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posted by  Zimmy on 11/3/2009 2:21:18 PM  |  status: Closed  |  Earned Karma: 716

RL circuit 1

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N/A N/A N/A N/A 11/4/2009 at 8:00:00 AM
Question Details:

A series RL circuit with the values R= 40 ohms and L= 8 Henry is given.

The input to this circuit is a periodic signal of current i(t) for all time:

i(t)= 10 sin ( 5 t).

1.1  Derive the output voltage across the series circuit as a function of time in standard form.

1.2  Convert the output voltage to laboratory form.

1.3  Convert the output voltage to exponential form.

1.4  Derive the average power delivered to the resistor in one cycle of the input signal.

1.5  Derive the average power delivered to the series element in one cycle.

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Sage
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(IIT Kharagpur)
posted by Chelsea_fan(MNK) on 11/4/2009 12:21:43 AM  |  status: Live
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Response Details:
here i(t)=10sin(5t)
converting to frequency domain,
i=10/_-90

ω=5
f=5/2π=.796 Hz
T=1/f=1.2563 sec

8 H=>j5*8=j40
                                             


Xtotal=40+40j=[√402+402]tan-140/40 =40√2/_45

Vtot=i*Xtotal=[10/_-90*]40√2/_45=400√2/_-45

Vtot=400√2/_-45

we know ,
r/_θ=re^jθ

so Vtot can be written as,
Vtot=400√2e-j45

converting to time domain,
Vtot=400√2cos(5t-45)=400√2sin(5t-45+90)=400√2sin(45+5t)

Vtot=400√2sin(45+5t)
power delivered to resistor ,
Pavg=1/T

Pavg=1/1.2563
putting i=10sin5t and integrating we get,
Pavg=1965.3 W





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