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A cicuit consisting of a resistor 12ohms an inductor 2H and…

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A cicuit consisting of a...
A cicuit consisting of a resistor 12ohms an inductor 2H and a capacitor of 0.05 F is connected in series.
Use Laplace transforms to obtain an expression for the current flowing in the circuit in terms of the Laplace Operator s and a general input voltage V(S).
Use MATHCAD to obtain expressions for the current flowing in the circuit for the following input voltages:-
a) A step input of 10 volts
b) A voltage which rises linearly from zero to 10 volts.
c) A sinusoidal voltage which is given by v=10sin5t
Submitted: 8 years ago.Category: Homework
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6/25/2010
Tutor: Dr. PAUL, Professor replied 8 years ago
Dr. PAUL
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The solution to part (a) is as follows:

 

 

In the series RLC circuit, assume that at t = 0 that there is no initial charge on the capacitor. If the switch is closed required is the resulting current. The time-domain equation of the given circuit is:

Ri + L(di/dt) +(1/C){Integral from 0 to t}i(t)dt = V

Because: i(0+) = 0, and since V = a step input, the Laplace transform of the differential equation is:

RI(s) + sLI(s) + (1/sC)I(s) = (V/s)

[sR + s^2L + (1/C)]I(s) = V

I(s) = V/[s^2L + sR + (1/C)]

If V is a step input to 10 Volts, R = 12 ohms, L = 2 H and C = 0.05 F then:

I(s) = 10/[2s^2 + 12s + 20]

I(s) = 5/[s^2 + 6s + 10]

I(s) = 5/[(s + 3 - j)(s + 3 + j)]

This is now expanded by partial fractions:

I(s) ={ j(5/2)/[s + 3 - j]} - {j(5/2)/[s + 3 + j]}

The inverse Laplace transform of this equation yields:

i(s) = j(5/2){exp[(-3 -j)t] - exp[(-3 + j)t]}

 

The solution to parts (b) and (c) will follow shortly

 

Dr. Paul

 

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Customer reply replied 8 years ago
Can you send rest of answer as soon as possible please?
Tutor: Dr. PAUL, Professor replied 8 years ago

(c) Assume that V(t) = ASin (omega*t) = 10Sin5t

According to the transform tables, the transform of the source is:

V(s) = [(10)(5)]/[s^2 + (5)^2]

Therefore in the s domain the equation now becomes:

RI(s) + sLI(s) + (1/sC)I(s) = [(50)]/[s^2 + 25]

sRI(s) + s^2LI(s) + (1/C)I(s) = [(50s)]/[s^2 + 25]

I(s) = [(50s)]/[(s^2 + 25)( s^2L + sR + (1/C))]

Separating these terms using partial fractions and taking the inverse Laplace transform of each term, the result simplifies to:

i(t) = (V/Z^2)[RSin(omega*t) - XCos(omega*t)]

where: X = L(omega) - (1/[C(omega)])

X = 2(5) - [1/((0.05)(5))] = 10 - 4 = 6

Z = sqrt(R^2 + X^2)

Z = sqrt(12^2 + 6^2) = sqrt(144 + 36) = 13.4

The equation for the current in the circuit as a function of time is therefore:

i(t) = (10/180)[12Sin(5t) - 6Cos(5t)]

This solution can also be written in the form:

i(t) = (V/Z)Sin[omega*t - Arctan(X/R)]

i(t) = (10/13.6)Sin[5t - Arctan(6/12)]

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Customer reply replied 8 years ago
That's fine, have you got the answer to part b)?
Tutor: Dr. PAUL, Professor replied 8 years ago

b) A voltage which rises linearly from zero to V volts, where V = 10 Volts.

 

Define the voltage in the circuit as the following piecewise continuous function:

 

V(t) = 0 (t = 0)

V(t) = mt (m = constant, 0 < t < a)

V(t) = V (t > a)

 

Therefore the Laplace Transform of V(t) = [Integral from 0 to Infinity]{exp(-st)V(t) dt}

T{V(t)} = [Integral from 0 to a]{exp(-st)mt dt} + [Integral from a to infinity]{exp(-st)V dt}

T{V(t)} = m[(-t exp(-st)/s) + (exp(-st)/s^2)][Evaluated from 0 to a]

+ [V(exp(-st)/(-s)][Evaluated from a to infinity]

T{V(t)} = m{[(1 - as)exp(-sa) - 1]/s^2} + V[exp(-sa)/s]

 

Therefore in the s domain, the equation is:

 

[sR + s^2L + (1/C)]I(s) = m{[(1 - as)exp(-sa) - 1]/s^2} + V[exp(-sa)/s]

I(s) = {m{[(1 - as)exp(-sa) - 1]/s^2} + V[exp(-sa)/s]} /[sR + s^2L + (1/C)]

If V = 10 Volts, R = 12 ohms, L = 2 H and C = 0.05 F then:

I(s) = {m{[(1 - as)exp(-sa) - 1]/s^2} + V[exp(-sa)/s]} /[2s^2 + 12s + 20]

 

Finding i(t) is now a matter of finding the inverse transform of the above equation.

 

This is as far as I could get with my present knowledge. Perhaps MathCAD could be used to find the inverse transform of this equation in order to find i(t). I don't presently have a copy of MathCAD to do this myself.

 

 

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Customer reply replied 8 years ago

Neither do i have a copy of MathCad that's why i asked you!

Tutor: Dr. PAUL, Professor replied 8 years ago
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Customer reply replied 8 years ago
Ok thanks
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