Question

1. Write an equation that describes voltage supplied by a wall outlet.

(f = 60 Hz, Vrms = 120 V)

2. What is the instantaneous voltage of a sine wave at half of a period?

Describe Alternating Current (AC). Explain how resistors, capacitors, and inductors behave in an AC circuit. WRITEUP REQUIREMENTS Informal: Questions - 6 pts, Data Tables - 3 pts, Survey Questions from Syllabus - 1 pt INTRODUCTION V Up to this point, we have been using supply voltages and currents that are constant in time; this is known as direct current (DC). Electric power is often generated on a large scale by moving a magnetic field in and out of a coiled conductor to generate an EMF. This creates an alternating current (AC) that is most generally described as an electric current that changes direction over time. More simply, AC is a supply voltage that varies with time (V = V(O). An AC signal can be described by its waveform, frequency, and amplitude. The waveform of an AC signal describes its functional form (e.g. sine wave, square wave, sawtooth). Frequency describes the rate at which current changes direction in the units of cycles per second, or hertz (Hz). Frequency is often measured by evaluating the period T, which is the time per cycle. The relationship between period and frequency is given by: Amplitude (arb) 27 time (arb) Figure 9-1. Example of AC sine wave with the equivalent DC voltage. T= (9-1) With AC, there are three different ways voltage is described: peak voltage (V pk), peak-to-peak voltage (Vpp), and root-mean-square voltage (Vrms). Peak voltage is the difference between the maximum voltage and the average voltage. Peak-to-peak voltage is the difference between the maximum and minimum voltage. RMS voltage describes the amount of work an AC source can do, and is the DC equivalent of the AC voltage source. It effectively measures how often the waveform spends at a relatively high potential. There is also instantaneous voltage which refers to the voltage at a specific time in an AC cycle. An example of the sine wave described by Eq. (9-2) is shown in Figure 9-1. V(t) = V sin GF) (9-2) Where V, is the peak voltage. It is important to understand how to convert between the different voltages. For a sine/cosine wave, the conversion is Vpk = Yepp = V2 Vrms (9-3)

Answer 1

Solution : f = 60HZ Voms = 120 V I given given. W- 2016 f - 315 ܀ ܀ W= 271 X60 : 120 TT Vims = Vo 2 vo y Vo = 120.72 volt so 1 a ) V = Vo sinut V = 120 Ja sin (12011)t b) When t = I 21 - TL 2 So, V = 120/7 sin (120 *9.2015) v = 120 12 sin ti V = 0 volt,