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Chapter 6: Q. 89 (page 411)

Alternating Current (ac) Generators The voltage V, in volts, produced by an ac generator at time t, in seconds, is

V(t)=220sin(120蟺迟)

(a) What is the amplitude? What is the period?
(b) Graph V over two periods, beginning att=0.
(c) If a resistance of R=10ohms is present, what is the currentI ?
[Hint: Use Ohm's Law,role="math" localid="1646392386825" [V=IR]
(d) What is the amplitude and period of the current I?
(e) Graph Iover two periods, beginning at t=0.

Short Answer

Expert verified

(a) Amplitude and period of the given function 220and16respectively

(b) Graph V over two periods, beginning at t=0

(c) Current I=22sin(120蟺迟)

(d) current amplitude and period 22and160respectively

(e) Graph of the function

Step by step solution

01

Step 1.Given information

The given functionV=220sin(120蟺迟)

02

Step 2.(a) What is the amplitude? What is the period? 

For the given current function V(t)=220sin(120蟺迟), we can find the amplitude and period by comparing it with y=Asin(蝇t)So,A=220 and 蝇=120蟺and hence we can find the period

T=2蟺蝇=2蟺12蟺=16

03

Step 3.(b) Graph V over two periods, beginning at t=0

04

Step 4.(c) If a resistance of R=10ohms is present, what is the current I ? 

Substitute 220sin(120蟺迟) for Vand 10 for R in the equation V=IR

220sin(120蟺迟)=I.1010.I=220sin(120蟺迟)

Divide both sides of the equation by 10 .

10.I10=220sin(120蟺迟)10I=22sin(120蟺迟)

05

Step 5.(d) What is the amplitude and period of the current I ? 

Comparing I(t)=22sin(120蟺迟)to the standard form y=Asin(蝇t). we get A=22,蝇=120蟺 , androle="math" localid="1646393840670" T=2蟺120蟺=16
Therefore, the period and amplitude of the function I(t)=22sin(120蟺迟)are 160and 22 , respectively.
06

Step 6.Graph I over two periods, beginning at t=0

By using amplitude and period Ploting graph as

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