, decreasing, or (instantaneously) not changing, and the direction of the induced emf in the inductor (or that it is zero).
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