4
Exercises
E4.1 CHAPTER The voltage across the circuit is given by Equation 4.8: v C (t ) = Vi exp( t / RC )
in which Vi is the initial voltage. At the time t1% for which the voltage reaches 1% of the initial value, we have 0.01 = exp( t1% / RC ) Taking the natural logarithm of both sides of the equation, we obtain ln(0.01) = 4.605 = t1% / RC Solving and substituting values, we find t1% = 4.605RC = 23.03 ms.
E4.2
The exponential transient shown in Figure 4.4 is given by v C (t ) = Vs Vs exp( t / ) Taking the derivative with respect to time, we have dv C (t ) Vs = exp( t / ) Evaluating at t = 0, we find that the initial slope is VS / . Because this matches the slope of the straight line shown in Figure 4.4, we have shown that a line tangent to the exponential transient at the origin reaches the final value in one time constant.
dt
E4.3
(a) In dc steady state, the capacitances act as open circuits and the inductances act as short circuits. Thus the steady-state (i.e., t approaching infinity) equivalent circuit is:
From this circuit, we see that ia = 2 A. Then ohms law gives the voltage as v a = Ria = 50 V.
1
(b) The dc steady-state equivalent circuit is:
Here the two 10- resistances are in parallel with an equivalent resistance of 1/(1/10 + 1/10) = 5 . This equivalent resistance is in series with the 5- resistance. Thus the equivalent resistance seen by the source is 10 , and i1 = 20 / 10 = 2 A. Using the current division principle, this current splits equally between the two 10- resistances, so we have i2 = i3 = 1 A. E4.4 (a) = L / R2 = 0.1 / 100 = 1 ms (b) Just before the switch opens, the circuit is in dc steady state with an inductor current of Vs / R1 = 1.5 A. This current continues to flow in This current must flow (upward) through R2 so the initial value of the voltage is v (0 +) = R2i (0 +) = 150 V. (c) We see that the initial magnitude of v(t) is ten times larger than the source voltage. (d) The voltage is given by
the inductor immediately after the switch opens so we have i (0 +) = 1.5 A .
v (t ) =
Vs L exp( t / ) = 150 exp( 1000t ) R1
Let us denote the time at which the voltage reaches half of its initial magnitude as tH. Then we have 0.5 = exp( 1000tH ) Solving and substituting values we obtain tH = 10 3 ln(0.5) = 10 3 ln(2) = 0.6931 ms
2
E4.5
First we write a KCL equation for t 0. t v (t ) 1 + v (x )dx + 0 = 2
R
L
0
Taking the derivative of each term of this equation with respect to time and multiplying each term by R, we obtain: dv (t ) R + v (t ) = 0 The solution to this equation is of the form: v (t ) = K exp( t / ) in which = L / R = 0.2 s is the time constant and K is a constant that must be chosen to fit the initial conditions in the circuit. Since the initial (t = 0+) inductor current is specified to be zero, the initial current in the resistor must be 2 A and the initial voltage is 20 V: v (0+) = 20 = K Thus, we have v (t ) = 20 exp( t / ) 1
t
dt
L
iR = v / R = 2 exp( t / )
t
1 iL (t ) = v (x )dx = [ 20 exp( x / )] = 2 2 exp( t / ) L0 2 0
E4.6
Prior to t = 0, the circuit is in DC steady state and the equivalent circuit is
Thus we have i(0-) = 1 A. However the current through the inductor cannot change instantaneously so we also have i(0+) = 1 A. With the switch open, we can write the KVL equation: di (t ) + 200i (t ) = 100 The solution to this equation is of the form i (t ) = K 1 + K 2 exp( t / ) in which the time constant is = 1 / 200 = 5 ms. In steady state with the switch open, we have i () = K 1 = 100 / 200 = 0.5 A. Then using the initial
dt
3
current, we have i (0+) = 1 = K 1 + K 2 , from which we determine that
K 2 = 0.5. Thus we have
i (t ) = 1.0 A for t < 0
= 0.5 + 0.5 exp( t / ) for t > 0.
v (t ) = L
= 0 V for t < 0 = 100 exp( t / ) for t > 0.
di (t ) dt
E4.7
As in Example 4.4, the KVL equation is t 1 Ri (t ) + i (x )dx + v C (0+) 2 cos(200t ) = 0
C
0
Taking the derivative and multiplying by C, we obtain di (t ) RC + i (t ) + 400C sin(200t ) = 0 Substituting values and rearranging the equation becomes di (t ) 5 10 3 + i (t ) = 400 10 6 sin(200t ) The particular solution is of the form i p (t ) = A cos(200t ) + B sin(200t ) Substituting this into the differential equation and rearranging terms results in 5 10 3 [ 200A sin(200t ) + 200B cos(200t )] + A cos(200t ) + B sin(200t )
= 400 10 6 sin(200t )
dt
dt
Equating the coefficients of the cos and sin terms gives the following equations: A + B = 400 10 6 and B + A = 0 from which we determine that A = 200 10 6 and B = 200 10 6 . Furthermore, the complementary solution is iC (t ) = K exp( t / ) , and the
complete solution is of the form
i (t ) = 200 cos(200t ) 200 sin(200t ) + K exp( t / ) A
At t = 0+, the equivalent circuit is
4
from which we determine that i (0 +) = 2 / 5000 = 400 A. Then evaluating our solution at t = 0+, we have i (0 +) = 400 = 200 + K , from which we determine that K = 200 A. Thus the complete solution is i (t ) = 200 cos(200t ) 200 sin(200t ) + 200 exp(t / ) A
E4.8
The KVL equation is t 1 Ri (t ) + i (x )dx + v C (0+) 10 exp(t ) = 0
C
0
Taking the derivative and multiplying by C, we obtain di (t ) RC + i (t ) + 10C exp( t ) = 0 Substituting values and rearranging, the equation becomes di (t ) 2 + i (t ) = 20 10 6 exp( t )
dt
dt
The particular solution is of the form i p (t ) = A exp( t ) Substituting this into the differential equation and rearranging terms results in 2A exp(t ) + A exp( t ) = 20 10 6 exp( t ) Equating the coefficients gives A = 20 10 6. Furthermore, the complementary solution is iC (t ) = K exp( t / 2) , and the complete solution is of the form
i (t ) = 20 exp(t ) + K exp( t / 2) A
At t = 0+, the equivalent circuit is
5
from which we determine that i (0+) = 5 / 10 6 = 5 A. Then evaluating our solution at t = 0+, we have i (0 +) = 5 = 20 + K , from which we determine that K = 15 A. Thus the complete solution is i (t ) = 20 exp( t ) 15 exp( t / 2) A
E4.9
(a) 0 =
1
LC
=
1 10
3
10
7
= 10 5
=
2RC
1
= 2 10 5
=
=2 0
(b) At t = 0+, the KCL equation for the circuit is v ( 0 +) 0. 1 = + iL (0 +) + Cv (0 +) (1) However, v (0 +) = v (0 ) = 0 , because the voltage across the capacitor cannot change instantaneously. Furthermore, iL (0+) = iL (0) = 0 , because the current through the inductance cannot change value instantaneously. Solving Equation (1) for v (0 +) and substituting values, we find that v (0+) = 10 6 V/s. (c) To find the particular solution or forced response, we can solve the circuit in steady-state conditions. For a dc source, we treat the capacitance as an open and the inductance as a short. Because the inductance acts as a short v p (t ) = 0. (d) Because the circuit is overdamped ( > 1), the homogeneous solution is the sum of two exponentials. The roots of the characteristic solution are given by Equations 4.72 and 4.73:
2 s 1 = 2 0 = 373.2 10 3 2 s 2 = + 2 0 = 26.79 103
R
Adding the particular solution to the homogeneous solution gives the general solution:
6
v (t ) = K 1 exp(s1t ) + K 2 exp(s 2t )
Now using the initial conditions, we have v ( 0 +) = 0 = K 1 + K 2 v (0+) 10 = 6 = K 1s1 + K 2s 2
Solving we find K 1 = 2.887 and K 2 = 2.887. Thus the solution is:
v (t ) = 2.887[exp(s 2t ) exp(s1t )]
1 = 1 10 3 10 7 = 10 5
E4.10
(a) 0 =
LC
=
2RC
1
= 10 5
=
=1 0
(b) The solution for this part is the same as that for Exercise 4.9b in which we found thatv (0+) = 10 6 V/s. (c) The solution for this part is the same as that for Exercise 4.9c in which we found v p (t ) = 0. (d) The roots of the characteristic solution are given by Equations 4.72 and 4.73:
2 s1 = 2 0 = 10 5 2 s 2 = + 2 0 = 10 5
Because the circuit is critically damped ( = 1), the roots are equal and the homogeneous solution is of the form: v (t ) = K 1 exp(s1t ) + K 2t exp(s1t ) Adding the particular solution to the homogeneous solution gives the general solution: v (t ) = K 1 exp(s1t ) + K 2t exp(s1t ) Now using the initial conditions we have v ( 0 +) = 0 = K 1 v (0+) = 10 6 = K 1s1 + K 2 Solving we find K 2 = 10 6 Thus the solution is:
v (t ) = 10 6t exp( 10 5t )
1 = 1 10 3 10 7
E4.11
(a) 0 =
LC
= 10 5
=
2RC
1
= 2 10 4
=
= 0.2 0
(b) The solution for this part is the same as that for Exercise 4.9b in which we found that v (0+) = 10 6 V/s. (c) The solution for this part is the same as that for Exercise 4.9c in
7
which we found v p (t ) = 0. (d) Because we have ( < 1), this is the underdamped case and we have
2 n = 0 2 = 97.98 10 3
Adding the particular solution to the homogeneous solution gives the general solution: v (t ) = K 1 exp( t ) cos( nt ) + K 2 exp( t ) sin( nt ) Now using the initial conditions we have v ( 0 +) = 0 = K 1 v (0+) = 10 6 = K 1 + n K 2 Solving we find K 2 = 10.21 Thus the solution is:
v (t ) = 10.21 exp( 2 10 4t ) sin(97.98 103t ) V
E4.12
The commands are: syms ix t R C vCinitial w ix = dsolve('(R*C)*Dix + ix = (w*C)*2*cos(w*t)', 'ix(0)=-vCinitial/R'); ians =subs(ix,[R C vCinitial w],[5000 1e-6 1 200]); pretty(vpa(ians, 4)) ezplot(ians,[0 80e-3]) An m-file named Exercise_4_12 containing these commands can be found in the MATLAB folder on the OrCAD disk. The commands are: syms vc t % Case I R = 300: vc = dsolve('(1e-8)*D2vc + (1e-6)*300*Dvc+ vc =10', ... 'vc(0) = 0','Dvc(0)=0'); vpa(vc,4) ezplot(vc, [0 1e-3]) hold on % Turn hold on so all plots are on the same axes % Case II R = 200: vc = dsolve('(1e-8)*D2vc + (1e-6)*200*Dvc+ vc =10',... 'vc(0) = 0','Dvc(0)=0'); vpa(vc,4) ezplot(vc, [0 1e-3]) % Case III R = 100: vc = dsolve('(1e-8)*D2vc + (1e-6)*100*Dvc+ vc =10',... 'vc(0) = 0','Dvc(0)=0'); vpa(vc,4) 8
E4.13
ezplot(vc, [0 1e-3]) An m-file named Exercise_4_13 containing these commands can be found in the MATLAB folder on the OrCAD disk.
Answers for Selected Problems
P4.2* P4.3* P4.4* P4.5* P4.6* P4.21* P4.22* P4.23*
The leakage resistance must be greater than 11.39 M.
v C (t ) = 10 20 exp( t (2 10 3 ) ) V t2 = 0.03466 seconds R = 4.328 M v (t ) = V1 exp[(t t1 ) / RC ] for t t1 i1 = 0 i3 = i2 = 2 A t99 = 46.05 ms
t0 = 2 ln(2) = 1.386 ms
v C ,steady state = 10 V
v R (t ) = 0 t < 0
= 10 exp( t / )
V for t 0
P4.33*
i (t ) = 0
= 1 exp( 20t ) A for t 0
for t < 0
9
P4.34*
iL (t ) = 0.3 - 0.5exp (- 2 10 5t ) A v (t ) = 0 for t < 0
= 1000 exp( 2 10 5t ) A
for t > 0
for t > 0
P4.36* P4.45* P4.46* P4.47* P4.61*
R 399.6
iL (t ) = exp( t ) + exp( Rt L ) for t 0 v C (t ) = 10 6 exp( t ) 10 6 exp( 3t ) t > 0 v (t ) = 25 exp(t / ) + 25 cos(10t ) 25 sin(10t ) t 0 s1 = 0.2679 10 4 s 2 = 3.732 10 4 v C (t ) = 50 53.87 exp(s1t ) + 3.867 exp(s 2t ) s 1 = 10 4 v C (t ) = 50 50 exp(s1t ) (50 10 4 )t exp(s1t )
P4.62*
P4.63*
= 0.5 10 4 n = 8.660 10 3
10
v C (t ) = 50 50 exp( t ) cos(nt ) (28.86) exp( t ) sin(nt )
Practice Test T4.1
(a) Prior to the switch opening, the circuit is operating in DC steady state, so the inductor acts as a short circuit, and the capacitor acts as an open circuit. i1 (0) = 10 / 1000 = 10 mA i2 (0) = 10 / 2000 = 5 mA
v C (0) = 10 V
i3 (0) = 0
iL (0) = i1 (0) + i2 (0) + i3 (0) = 15 mA
v C (0+) = v C (0) = 10 V. Also, we have i1 (0+) = iL (0+) = 15 mA, i2 (0+) = v C (0+) / 5000 = 2 mA, and i3 (0+) = i2 (0+) = 2 mA.
(b) Because infinite voltage or infinite current are not possible in this circuit, the current in the inductor and the voltage across the capacitor cannot change instantaneously. Thus, we have iL (0+) = iL (0) = 15 mA and
(c) The current is of the form iL (t ) = A + B exp(t / ). Because the inductor acts as a short circuit in steady state, we have iL () = A = 10 / 1000 = 10 mA
At t = 0+, we have iL (0+) = A + B = 15 mA, so we find B = 5 mA. The time constant is = L / R = 2 10 3 / 1000 = 2 10 6 s. Thus, we have iL (t ) = 10 + 5 exp( 5 10 5t ) mA.
(d) This is a case of an initially charged capacitance discharging through a resistance. The time constant is = RC = 5000 10 6 = 5 10 3 s. Thus we have v C (t ) = Vi exp( t / ) = 10 exp( 200t ) V.
T4.2
di (t ) + i (t ) = 5 exp( 3t ) dt (b) The time constant is = L / R = 2 s and the complementary solution is of the form ic (t ) = A exp( 0.5t ) .
(a) 2 (c) The particular solution is of the form i p (t ) = K exp(3t ) . Substituting into the differential equation produces
11
6K exp( 3t ) + K exp( 3t ) 5 exp(3t ) from which we have K = 1. (d) Adding the particular solution and the complementary solution, we have i (t ) = A exp(0.5t ) exp( 3t ) However, the current must be zero in the inductor prior to t = 0 because of the open switch, and the current cannot change instantaneously in this circuit, so we have i (0 +) = 0. This yields A = 1. Thus, the solution is
i (t ) = exp( 0.5t ) exp(3t ) A
T4.3
(a) Applying KVL to the circuit, we obtain
L
For the capacitance, we have
di (t ) + Ri (t ) + v C (t ) = 15 dt
dv C (t ) dt
(1)
i (t ) = C
(2)
Using Equation (2) to substitute into Equation (1) and rearranging, we have
d 2v C (t ) dv (t ) + (R L ) C + (1 LC ) C (t ) = 15 LC v 2 dt dt d 2v C (t ) dv (t ) + 2000 C + 25 10 6v C (t ) = 375 10 6 2 dt dt
(3)
(b) We try a particular solution of the form v Cp (t ) = A , resulting in
A = 15 . Thus, v Cp (t ) = 15 . (An alternative method to find the particular
solution is to solve the circuit in dc steady state. Since the capacitance acts as an open circuit, the steady-state voltage across it is 15 V.) (c) We have
0 =
1
LC
= 5000 and =
R = 1000 2L
Since we have < 0 , this is the underdamped case. The natural frequency is given by:
2 n = 0 2 = 4899
The complementary solution is given by: v Cc (t ) = K 1 exp( 1000t ) cos(4899t ) + K 2 exp( 1000t ) sin(4899t )
12
(d) The complete solution is v C (t ) = 15 + K 1 exp( t ) cos(nt ) + K 2 exp( t ) sin(nt ) The initial conditions are
v C (0 ) = 0
Thus, we have
and
i (0 ) = 0 = C
dv C (t ) dt
t =0
v C (0 ) = 0 = 15 + K 1 dv C (t ) = 0 = K 1 + n K 2 dt t = 0 Solving, we find K 1 = 15 and K 2 = 3.062 . Finally, the solution is
v C (t ) = 15 15 exp( 1000t ) cos(4899t ) (3.062) exp( 1000t ) sin(4899t ) V
T4.4
One set of commands is syms vC t S = dsolve('D2vC + 2000*DvC + (25e6)*vC = 375e6',... 'vC(0) = 0, DvC(0) = 0'); simple(vpa(S,4)) These commands are stored in the m-file named T_4_4 on the OrCAD disk.
13