FIRST-ORDER ACTIVE CIRCUITS

CHAPTER 17 Fourier Method of Waveform Analysis 420

7.16 FIRST-ORDER ACTIVE CIRCUITS

Active circuits containing op amps are less susceptible to loading effects when interconnected with other circuits. In addition, they offer a wider range of capabilities with more ease of realization than passive circuits. In our present analysis of linear active circuits we assume ideal op amps; that is; (1) the current drawn by the op amp input terminals is zero and (2) the voltage difference between the inverting and noninverting terminals of the op amp is negligible (see Chapter 5). The usual methods of analysis are then applied to the circuit as illustrated in the following examples.

Fig. 7-19

EXAMPLE 7.15 Highpass filter. The op amp in the circuit of Fig. 7-44 is ideal. Find the unit-step response of the circuit; that is,v₂forv₁¼uðtÞ:

The inverting input terminal of the op amp is at virtual ground and the capacitor has zero voltage att¼0^þ. The 1-V step input therefore generates an exponentially decaying currentithroughR₁C(from left to right, with a time constantR₁Cand initial value of 1=R₁).

i¼ 1

R₁e^t=ðR1CÞuðtÞ

All of the preceding current passes throughR₂(the op amp draws no current), generatingv₂¼ R₂iat the output terminal. The unit-step response is therefore

v₂¼ R₂

R₁e^t=ðR1CÞuðtÞ

EXAMPLE 7.16 In the circuit of Fig. 7-44 derive the differential equation relatingv₂ tov₁. Find its unit-step response and compare with the answer in Example 7.15.

Since the inverting input terminal of the op amp is at virtual ground and doesn’t draw any current, the currenti passing throughC,R₁, andR₂from left to right isv₂=R₂. Letv_Abe the voltage of the node connectingR₁andC.

Then, the capacitor voltage isv₁v_A(positive on the left side). The capacitor current and voltage are related by v₂

R₂¼dðv₁v_AÞ dt

To eliminatev_A, we note that the segment made ofR₁,R₂, and the op amp form an inverting amplifier with v₂¼ ðR₂=R₁Þv_A, from whichv_A¼ ðR₁=R₂Þv₂. Substituting forv_A, we get

v₂þR₁Cdv₂

dt ¼ R₂Cdv₁ dt To find the unit-step response, we first solve the following equation:

v₂þR₁Cdv₂

dt ¼ R₂C t>0 0 t<0

The solution of the preceding equation isR₂Cð1e^t=ðR1CÞÞuðtÞ. The unit-step response of the circuit is the time- derivative of the preceding solution.

Table 7-2

fðtÞ v_pðtÞ

1 1

a

t t

a 1 a² e^st;ðs6¼ aÞ e^st

sþa

e^at te^at

cos!t Acosð!tÞ where A¼ 1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a²þ!²

p and tan¼! a e^btcos!t Ae^btcosð!tÞ where A¼ 1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðabÞ²þ!²

q and tan¼ !

ab

v₂ðtÞ ¼ R₂

R₁e^t=ðR1CÞuðtÞ

Alternate Approach

The unit step response may also be found by the Laplace transform method (see Chapter 16).

EXAMPLE 7.17 Passive phase shifter. Find the relationship betweenv₂andv₁in the circuit of Fig. 7-45(a).

Let node D be the reference node. Apply KCL at nodes A and B to find KCL at node A: Cdv_A

dt þðv_Av₁Þ

R ¼0

KCL at node B: Cdðv_Bv₁Þ dt þv_B

R¼0

Subtracting the second equation from the first and noting thatv₂¼v_Av_Bwe get v₂þRCdv₂

dt ¼v₁RCdv₁ dt

EXAMPLE 7.18 Active phase shifter. Show that the relationship betweenv₂andv₁in the circuit of Fig. 7-45(b) is the same as in Fig. 7-45(a).

Apply KCL at the inverting (node A) and non-inverting (node B) inputs of the op amp.

KCL at node A: ðv_Av₁Þ

R₁ þðv_Av₂Þ R₁ ¼0 KCL at node B: ðv_Bv₁Þ

R þCdv_B dt ¼0

From the op amp we havev_A¼v_Band from the KCL equation for node A, we havev_A¼ ðv₁þv₂Þ=2. Substituting the preceding values in the KCL at node B, we find

v₂þRCdv₂

dt ¼v₁RCdv₁ dt

Solved Problems

7.1 At t¼0, just before the switch is closed in Fig. 7-20, v_C¼100 V. Obtain the current and charge transients.

With the polarities as indicated on the diagram, v_R¼v_C for t>0, and 1=RC¼62:5 s¹. Also, v_Cð0^þÞ ¼v_Cð0Þ ¼100 V. Thus,

v_R¼v_C¼100e^62:5t ðVÞ i¼v_R

R¼0:25e^62:5t ðAÞ q¼Cv_C¼4000e^62:5t ðmCÞ Fig. 7-20

7.2 In Problem 7.1, obtain the power and energy in the resistor, and compare the latter with the initial energy stored in the capacitor.

p_R¼v_Ri¼25e^125t ðWÞ w_R¼

ðt 0

p_Rdt¼ ðt

0

25e^125tdt¼0:20ð1e^125tÞ ðJÞ The initial stored energy is

W₀¼¹₂CV₀²¼¹₂ð4010⁶Þð100Þ²J¼0:20¼w_Rð1Þ

In other words, all the stored energy in the capacitor is eventually delivered to the resistor, where it is converted into heat.

7.3 AnRC transient identical to that in Problems 7.1 and 7.2 has a power transient p_R¼360e^t=0:00001 ðWÞ

Obtain the initial charge Q₀, ifR¼10. p_R¼P₀e^2t=RC or 2

RC¼10⁵ or C¼2mF w_R¼

ðt 0

p_Rdt¼3:6ð1e^t=0:00001Þ ðmJÞ Then,w_Rð1Þ ¼3:6 mJ¼Q²₀=2C, from whichQ₀¼120mC.

7.4 The switch in theRLcircuit shown in Fig. 7-21 is moved from position1to position2att¼0.

Obtainv_Randv_L with polarities as indicated.

The constant-current source drives a current through the inductance in the same direction as that of the transient currenti. Then, fort>0,

i¼I₀e^Rt=L¼2e^25t ðAÞ v_R¼Ri¼200e^25t ðVÞ v_L¼ v_R¼ 200e^25t ðVÞ

7.5 For the transient of Problem 7.4 obtainp_R andp_L. p_R¼v_Ri¼400e^50t ðWÞ p_L¼v_Li¼ 400e^50t ðWÞ

Negative power for the inductance is consistent with the fact that energy is leaving the element. And, since this energy is being transferred to the resistance,p_Ris positive.

Fig. 7-21

7.6 A seriesRCcircuit withR¼5 kandC¼20mF has a constant-voltage source of 100 V applied att¼0; there is no initial charge on the capacitor. Obtaini,v_R,v_C, andq, fort>0.

The capacitor charge, and hencev_C, must be continuous att¼0:

v_Cð0^þÞ ¼v_Cð0Þ ¼0

Ast! 1,v_C!100 V, the applied voltage. The time constant of the circuit is¼RC¼10¹s. Hence, from Section 6.10,

v_C¼ ½v_Cð0^þÞ v_Cð1Þe^t=þv_Cð1Þ ¼ 100e^10tþ100 ðVÞ

The other functions follow from this. If the element voltages are both positive where the current enters,v_Rþv_C¼100V, and so

v_R¼100e^10t ðVÞ i¼v_R

R¼20e^10t ðmAÞ

q¼Cv_C¼2000ð1e^10tÞ ðmCÞ

7.7 The switch in the circuit shown in Fig. 7-22(a) is closed att¼0, at which moment the capacitor has chargeQ₀¼500mC, with the polarity indicated. Obtainiandq, fort>0, and sketch the graph ofq.

The initial charge has a corresponding voltageV₀¼Q₀=C¼25 V, whencev_Cð0^þÞ ¼ 25 V. The sign is negative because the capacitor voltage, in agreement with the positive direction of the current, would beþ on the top plate. Alsov_Cð1Þ ¼ þ50 V and¼0:02 s. Thus, as in Problem 7.6,

v_C¼ 75e^50tþ50 ðVÞ from which

q¼Cv_C¼ 1500e^50tþ1000 ðmCÞ i¼dq

dt¼75e^50t ðmAÞ

The sketch in Fig. 7-22(b) shows that the charge changes from 500mC of one polarity to 1000mC of the opposite polarity.

7.8 Obtain the currenti, for all values oft, in the circuit of Fig. 7-23.

Fort<0, the voltage source is a short circuit and the current source shares 2 A equally between the two 10-resistors:

Fig. 7-22

iðtÞ ¼ið0Þ ¼ið0^þÞ ¼1 A

Fort>0, the current source is replaced by an open circuit and the 50-V source acts in theRLseries circuitðR¼20Þ. Consequently, ast! 1,i! 50=20¼ 2:5 A. Then, by Sections 6.10 and 7.3,

iðtÞ ¼ ½ðið0^þÞ ið1Þe^Rt=Lþið1Þ ¼3:5e^100t2:5 ðAÞ

By means of unit step functions, the two formulas may be combined into a single formula valid for allt:

iðtÞ ¼uðtÞ þ ð3:5e^100t2:5ÞuðtÞ ðAÞ

7.9 In Fig. 7-24(a), the switch is closed att¼0. The capacitor has no charge fort<0. Findi_R,i_C, v_C, andv_s for all times ifi_s¼2 mA.

Fort<0,i_R¼2 mA,i_C¼v_C¼0, andv_s¼ ð2 mAÞð5000Þ ¼10 V.

Fort>0, the time constant is¼RC¼10 ms and

i_Rð0^þÞ ¼0;i_Rð1Þ ¼2 mA, andi_R¼2ð1e^100tÞ ðmAÞ [See Fig. 7-24ðbÞ:

v_Cð0^þÞ ¼0;v_Cð1Þ ¼ ð2 mAÞð5 kÞ ¼10 V, andv_C¼10ð1e^100tÞ ðVÞ [See Fig. 7-24ðcÞ:

i_Cð0^þÞ ¼2 mA;i_Cð1Þ ¼0, andi_C¼2e^100t ðmAÞ [See Fig. 7-24ðdÞ:

v_sð0^þÞ ¼0;v_sð1Þ ¼ ð2 mAÞð5 kÞ ¼10 V, andv_s¼10ð1e^100tÞ ðVÞ [See Fig. 7-24ðeÞ:

7.10 In Fig. 7-25, the switch is opened att¼0. Find i_R,i_C,v_C, and v_s.

Fort<0, the circuit is at steady state withi_R¼6ð4Þ=ð4þ2Þ ¼4 mA,i_C¼0, andv_C¼v_s¼4ð2Þ ¼8 V.

During the switching at t¼0, the capacitor voltage remains the same. After the switch is opened, at t¼0^þ, the capacitor has the same voltagev_Cð0^þÞ ¼v_Cð0Þ ¼8 V.

Fort>0, the capacitor discharges in the 5-kresistor, produced from the series combination of the 3-k and 2-k resistors. The time constant of the circuit is ¼ ð2þ3Þð10³Þð210⁶Þ ¼0:01 s. The currents and voltages are

v_C¼8e^100t ðVÞ

i_R¼ i_C¼v_C=5000¼ ð8=5000Þe^100t¼1:6e^100t ðmAÞ v_s¼ ð6 mAÞð4 kÞ ¼24 V

since, fort>0, all of the 6 mA goes through the 4-kresistor.

7.11 The switch in the circuit of Fig. 7-26 is closed on position1att¼0 and then moved to2after one time constant, att¼¼250ms. Obtain the current fort>0.

It is simplest first to find the charge on the capacitor, since it is known to be continuous (att¼0 and at t¼), and then to differentiate it to obtain the current.

For 0t,qmust have the form

q¼Ae^t=þB Fig. 7-23

Fig. 7-24

Fig. 7-25

Fig. 7-26

From the assumptionqð0Þ ¼0 and the condition ið0^þÞ ¼dq

dt ^0þ ¼ 20 V

500¼40 mA

we find thatA¼ B¼ 10mC, or

q¼10ð1e^4000tÞ ðmCÞ ð0tÞ ð20Þ From (20),qðÞ ¼10ð1e¹ÞmC; and we know thatqð1Þ ¼ ð0:5mFÞð40 VÞ ¼ 20mC.

Hence,q, is determined fortas

q¼ ½qðÞ qð1Þe^ðtÞ=þqð1Þ ¼71:55e^4000t20 ðmCÞ ð21Þ Differentiating (20) and (21),

i¼dq

dt¼ 40e^4000t ðmAÞ ð0<t< Þ 286:2e^4000t ðmAÞ ðt> Þ

See Fig. 7-27.

7.12 A seriesRLcircuit has a constant voltageV applied att¼0. At what time doesv_R¼v_L? The current in anRLcircuit is a continuous function, starting at zero in this case, and reaching the final valueV=R. Thus, fort>0,

i¼V

Rð1e^t=Þ and v_R¼Ri¼Vð1e^t=Þ

where¼L=Ris the time constant of the circuit. Sincev_Rþv_L¼V, the two voltages will be equal when v_R¼¹₂V

Vð1e^t=Þ ¼¹₂V e^t= ¼¹₂

t ¼ln 2 that is, whent¼0:693. Note that this time is independent ofV.

7.13 A constant voltage is applied to a seriesRLcircuit att¼0. The voltage across the inductance is 20 V at 3.46 ms and 5 V at 25 ms. ObtainRifL¼2H.

Using the two-point method of Section 7-6.

Fig. 7-27

¼ t₂t₁

lnv₁lnv₂¼ 253:46

ln 20ln 5¼15:54 ms R¼L

¼ 2

15:5410³¼128:7 and so

7.14 In Fig. 7-28, switchS₁ is closed att¼0. SwitchS₂ is opened att¼4 ms. Obtainifort>0.

As there is always inductance in the circuit, the current is a continuous function at all times. In the interval 0t4 ms, with the 100shorted out and a time constant¼ ð0:1 HÞ=ð50Þ ¼2 ms,istarts at zero and builds toward

100 V 50 ¼2 A

even though it never gets close to that value. Hence, as in Problem 7.12

i¼2ð1e^t=2Þ ðAÞ ð0t4Þ ð22Þ

whereintis measured in ms. In particular,

ið4Þ ¼2ð1e²Þ ¼1:729 A

In the intervalt4 ms,istarts at 1.729 A and decays toward 100=150¼0:667 A, with a time constant 0:1=150¼²₃ms. Therefore, withtagain in ms,

i¼ ð1:7290:667Þe^{ðt4Þ=ð2=3Þ}þ0:667¼428:4e^3t=2þ0:667 ðAÞ ðt4Þ ð23Þ

7.15 In the circuit of Fig. 7-29, the switch is closed at t¼0, when the 6-mF capacitor has charge Q₀¼300mC. Obtain the expression for the transient voltagev_R.

The two parallel capacitors have an equivalent capacitance of 3mF. Then this capacitance is in series with the 6mF, so that the overall equivalent capacitance is 2mF. Thus,¼RC_eq¼40ms.

Att¼0^þ, KVL givesv_R¼300=6¼50 V; and, ast! 1,v_R!0 (sincei!0). Therefore, v_R¼50e^t=¼50e^t=40 ðVÞ

in whichtis measured inms.

Fig. 7-28

Fig. 7-29 Fig. 7-30

7.16 In the circuit shown in Fig. 7-30, the switch is moved to position2att¼0. Obtain the currenti₂ at t¼34:7 ms.

After the switching, the three inductances have the equivalent L_eq¼10

6 þ5ð10Þ 15 ¼5 H Then¼5=200¼25 ms, and so, withtin ms,

i¼6e^t=25 ðAÞ i₂¼ 5

15 i¼2e^t=25 ðAÞ i₂ð34:7Þ ¼2e^34:7=25A¼0:50 A

and

7.17 In Fig. 7-31, the switch is closed at t¼0. Obtain the current i and capacitor voltagev_C, for t>0.

As far as the natural response of the circuit is concerned, the two resistors are in parallel; hence, ¼R_eqC¼ ð5Þð2mFÞ ¼10ms

By continuity,v_Cð0^þÞ ¼v_Cð0Þ ¼0. Furthermore, ast! 1, the capacitor becomes an open circuit, leav- ing 20in series with the 50 V. That is,

ið1Þ ¼50

20¼2:5 A v_Cð1Þ ¼ ð2:5 AÞð10Þ ¼25 V Knowing the end conditions onv_C, we can write

v_C¼ ½v_Cð0^þÞ v_Cð1Þe^t=þv_Cð1Þ ¼25ð1e^t=10Þ ðVÞ whereintis measured inms.

The current in the capacitor is given by i_C¼Cdv_C

dt ¼5e^t=10 ðAÞ and the current in the parallel 10-resistor is

i₁₀¼ v_C

10¼2:5ð1e^t=10Þ ðAÞ i¼i_Cþi₁₀¼2:5ð1þe^t=10Þ ðAÞ Hence,

The problem might also have been solved by assigning mesh currents and solving simultaneous differ- ential equations.

7.18 The switch in the two-mesh circuit shown in Fig. 7-32 is closed att¼0. Obtain the currentsi₁ andi₂, fort>0.

Fig. 7-31

10ði₁þi₂Þ þ5i₁þ0:01di₁

dt ¼100 ð24Þ

10ði₁þi₂Þ þ5i₂¼100 ð25Þ

From (25),i₂¼ ð10010i₁Þ=15. Substituting in (24), di₁

dt þ833i₁¼3333 ð26Þ

The steady-state solution (particular solution) of (26) isi₁ð1Þ ¼3333=833¼4:0 A; hence i₁¼Ae^833tþ4:0 ðAÞ

The initial conditioni₁ð0Þ ¼i₁ð0^þÞ ¼0 now givesA¼ 4:0 A, so that

i₁¼4:0ð1e^833tÞ ðAÞ and i₂¼4:0þ2:67e^833t ðAÞ Alternate Method

When the rest of the circuit is viewed from the terminals of the inductance, there is equivalent resistance R_eq¼5þ5ð10Þ

15 ¼8:33 Then 1=¼R_eq=L¼833 s¹. Att¼ 1, the circuit resistance is

R_T¼10þ5ð5Þ

10 ¼12:5

so that the total current isi_T¼100=12:5¼8 A. And, att¼ 1, this divides equally between the two 5- resistors, yielding a final inductor current of 4 A. Consequently,

i_L¼i₁¼4ð1e^833tÞ ðAÞ

7.19 A seriesRLcircuit, with R¼50andL¼0:2 H, has a sinusoidal voltage v¼150 sinð500tþ0:785Þ ðVÞ

applied att¼0. Obtain the current fort>0.

The circuit equation fort>0 is di

dtþ250i¼750 sinð500tþ0:785Þ ð27Þ

The solution is in two parts, the complementary function (i_c) and the particular solution ði_pÞ, so that i¼i_cþi_p. The complementary function is the general solution of (27) when the right-hand side is replaced by zero:i_c¼ke^250t. Themethod of undetermined coefficientsfor obtainingi_pconsists in assuming that

i_p¼Acos 500tþBsin 500t

since the right-hand side of (27) can also be expressed as a linear combination of these two functions. Then Fig. 7-32

di_p

dt ¼ 500Asin 500tþ500Bcos 500t

Substituting these expressions fori_panddi_p=dtinto (27) and expanding the right-hand side,

500Asin 500tþ500Bcos 500tþ250Acos 500tþ250Bsin 500t¼530:3 cos 500tþ530:3 sin 500t Now equating the coefficients of like terms,

500Aþ250B¼530:3 and 500Bþ250A¼530:3 Solving these simultaneous equations,A¼ 0:4243 A,B¼1:273 A.

i_p¼ 0:4243 cos 500tþ1:273 sin 500t¼1:342 sinð500t0:322Þ ðAÞ i¼i_cþi_p¼ke^250tþ1:342 sinð500t0:322Þ ðAÞ

and

Att¼0,i¼0. Applying this condition,k¼0:425 A, and, finally, i¼0:425e^250tþ1:342 sinð500t0:322Þ ðAÞ

7.20 For the circuit of Fig. 7-33, obtain the current i_L, for all values of t.

Fort<0, the 50-V source results in inductor current 50=20¼2:5 A. The 5-A current source is applied fort>0. Ast! 1, this current divides equally between the two 10-resistors, whencei_Lð1Þ ¼ 2:5 A.

The time constant of the circuit is

¼0:210³H

20 ¼ 1

100ms and so, withtin ms and usingi_Lð0^þÞ ¼i_Lð0Þ ¼2:5 A,

i_L¼ ½i_Lð0^þÞ i_Lð1Þe^t=þi_Lð1Þ ¼5:0e^100t2:5 ðAÞ Finally, using unit step functions to combine the expressions fort<0 andt>0,

i_L¼2:5uðtÞ þ ð5:0e^100t2:5ÞuðtÞ ðAÞ

7.21 The switch in Fig. 7-34 has been in position1for a long time; it is moved to2att¼0. Obtain the expression fori, fort>0.

With the switch on1,ið0Þ ¼50=40¼1:25 A. With an inductance in the circuit,ið0Þ ¼ið0^þÞ. Long after the switch has been moved to2,ið1Þ ¼10=40¼0:25 A. In the above notation,

B¼ið1Þ ¼0:25 A A¼ið0^þÞ B¼1:00 A and the time constant is¼L=R¼ ð1=2000Þs. Then, fort>0,

i¼1:00e^2000tþ0:25 ðAÞ Fig. 7-33

7.22 The switch in the circuit shown in Fig. 7-35 is moved from1to2att¼0. Findv_Candv_R, for t>0.

With the switch on 1, the 100-V source results in v_Cð0Þ ¼100 V; and, by continuity of charge, v_Cð0^þÞ ¼v_Cð0Þ. In position2, with the 50-V source of opposite polarity,v_Cð1Þ ¼ 50 V. Thus,

B¼v_Cð1Þ ¼ 50 V A¼v_Cð0^þÞ B¼150 V

¼RC¼ 1

200s v_C¼150e^200t50 ðVÞ and

Finally, KVL givesv_Rþv_Cþ50¼0, or

v_R¼ 150e^200t ðVÞ

7.23 Obtain the energy functions for the circuit of Problem 7.22.

w_C¼¹₂Cv²_C¼1:25ð3e^200t1Þ² ðmJÞ w_R¼

ðt 0

v²_R

R dt¼11:25ð1e^400tÞ ðmJÞ

7.24 A seriesRC circuit, withR¼5 kandC¼20mF, has two voltage sources in series, v₁¼25uðtÞ ðVÞ v₂¼25uðtt⁰Þ ðVÞ

Obtain the complete expression for the voltage across the capacitor and make a sketch, ift⁰is positive.

The capacitor voltage is continuous. Fort0,v₁results in a capacitor voltage of 25 V.

For 0tt⁰, both sources are zero, so thatv_Cdecays exponentially from 25 V towards zero:

v_C¼25e^t=RC¼25e^10t ðVÞ ð0tt⁰Þ In particular,v_Cðt⁰Þ ¼25e^10t0 (V).

Fortt⁰,v_Cbuilds fromv_Cðt⁰Þtowards the final value 25 V established byv₂: v_C¼ ½v_Cðt⁰Þ v_Cð1Þe^ðtt0Þ=RCþv_Cð1Þ

¼25½1 ðe^10t01Þe^10t ðVÞ ðtt⁰Þ Thus, for allt,

v_C¼25uðtÞ þ25e^10t½uðtÞ uðtt⁰Þ þ25½1 ðe^10t01Þe^10tuðtt⁰Þ ðVÞ See Fig. 7-36.

Fig. 7-34 Fig. 7-35

Supplementary Problems

7.25 The capacitor in the circuit shown in Fig. 7-37 has initial chargeQ₀¼800mC, with polarity as indicated. If the switch is closed att¼0, obtain the current and charge, fort>0.

Ans: i¼ 10e^{25 000t} ðAÞ;q¼410⁴ð1þe^{25 000t}Þ ðCÞ

7.26 A 2-mF capacitor, with initial chargeQ₀¼100mC, is connected across a 100-resistor att¼0. Calculate the time in which the transient voltage across the resistor drops from 40 to 10 volts. Ans: 0:277 ms

7.27 In theRCcircuit shown in Fig. 7-38, the switch is closed on position1att¼0 and then moved to2after the passage of one time constant. Obtain the current transient for (a) 0<t< ; ðbÞ t> .

Ans: ðaÞ 0:5e^200t ðAÞ; ðbÞ 0:516e^200ðtÞ (A)

7.28 A 10-mF capacitor, with initial chargeQ₀, is connected across a resistor att¼0. Given that the power transient for the capacitor is 800e^4000t (W), findR,Q₀, and the initial stored energy in the capacitor.

Ans: 50;2000mC;0:20 J

7.29 A seriesRLcircuit, withR¼10andL¼1 H, has a 100-V source applied att¼0. Find the current for t>0. Ans: 10ð1e^10tÞ (A)

7.30 In Fig. 7-39, the switch is closed on position1att¼0, then moved to2att¼1 ms. Find the time at which the voltage across the resistor is zero, reversing polarity. Ans: 1:261 ms

7.31 A series RLcircuit, withR¼100 and L¼0:2 H, has a 100-V source applied at t¼0; then a second source, of 50 V with the same polarity, is switched in att¼t⁰, replacing the first source. Findt⁰such that the current is constant at 0.5 A fort>t⁰. Ans: 1:39 ms

Fig. 7-36

Fig. 7-37

Fig. 7-38

7.32 The circuit of Problem 7.31 has a 50-V source ofoppositepolarity switched in att¼0:50 ms, replacing the first source. Obtain the current for (a) 0<t<0:50 ms; ðbÞ t>0:50 ms.

Ans: ðaÞ 1e^500t ðAÞ; ðbÞ 0:721e500ðt0:0005Þ0:50 ðAÞ

7.33 A voltage transient, 35e^500t (V), has the value 25 V at t₁¼6:7310⁴s. Show that at t¼t₁þ the function has a value 36.8 percent of that att₁:

7.34 A transient that increases from zero toward a positive steady-state magnitude is 49.5 att₁¼5:0 ms, and 120 att₂¼20:0 ms. Obtain the time constant. Ans: 12:4 ms

7.35 The circuit shown in Fig. 7-40 is switched to position1att¼0, then to position2att¼3. Find the transient currentifor (a) 0<t<3; ðbÞ t>3.

Ans: ðaÞ 2:5e^{50 000t} ðAÞ; ðbÞ 1:58e66 700ðt0:00006Þ (A)

7.36 AnRLcircuit, withR¼300andL¼1 H, has voltagev¼100 cosð100tþ458Þ (V) applied by closing a switch att¼0. [A convenient notation has been used for the phase ofv, which, strictly, should be indicated as 100tþ ð=4Þ(rad).] Obtain the resulting current fort>0.

Ans: 0:282e^300tþ0:316 cosð100tþ26:68Þ (A)

7.37 The RC circuit shown in Fig. 7-41 has an initial charge on the capacitor Q₀¼25mC, with polarity as indicated. The switch is closed att¼0, applying a voltagev¼100 sinð1000tþ308Þ (V). Obtain the current fort>0. Ans: 153:5e^4000tþ48:4 sinð1000tþ1068Þ (mA)

7.38 What initial charge on the capacitor in Problem 7.37 would cause the current to go directly into the steady state without a transient? Ans: 13:37mC (þon top plate)

7.39 Write simultaneous differential equations for the circuit shown in Fig. 7-42 and solve fori₁andi₂. The switch is closed att¼0 after having been open for an extended period of time. (This problem can also be solved by applying known initial and final conditions to general solutions, as in Problem 7-17.)

Ans: i₁¼1:67e^6:67tþ5 ðAÞ;i₂¼ 0:555e^6:67tþ5 ðAÞ Fig. 7-39

Fig. 7-40 Fig. 7-41

7.40 For the RL circuit shown in Fig. 7-43, find the current i_L at the following times: (a) 1 ms, (b) 0^þ, (c) 0.3 ms, (d) 1. Ans: ðaÞ 2:00 A; ðbÞ 2:00 A; ðcÞ 2:78 A; ðdÞ 3:00 A

7.41 A series RC circuit, with R¼2 k and C¼40mF, has two voltage sources in series with each other, v₁¼50 V andv₂¼ 100uðtÞ(V). Find (a) the capacitor voltage att¼, (b) the time at which the capa- citor voltage is zero and reversing polarity. Ans: ðaÞ 13:2 V; ðbÞ 55:5 ms

7.42 Find the unit-impulse response of the circuit of Fig. 7-44; i.e.,v₂forv₁¼ðtÞ(a unit-area narrow voltage pulse).

Ans: v₂¼ R₂

R₁ ðtÞ 1

R₁Ce^t=ðR1CÞuðtÞ

7.43 In the circuits of Fig. 7-45, RC¼510⁷ and v₁ðtÞ ¼10þcosð1000tÞ þ3 cosð2000tÞ. Find v₂ðtÞ.

Assume tan when <18. Ans: v₂ðtÞ 10þ cos½1000ðt10⁶Þ þ3 cos½2000ðt10⁶Þ ¼ v₁ðt10⁶Þ

Fig. 7-42 Fig. 7-43

Fig. 7-44

Fig. 7-45

7.44 The input voltage in the circuits of 7-45 is a weighted sum of sinusoids with the highest frequencyf₀ Hz.

Assuming thatRC<1=ð360f₀Þ, findv₂ðtÞ. Ans: v₂ðtÞ v₁ðt2RCÞ

7.45 Find the relationship betweenv₂andv₁in the circuit of Fig. 7-46.

Ans: v₂þRCdv₂ dt ¼2v₁

7.46 In the circuit of Fig. 7-47, find the differential equation relatingv₂ tov₁. Compare with the circuit of Fig. 7-45(a) of Example 7.17.

Ans: v₂þRCdv₂ dt ¼1

2 v₁RC dv₁ dt

7.47 In the circuit of Fig. 7-48, find the relationship betweenv₂andv₁. Ans: v₂þR₁C₁dv₂

dt ¼ C₁

C₂ v₁R₂C₂dv₁ dt

Fig. 7-46

Fig. 7-47

Fig. 7-48

Fig. 7-49

7.48 In the circuit of Fig. 7-49, letk¼0. Findvandiafter the switch is closed att¼0.

Ans: v¼e^t;i¼10:5e^t

7.49 Show that the segment of the circuit enclosed by the dashed box in the circuit of Fig. 7-49 is equivalent to an inductor with valueL¼1=ð1kÞH. Hint: Write KVL between terminals AB of the dashed box.

7.50 The switch in the circuit of Fig. 7-49 is closed at t¼0. Find v at t>0 for the following values ofk:

(a) 0.5, (b) 1, (c) 2. Ans: ðaÞ v¼e^t=2; ðbÞ v¼1; ðcÞ v¼e^t

7.51 Findi, the current drawn from the battery, in Problem 7.50.

Ans: ðaÞ i¼10:5e^t=2; ðbÞ i¼0:5; ðcÞ i¼10:5e^t

161

Higher-Order Circuits

and Complex Frequency