2270                       PRACTICE FINAL EXAM SOLUTION Prob 1                              

 

1.      (50 points)

After having been open for a long time, the switch is closed at t = 0.

R1 = 12.5Ω                     R2 = 12.5Ω                 L = 6.25 mH

a.   Two capacitances are available: 250 nF and 2 nF. Specify the value of C that will make v(t) overdamped.

b.   Using the value of C found in (a), write a time-domain expression for v(t).

 

ans:  a)     C = 250 nF

         b)     v(t) = 13.3 (e-0.4Mt − e−1.6Mt) V

 

sol'n:  (a)   To make the response overdamped, we must have two real characteristic roots. We use the circuit for t > 0, consisting of C, R1, L, and vA in series. We may find the characteristic equation by looking it up in a textbook or by setting the impedance of R1, C, and L in series to zero.

The characteristic roots for the quadratic equation are

or

We want an overdamped response, (real roots α2 > ωo2).

Try each C value in turn.

C = 2 nF:

ωo = 8.9M rad/s > α2 (underdamped)

C = 250 nF:

ωo = 0.8M rad/s < α2 (overdamped)

We need C = 250 nF for an overdamped solution.

sol'n:  (b)  We use the exponential solution for the overdamped case:

Because the value of A3 is all that is left of v(t) as t → ∞, we first find the constant term, A3. (The other terms decay because the characteristic roots always have negative real parts in a passive RLC circuit. When the switch opens, the energy sloshing back and forth in the L and C will decay owing to power dissipated by the series resistor R1.)

As t → ∞, the circuit reaches equilibrium. C acts like an open circuit, L acts like a short circuit or wire.

Model:

Since L acts like a wire, there is no voltage drop across it.

Thus, A3 = v(t→∞) = 0.

We find coefficients A1 and A2 by matching initial conditions in the circuit. We find initial conditions by examining the circuit at t = 0−, when the circuit has reached equilibrium. We find the values of iL and vC, the energy variables, at t = 0− and use the same values at t = 0+ (since the energy in the circuit cannot change instantly).

Mathematically, our general form of solution for the overdamped case gives the following values for v(0+) and dv(t)/dt|t=0+:

Note:   We must always differentiate first and then plug in t = 0+. Otherwise, we always get zero.

Now we find the numerical values of v(0+) and dv(t)/dt|t=0+.

At t = 0−, C acts like an open circuit and L acts like a short circuit.

Model:

At time t = 0+, we have iL(0+) = iL(0−) = 4 A and vC(0+) = vC(0−) = 50 V. We solve the circuit at t = 0+, treating iL(0+) as a current source and vC(0+) as a voltage source.

We now solve for v(0+) and dv(t)/dt|t=0+. From these we find A1 and A2.

Model:

We may apply any standard method to solve the circuit, but we can solve the above circuit using a voltage loop.

The same equation applies for t > 0, and we may differentiate to find dv(t)/dt in terms of energy (or state) variables iL and vC.

The basic equations for L and C, rearranged, allow us to translate the derivatives on the right side of this equation into non-derivatives we can calculate numerically.

Applying these identities, we have

Only now that we have differentiated do we finally evaluate the derivative we seek at t = 0:

Since iC is in series with iL, we have iC(0+) = iL(0+) = 4A.

Now we find A1 and A2.

Concluding the algebra, we find the numerical values of the coefficients A1 and A2.

Using the values of α and ωo from above, we find the values of s1 and s2.

s1 = −1 M + 0.6 M = −0.4 M

s2 = −1 M − 0.6 M = −1.6 M

Plugging into the general form of underdamped solution completes our answer:

v(t) = 13.3 (e-0.4Mt − e−1.6Mt) V