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posted by  Zimmy on 10/31/2009 8:48:05 PM  |  status: Closed  |  Earned Karma: 716

RC circuit 2.

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N/A N/A N/A N/A 10/5/2009 at 4:00:00 PM
Question Details:
 

A parallel RC circuit has been connected to a voltage source for all time.

R=0.025 Ohms, C= 0.125 farad and  v(t)= 10 sin (5t).

2.1  Derive the current I(t), delivered by the source to the RC circuit as a function of time.

2.2  Convert the above current I(t) to the laboratory form.

2.3  Convert the current to exponential form.

2.4  Derive the average power delivered by the voltage source to the resistor.

2.5 Derive the average power delivered to the parallel element by the source.
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posted by Chelsea_fan(MNK) on 11/4/2009 11:46:47 PM  |  status: Live
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Response Details:
it is given, R=1/40 Ω and C=1/8 F....they are in parallel...
so equivalent impedance Zeq=1/40 *1/j*5*8]/[1/40 +1/j*5*8]  =1/40j+40  =[1/40√2]/_-45  Ω

a) current i(t)=v(t)/Zeq=10sin5t/[1/40√2]/_-45   = [400√2/_45]sin5t  A
i(t)=[400√2/_45]sin5t  A
   
i(t)=[400√2/_45]sin5t A



b)again i(t)=[400√2/_45]sin5t

so i(t)=[400√2/_45]*1/_-90 =400√2/_-45 A   since sin5t=1/_-90

i(t)=400√2/_-45 A


c)we know, r/_θ=re^jθ

so i(t)=400√2e-j45 A


d)here ω=5
         2πf=5
         f=5/2π
now time period T=1/f=1/5/2π=1.257 s

as resistor and capacitor are in parallel so voltage drop across them equal to source voltage

so Pavg=1/T

           integrating we get,
Pavg=1965.4W

Pavg=1965.4W

e)the power delivered to parallel element is

Pavg=1/T

integrating we get Pavg=2779.3 W
Pavg=2779.3 W
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