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PracticeProblems1 - ,fx'g Chapter 1 Review of Linear...

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Unformatted text preview: ,fx'g Chapter 1 Review of Linear Circuit Theory 1.2 1.4 Plot the v-i characteristic of the following circuit when Va = 5 V. ' R = 1 k0 i >—o a 4. Va 5 V V a) A 2 kn resistor is connected across the terminals a—a’ in the circuit shown above. Find the operating point of the resistor using the graphical method. Verify the result using Uhm’ Law. Find the Thevenin equivalent of the circuit shown below, as seen from the terminals a—a’. R2 = 20 kn 11 5mA R1=10 Id: In the following circuit, use superposition to find the Thevenin equivalent seen at the terminals b-b’ The circuit shown below contains a dependent source. Find the Thevenin equivalent of the circuit as seen at the terminals a—a’ The circuit shown below contains a dependent source. Find the Thevenin equivalent of the circuit as seen at the terminals b-b’. Use the "test source" method to find RTh- The circuit shown below contains a current—dependent current source. I Find the equivalent resistance seen looking into the terminals a-a’. Use the "test source" method. a 1.8 Find the Norton equivalent of the circuit shown below. R = 22 k0 a 9 V Va """"" x5 3, 1.9 Find the Norton equivalent of the following circuit: R2 = 5 k0 a 11 = 8 mA R1 = 12 kn a) 1.10 The circuit shown below contains a dependent source. Find the Norton equivalent of the circuit as seen at terminals a-a’. This circuit is the same one analyzed in Prob. 1.5. Measurements with a voltmeter are made 'at the terminals of a circuit that contains only resistors, dc current sources, and fixed voltage sources. The meter has an internal resistance of 1 MD. With no other loads applied, the measured voltage is 12 V. With an additional external 1 Mn resistance connected in parallel with the meter, the measured voltage is 10 V. Find the Thevenin equivalent of the resistive circuit as seen at the measured terminals. A circuit is powered from a 10-V voltage supply, as shown below. When the circuit drives a 10 kfl resistive load at 5 V, it draws 2 mA from this supply. Find the power dissipated in the load; find the power dissipated in the circuit. 2 mA Voltage supply SOLUTIONS T0 PROBLEMS Chapter 1 - Since the circuit is resistive (i.e., contains only a resistor and a constant voltage source), it’s v-i characteristic is a straight line. The v-i characteristic can be plotted by finding two points. For i = 0 (open circuit), v = Va = 5 V (v-axis intercept of the v—i characteristic). For v = 0 (short circuit), i = Va/R = 5 mA (i—axis in— tercept). The resulting v-i charac- teristic is shown below. A i (M) 0123456v(V) The v—i characteristic of a 2 kn resistor consists of a straight line with slope '(1 mA)/(2 V) = 0.5 mA/V passing through the origin: 73 "““i‘tiaA; , Thevenin circuit operating point O 1 2 3 4 5 6 V (V) The point of intersection, i.e., the operating point, is seen to be 3.3 V; 1.6 mA. An analytical solution con- firms this result. From Uhm’s Law: 5 V i = -——-—-—-——~ = 1.67 mA g 1 k0 + 2 kn v = (1.66 mA)(2 kfl) = 3.33 V Alternatively, from the voltage divider relation: 2 kn v = 5 V —--—~—-—~— = 3.33 V 1 k0 + 2 k0 The Thevenin voltage is equal to the voltage at the terminals a-a’ ' under open-circuit conditions (i.e., with no other elements connected to the terminals): R2 =.20 kfl' i = 0 I1 I1 0 50v 5 mA — Under open circuit conditions, all of , , the current 11 flows through R1, To find vTh' find the separate causing a voltage drop contributions of V1 and V2 to vac. IlRl = (5mA)(10kfl) = 50 v Here is the circuit with V1 set to to develop across R1. No voltage zero and V2 ”°""i drop is developed across R2, since R the current through it is zero. R1 2 . b Thus, vac = 50 V. The Thevenin + resistance is equal to the resistance v seen at the terminals a-a’ with 11 9 R4 0C1 set to zero (open circuit): .-b: R2 = 20 k0 From the voltage divider relation v R4 v = -—-—- 0C1 2 R2 + R4 Here is the circuit with V2 set to zero and V1 ”on": By inspection, RTh = R1 + R2 = 30 k0. 0 Alternative method for finding RThi Find the current flowing through a short-circuit applied to the termi- nals a-a’: R = 20 k0 . . 2 a Because a short curcuut appears directly across the series combina— iSC tion of R2 and R4 (voltage drop equal to zero), vgcz = 0. Thus VTh is 1.7mA equal to v0c1 alone: v R4 VTh ’ V0C1 — 2 R2 + R4 The current flowing through the ‘ short-circuit is also equal to the T0 flnd RTh: set b°th v1 and V2 t° current flowing through R2. From the zero (short circuits) and. f'"d th? current divider relation: resistance seen at the terminals b-b R1 ‘ = I -———-—-— R R2 'SC 1 R1 + R2 1 b 10 kfl = 5 mA —--— = 1.67 mA ., (= RTh 10 kfl + 20 kfl ‘ .b’ ' 50 V RTh=fl=——=30kll ' . 'SC 1-67 mA By Inspection, RTh = R4||R2. Note that the R3||R1 combination is bypassed by the V2=0 short circuit. 74 Alternatively, RTh could be found by applying a short circuit to the terminals b-b’ and using superposi— tion to find the contributions of V1 and V2 to isc. Find the open—circuit voltage at the terminals a—a’. From KVL around the input loop, (R1 + R2)l2 + R403 4- 1)i2 = 0 Solving for i2 yields i2 = 0. Consequently, the dependent source is set to zero as well (open circuit), and RTh in the remaining circuit becomes just R3. ° Alternative method: With V1 on, the current flowing through.a short circuit applied to i a l i l l l l i i i The dependent source pulls a current the a-a’ terminals becomes ~fii2, 5° /,m1 p32 up from ground through R3, so that RTh = 19$ = —fi32R3 = 3 1_W} that vac = —pi2R3. Find i2 by apply— isc -pi2 ing KVL to the llinput loop" of the circuit. Note that a total current of (p + 1)i2 flows through R4, with 1.6 Ste 1: Re resent v R and i2 entering R4 via R2 and fiiz enter— p p S, 1’ ing R4 via the dependent source. R2 by a Simpler Thevenln curcu1t consisting of one resistor and one From KVL: . V1 = (R1 + R2)i2 + R4(fi + 1)i2 voltage source. Solving for i2 yields R R . v1 1l| 2 |=-—-—--—-—-—— 2 R1 + R2 + (fl + 1)R4 vs R2 so that fiR _______ vac = -pi2R3 = v 3 R1 + R2 1 R1 + R2 + (p + 1)R4 where VTh E vgc. T To find RTh: set V1 to zero (short circuit) and find the resistance seen looking into the terminals a-a’. Next find VOC (i.e., VTh at .the /’NE terminals b-b’) by applying KVL to i :7 the llinput loop”. Note that no current flows through the open 5 75 circuit at the terminals where v1 is defined, hence no current flows through R1||R2, and the voltage drop across the resistance R1||R2 is zero. Applying KVL yields VS __Eg___ = V1 + 9mV1R3 R1+R2 or R 1 v1,= Vs 2 R1 + R2 (1 + nga) The open—circuit voltage thus becomes R2 9mR3 R1 + R2 (1 + 9mR3) To find RTh: set V3 to zero and apply a ”test" voltage source to b— b’. The value of RTh is determined by Vtest/ltest- Note that a test current source could also be used, but a test voltage source simplifies the computation because it directly fixes v1 to a known value. voc = 9mV1R3 = vs Since i1 = O, the voltage drop across R1l|R2 is zero, so that v1 = ‘Vtest- The current ltest has two compo- nents, one through R3 and one through the dependent source: . Vtest _ vtest 'test= R ’ 9mV1 ' + gmvtest 3 R3 The ratio of Vtest to itest can be computed from this equation, i.e., —1 Vt t 1 _ 1 RTh= es = ['R—+gm] =R3H— 'test 3 76 Find RTh by applying a test source to the terminals a-a’. In this case, a test current source works best because it directly fixes i1 to the value itest' R1 R2 i2 a Applying KCL to node "a" yields ltest = i2 - fli1 = i2 - pitest Solving for i2 results in i2 = (P + 1)ltest Adding up the drops across R1 and R2 yields Vtest = 31R1 + i2R2 . = 'testRl + (fl + 1)‘testR2 so that v Rn. = .test = R1 + (p + 1m ltest The value of R2 as seen from a—a’ has essentially been multiplied by the factor (p + 1) via the action of the dependent source. The Norton equivalent consists of a current source and parallel resistance. The value of the source is equal to the short-circuit current measured at terminals a-a’: R = 22 kfl a where isc = Va/R = (9 V)/(22 k0) = 0.41 mA/V. The Norton resistance is found by. setting the Va source to zero (short circuit), yielding RN = R = 22 kfl. The complete Norton equivalent circuit is shown below. The Norton current is found by applying a short-circuit to the a-a’ terminals: R2 = 5 kn a From current division: ' R1 I 12 kne A 5 65 A I = -"""—— = In = . m 5C R1 + R2 1 17 kn The Norton resistance is found by setting I1 to zero (open circuit) and observing the net resistance at the terminals a—a’: R2 = 5 kn a II = 0 R1 = 12 kfl ¢== RN J a RN = R1 + R2 = 5 kn + 12 kn = 17 kn. The complete Norton equivalent of the circuit is shown below. 77 The open-circuit voltage of this circuit was found in Prob. 1.5: . -fiR3 R1 + R2 + (fi+ 1)R4 Similarly, the Thevenin resistance, _computed with V1 set to zero, was found to be RTh = R3. These elements of the Thevenin equivalent circuit can be used to find the element values of the Norton equivalent circuit. Specifically, RN = R3, and voc -fiV1 R3 . R1 + R2 + (p + 1)R4 The Norton equivalent of the original Voc = -fi32R3 = V1 circuit can be represented in the following form: fiV1 3 R1 + R2 + (p + 1)R4 R3 3, Here is a summary of the measurements. With the meter alone, the current out of the resistive circuit is (12 V)/( 1 M0) = 12 pA.: 12 pA Resistive 12 V circuit - With an additional 1 Mn resistor connected, the current out of the resistive circuit becomes 10 V (1 M“)! l(1 M“) 20 pA = 20 pA Resistive {O V circuit 'For the general Thevenin equivalent shown below, v = VTh - iRTh. RTh i+ + VTh v Applying the known data results in: 12 V = VTh - (12 FA)RTh: and 10 V = VTh - (20 pA)RTh. Simultaneous solution of these equa- tions yields VTh = 15 V; RTh = 250 kfl The power extracted from the supply is equal to (10 V)(2 mA) = 20 mW. The power dissipated in the load is (5 V)2/(1o kn) = 2.5 mW. (Alterna- tively, (5 V)(Op5 mA) = 2.5 mW). The power dissipated in the circuit is equal to the difference: 20 mW - 2.5 mW = 17.5 mW If the op-amp is ideal, the current into the v+ terminal will be zero, and no voltage drop will occur across R2. The voltage at v+ will thus equal VIN. The remaining 78 circuit will multiply v+ by the non- inverting gain, so that the output becomes R3 + R1 19 kn ——————-—v = R1 1” 3 kn VOUT = VIN = 6.3VIN If the op—amp is ideal, the current into the v+ terminal will be zero. This current is identical to the iIN flowing through R2, hence lIN = O and Rin = N. [:::::] An ideal op-amp saturates when VOUT reaches VCC or VEE- As shown in Prob. 2.1, the circuit has voltage gain Av = 6.3, hence VDUT will equal VCC for V C 10 V VIN 2 -— = *-- 5 1.6 V Av 6.3 Similarly, the op-amp will saturate at VEE for VE -15 V VIN S ——— = --—- 2 -2.4 V Av 6.3 Since the op—amp is ideal, the current into the v+ terminal will be zero. As a result, no current flows through R2, and the voltage drop across R2 becomes zero. With v+ held at ground potential by R2, the cir- cuit functions as a simple inverting amplifier with ' -R3 -30 k0 VOUT- avmnkn VIN = -7.5 VIN AVD 0.68 V - 0.70 V = —1.5 mV/°C AT _ 10 0c ...
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