Alternating Current

harpazo

harpazo

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The voltage V, in volts, produced by an ac generator at time t, in seconds, is V(t) = 120 sin(120pi•t).

(a) What is the amplitude? What is the period?
(b) Graph V over two periods, beginning at t = 0.
(c) If a resistance of R = 20 ohms is present, what is the
current I?
[Hint: Use Ohm’s Law, V = IR.]
(d) What are the amplitude and period of the current I?
(e) Graph I over two periods, beginning at t = 0.

Part a

The amplitude is 120.

P = 2pi/B

P = 2pi/(120•pi)

P = 1/60 = period

For part b, I know what is meant by two cycles. What does it mean beginning at t = 0?

For part c, I must use V = IR. I replace R with 20 and must find I. However, to do so, I gotta know the value of V, right? What is the value of V?

For part d, what equation gives me the amplitude and period of the currentbI?

For part e, which equation must I graph two cycles for beginning at t = 0?
 
harpazo said:
The voltage V, in volts, produced by an ac generator at time t, in seconds, is V(t) = 120 sin(120pi•t).

(a) What is the amplitude? What is the period?
(b) Graph V over two periods, beginning at t = 0.
(c) If a resistance of R = 20 ohms is present, what is the
current I?
[Hint: Use Ohm’s Law, V = IR.]
(d) What are the amplitude and period of the current I?
(e) Graph I over two periods, beginning at t = 0.

Part a

The amplitude is 120.

P = 2pi/B

P = 2pi/(120•pi)

P = 1/60 = period

For part b, I know what is meant by two cycles. What does it mean beginning at t = 0?

For part c, I must use V = IR. I replace R with 20 and must find I. However, to do so, I gotta know the value of V, right? What is the value of V?

For part d, what equation gives me the amplitude and period of the currentbI?

For part e, which equation must I graph two cycles for beginning at t = 0?
Click to expand...
For (b), they want the graph to start at t=0, where the horizontal axis is t and the vertical axis is V. That is, start the graph at the vertical axis, not somewhere else.

For (c), the answer will be a function of t, not a number. Replace V with the expression you were given. Then (d) and (e) will be clear, because you will have an equation to work with.
 
A couple of comments:
harpazo said:
The voltage V, in volts, produced by an ac generator at time t, in seconds, is V(t) = 120 sin(120pi•t).

(a) What is the amplitude? What is the period?
(b) Graph V over two periods, beginning at t = 0.
(c) If a resistance of R = 20 ohms is present, what is the
current I?
[Hint: Use Ohm’s Law, V = IR.]
(d) What are the amplitude and period of the current I?
(e) Graph I over two periods, beginning at t = 0.

Part a

The amplitude is 120.

P = 2pi/B

P = 2pi/(120•pi)

P = 1/60 = period
Click to expand...
The problem asked for the period. But you did not say that "P" was the period!
Also where you have "P= 2pi/B" you haven't said what "B" is!
The period of "sin(x)" is 2pi so T will be the period when 120pi T= 2pi so T= 2pi/(120pi)= 1/60 second. You were told that the time was in second. If I were your teacher I would mark down for not say "1/60 second".

For part b, I know what is meant by two cycles. What does it mean beginning at t = 0?
Click to expand...
It means your graph should go from t= 0 to t= 2(1/60)= 1/30.

For part c, I must use V = IR. I replace R with 20 and must find I. However, to do so, I gotta know the value of V, right? What is the value of V?
Click to expand...
The answer is not a number, it is a function of t. You are told that V= 120 sin(120pi•t). So I= (120/20) sin(120pi t)= 6sin(120pi t).

For part d, what equation gives me the amplitude and period of the currentbI?
Click to expand...
What are the amplitude and period of I= 6 sin(120pi t)?
\(\quad\) EDITED

For part e, which equation must I graph two cycles for beginning at t = 0?
Click to expand...
You graph I= 6 sin(120pi t) from t= 0 to twice the period.
 
HallsofIvy said:
A couple of comments:

The problem asked for the period. But you did not say that "P" was the period!
Also where you have "P= 2pi/B" you haven't said what "B" is!
The period of "sin(x)" is 2pi so T will be the period when 120pi T= 2pi so T= 2pi/(120pi)= 1/60 second. You were told that the time was in second. If I were your teacher I would mark down for not say "1/60 second".


It means your graph should go from t= 0 to t= 2(1/60)= 1/30.


The answer is not a number, it is a function of t. You are told that V= 120 sin(120pi•t). So I= (120/20) sin(120pi t)= 6sin(120pi t).


What are the amplitude and period of I= 6 sin(230pi t)?


You graph I= 6 sin(120pi t) from t= 0 to twice the period.
Click to expand...

1. Thank you.

2. For 2pi/B, B is the coefficient of the variable in the argument of the function.

3. What are the amplitude and period of I= 6 sin(230pi t)?

Amplitude = 6

Let P = period.

P = 2pi/B

P = 2pi/230•pi

P = 2/230

P = 1/115 seconds

Yes?
 
Dr.Peterson said:
For (b), they want the graph to start at t=0, where the horizontal axis is t and the vertical axis is V. That is, start the graph at the vertical axis, not somewhere else.

For (c), the answer will be a function of t, not a number. Replace V with the expression you were given. Then (d) and (e) will be clear, because you will have an equation to work with.
Click to expand...

Excellent reply. I will show my work tonight or tomorrow. Thank you. This is all I need.
 
harpazo said:
1. Thank you.

2. For 2pi/B, B is the coefficient of the variable in the argument of the function.

3. What are the amplitude and period of I= 6 sin(230pi t)?

Amplitude = 6

Let P = period.

P = 2pi/B

P = 2pi/230•pi

P = 2/230

P = 1/115 seconds

Yes?
Click to expand...
I suspect that "230" was a typo for "120" on Halls' part! Did you consider asking about it?
 
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