chapter 05 - Chapter 5 Circuit Theorems Exercises Ex 5.3-1...

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Unformatted text preview: Chapter 5 Circuit Theorems Exercises Ex 5.3-1 R = 10 Ω and is = 1.2 A. Ex 5.3-2 R = 10 Ω and is = −1.2 A. Ex 5.3-3 R = 8 Ω and vs = 24 V. Ex 5.3-4 R = 8 Ω and vs = −24 V. Ex 5.4-1 vm = 20 10 2 (15) + 20 − ( 2 ) = 6 + 20(− ) = −2 V 10 + 20 + 20 5 10 + (20 + 20) Ex 5.4-2 im = 25 3 − ( 5) = 5 − 3 = 2 A 3+ 2 2+3 Ex 5.4-3 3 3 vm = 3 ( 5) − (18 ) = 5 − 6 = −1 A 3 + (3 + 3) 3 + (3 + 3) 5-1 Ex 5.5-1 5-2 Ex 5.5-2 ia = 2 i a − 12 ⇒ i a = −3 A 6 voc = 2 i a = −6 V 12 + 6 i a = 2 i a 3 i sc = 2 i a Rt = ⇒ i a = −3 A ⇒ i sc = 2 ( −3 ) = − 2 A 3 −6 =3Ω −2 Ex 5.6-1 5-3 Ex 5.6-2 ia = 2 i a − 12 ⇒ i a = −3 A 6 voc = 2 i a = −6 V 12 + 6 i a = 2 i a 3 i sc = 2 i a Rt = ⇒ i a = −3 A ⇒ i sc = 2 ( −3 ) = − 2 A 3 −6 =3Ω −2 5-4 Ex 5.6-3 12 × 24 12 × 24 = = 8Ω 12 + 24 36 24 voc = ( 30 ) = 20 V 12 + 24 Rt = i= 20 8+ R Ex 5.7-1 voc = 6 (18) = 12 V 6+3 Rt = 2 + ( 3) ( 6 ) = 4 Ω 3+ 6 For maximum power, we require R L = Rt = 4 Ω Then 2 pmax = voc 122 = =9 W 4 Rt 4 ( 4 ) 5-5 Ex 5.7-2 1 50 3 i sc = ( 5.6 ) = ( 5.6 ) = 5 A 11 1 50 + 1 + 5 + + 3 150 30 150 ( 30 ) Rt = 3 + = 3 + 25 = 28 Ω 150 + 30 2 R t i sc ( 28 ) 52 = 175 W pmax = = 4 4 Ex 5.7-3 10 R L 100 R L p=iv= (10 ) = 2 Rt + R L Rt + R L ( Rt + R L ) The power increases as Rt decreases so choose Rt = 1 Ω. Then pmax = i v = 100 ( 5 ) (1 + 5) 2 = 13.9 W Ex 5.7-4 From the plot, the maximum power is 5 W when R = 20 Ω. Therefore: Rt = 20 Ω and 2 pmax v = oc 4 Rt ⇒ voc = pmax 4 Rt = 5 ( 4 ) 20 = 20 V 5-6 Problems Section 5-3: Source Transformations P5.3-1 (a) 5-7 ∴ Rt = 2 Ω (b) (c) −9 − 4i − 2i + (−0.5) = 0 −9 + (−0.5) i= = −1.58 A 4+2 v = 9 + 4 i = 9 + 4(−1.58) = 2.67 V ia = i = − 1.58 A vt = − 0.5 V (checked using LNAP 8/15/02) P5.3-2 Finally, apply KVL: −10 + 3 ia + 4 ia − 16 =0 3 ∴ ia = 2.19 A (checked using LNAP 8/15/02) 5-8 P5.3-3 Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors: Equivalents for series resistors, series voltage source at left; series resistors, then source transformation at top: Source transformation at left; series resistors at right: Parallel resistors, then source transformation at left: 5-9 Finally, apply KVL to loop − 6 + i (9 + 19) − 36 − vo = 0 i = 5 / 2 ⇒ vo = −42 + 28 (5 / 2) = 28 V (checked using LNAP 8/15/02) P5.3-4 − 4 − 2000 ia − 4000 ia + 10 − 2000 ia − 3 = 0 ∴ ia = 375 µ A (checked using LNAP 8/15/02) 5-10 P5.3-5 −12 − 6 ia + 24 − 3 ia − 3 = 0 ⇒ ia = 1 A (checked using LNAP 8/15/02) P5.3-6 A source transformation on the right side of the circuit, followed by replacing series resistors with an equivalent resistor: Source transformations on both the right side and the left side of the circuit: 5-11 Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources with an equivalent current source: Finally, va = 50 (100 ) 100 ( 0.21) = ( 0.21) = 7 V 50 + 100 3 (checked using LNAP 8/15/02) 5-12 Section 5-4 Superposition P5.4–1 Consider 6 A source only (open 9 A source) Use current division: v1 15 =6 ⇒ v1 = 40 V 20 15 + 30 Consider 9 A source only (open 6 A source) Use current division: v2 10 =9 ⇒ v2 = 40 V 20 10 + 35 ∴ v = v1 + v2 = 40 + 40 = 80 V (checked using LNAP 8/15/02) P5.4-2 Consider 12 V source only (open both current sources) KVL: 20 i1 + 12 + 4 i1 + 12 i1 = 0 ⇒ i1 = −1 / 3 mA Consider 3 mA source only (short 12 V and open 9 mA sources) Current Division: 4 16 i2 = 3 = 3 mA 16 + 20 5-13 Consider 9 mA source only (short 12 V and open 3 mA sources) Current Division: 12 i3 = −9 = −3 mA 24 + 12 ∴ i = i1 + i2 + i3 = − 1 / 3 + 4 / 3 − 3 = − 2 mA (checked using LNAP 8/15/02) P5.4–3 Consider 30 mA source only (open 15 mA and short 15 V sources). Let i1 be the part of i due to the 30 mA current source. 2 ia = 30 = 6 mA ⇒ 2+8 6 i1 = ia = 2 mA 6 + 12 Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of i due to the 15 mA current source. 4 ib = 15 = 6 mA ⇒ 4+6 6 i2 = ib = 2 mA 6 + 12 5-14 Consider 15 V source only (open both current sources). Let i3 be the part of i due to the 15 V voltage source. 6 || 6 3 i3 = − 2.5 ( 6 || 6 ) + 12 = − 10 3 + 12 = −0. 5 mA Finally, i = i1 + i2 + i3 = 2 + 2 − 0. 5 = 3.5 mA (checked using LNAP 8/15/02) P5.4–4 Consider 10 V source only (open 30 mA source and short the 8 V source) Let v1 be the part of va due to the 10 V voltage source. 100 || 100 (10 ) (100 || 100 ) + 100 v1 = = Consider 8 V source only (open 30 mA source and short the 10 V source) 50 10 (10 ) = V 150 3 Let v2 be the part of va due to the 8 V voltage source. v1 = = 100 || 100 (8) (100 || 100 ) + 100 50 8 (8) = V 150 3 5-15 Consider 30 mA source only (short both the 10 V source and the 8 V source) Let v2 be the part of va due to the 30 mA current source. v3 = (100 || 100 || 100)(0.03) = va = v1 + v2 + v3 = Finally, 100 (0.03) = 1 V 3 10 8 + +1 = 7 V 33 (checked using LNAP 8/15/02) P5.4-5 Consider 8 V source only (open the 2 A source) Let i1 be the part of ix due to the 8 V voltage source. Apply KVL to the supermesh: 6 ( i1 ) + 3 ( i 1 ) + 3 ( i 1 ) − 8 = 0 i1 = Consider 2 A source only (short the 8 V source) 82 =A 12 3 Let i2 be the part of ix due to the 2 A current source. Apply KVL to the supermesh: 6 ( i 2 ) + 3 ( i 2 + 2 ) + 3 i2 = 0 i2 = Finally, i x = i1 + i 2 = −6 1 =− A 12 2 211 −=A 326 5-16 Section 5-5: Thèvenin’s Theorem P5.5-1 (checked using LNAP 8/15/02) 5-17 P5.5-2 The circuit from Figure P5.5-2a can be reduced to its Thevenin equivalent circuit in four steps: (a) (b) ( c) (d) Comparing (d) to Figure P5.5-2b shows that the Thevenin resistance is Rt = 16 Ω and the open circuit voltage, voc = −12 V. 5-18 P5.5-3 The circuit from Figure P5.5-3a can be reduced to its Thevenin equivalent circuit in five steps: (a) (b) ( c) (d) ( e) Comparing (e) to Figure P5.5-3b shows that the Thevenin resistance is Rt = 4 Ω and the open circuit voltage, voc = 2 V. (checked using LNAP 8/15/02) 5-19 P5.5-4 Find Rt: Rt = 12 (10 + 2 ) =6Ω 12 + (10 + 2 ) Write mesh equations to find voc: Mesh equations: 12 i1 + 10 i1 − 6 ( i2 − i1 ) = 0 6 ( i2 − i1 ) + 3 i 2 − 18 = 0 28 i1 = 6 i 2 9 i 2 − 6 i 1 = 18 36 i1 = 18 ⇒ i1 = i2 = Finally, 1 A 2 14 1 7 = A 3 2 3 7 1 voc = 3 i 2 + 10 i1 = 3 + 10 = 12 V 3 2 (checked using LNAP 8/15/02) 5-20 P5.5-5 Find voc: Notice that voc is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage: va = −voc Apply KCL at node a: 6 − voc voc 3 − + − voc = 0 + 8 4 4 −6 + voc + 2 voc − 6 voc = 0 ⇒ voc = −2 V Find Rt: We’ll find isc and use it to calculate Rt. Notice that the short circuit forces va = 0 Apply KCL at node a: 6−0 0 3 − + + − 0 + i sc = 0 844 i sc = Rt = 63 =A 84 voc −2 8 = =− Ω i sc 3 4 3 (checked using LNAP 8/15/02) 5-21 P5.5-6 Find voc: 2 va − va va = + 3 + 0 ⇒ va = 18 V 3 6 The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V Apply KCL at the top, middle node: Find isc: 2 va − va va v = + 3 + a ⇒ va = −18 V 3 6 3 va −18 i sc = = = −6 V Apply Ohm’s law to the right-hand 3 Ω resistor : 3 3 v 18 Finally: R t = oc = = −3 Ω i sc −6 Apply KCL at the top, middle node: (checked using LNAP 8/15/02) 5-22 P5.5-7 (a) −vs + R1 ia + ( d + 1) R 2 ia = 0 ia = vs R1 + ( d + 1) R 2 v oc = ( d + 1) R 2vs R1 + ( d + 1) R 2 vs R1 ia = i sc = ( d + 1) ia = −ia − d ia + ( d + 1) vs R1 vT − iT = 0 R2 R1 ia = −vT iT = ( d + 1) vT vT R 2 ( d + 1) + R1 + = × vT R1 R 2 R1 R 2 Rt = R1 R 2 vT = iT R1 + ( d + 1) R 2 (b) Let R1 = R2 = 1 kΩ. Then 625 Ω = R t = and 5 = voc = 1000 1000 ⇒ d= − 2 = −0.4 A/A 625 d +2 ( d + 1) vs d +2 ⇒ vs = −0.4 + 2 5 = 13.33 V −0.4 + 1 (checked using LNAP 8/15/02) 5-23 P5.5-8 From the given data: 2000 voc R t + 2000 voc = 1.2 V ⇒ 4000 R t = −1600 Ω voc 2= R t + 4000 6= When R = 8000 Ω, v= R voc Rt + R v= 8000 (1.2 ) = 1.5 V −1600 + 8000 P5.5-9 From the given data: voc R t + 2000 voc = 24 V ⇒ voc R t = 4000 Ω 0.003 = R t + 4000 0.004 = i= voc Rt + R (a) When i = 0.002 A: 24 0.002 = ⇒ R = 8000 Ω 4000 + R (b) Maximum i occurs when R = 0: 24 = 0.006 = 6 mA ⇒ i ≤ 6 mA 4000 P5.5-10 The current at the point on the plot where v = 0 is the short circuit current, so isc = 20 mA. The voltage at the point on the plot where i = 0 is the open circuit voltage, so voc = −3 V. The slope of the plot is equal to the negative reciprocal of the Thevenin resistance, so 1 0 − 0.002 − = ⇒ R t = −150 Ω Rt −3 − 0 5-24 P5.5-11 −12 + 6000 ia + 2000 ia + 1000 ia = 0 ia = 4 3000 A voc = 1000 ia = 4 V 3 ia = 0 due to the short circuit −12 + 6000 isc = 0 ⇒ isc = 2 mA 4 voc Rt = = 3 = 667 Ω isc .002 4 3 ib = 667 + R ib = 0.002 A requires 4 3 − 667 = 0 R= 0.002 (checked using LNAP 8/15/02) 5-25 P5.5-12 10 = i + 0 ⇒ i = 10 A voc + 4 i − 2 i = 0 ⇒ voc = −2 i = −20 V i + i sc = 10 ⇒ i = 10 − i sc 4 i + 0 − 2 i = 0 ⇒ i = 0 ⇒ i sc = 10 A Rt = −2 = iL = voc −20 = = −2 Ω isc 10 −20 ⇒ RL = 12 Ω RL − 2 (checked using LNAP 8/15/02) 5-26 Section 5-6: Norton’s Theorem P5.6-1 When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor in Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will cool off. P5.6-2 (checked using LNAP 8/16/02) 5-27 P5.6-3 P5.6-4 To determine the value of the short circuit current, isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 5.6-4a after adding the short circuit and labeling the short circuit current. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (a), mesh current i2 is equal to the current in the short circuit. Consequently, i2 = isc . The controlling current of the CCVS is expressed in terms of the mesh currents as i a = i1 − i 2 = i1 − isc Apply KVL to mesh 1 to get 3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 7 i1 − 4 i 2 = 10 (1) Apply KVL to mesh 2 to get 5 i 2 − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = 0 ⇒ i1 = 11 i2 6 Substituting into equation 1 gives 11 7 i 2 − 4 i 2 = 10 ⇒ i 2 = 1.13 A ⇒ i sc = 1.13 A 6 5-28 Figure (a) Calculating the short circuit current, isc, using mesh equations. To determine the value of the Thevenin resistance, Rt, first replace the 10 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as v Rt = T iT In Figure (b), the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (b), mesh current i2 is equal to the negative of the current source current. Consequently, i2 = i T . The controlling current of the CCVS is expressed in terms of the mesh currents as i a = i1 − i 2 = i1 + i T Apply KVL to mesh 1 to get 3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) = 0 ⇒ 7 i1 − 4 i 2 = 0 ⇒ i1 = 4 i2 7 (2) Apply KVL to mesh 2 to get 5 i 2 + vT − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = −vT Substituting for i1 using equation 2 gives 4 −6 i 2 + 11 i 2 = −vT 7 Finally, Rt = ⇒ 7.57 i 2 = −vT vT −vT −vT = = = 7.57 Ω −iT iT i2 5-29 Figure (b) Calculating the Thevenin resistance, R t = vT , using mesh equations. iT To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure 4.6-4a after adding the open circuit and labeling the open circuit voltage. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (c), mesh current i2 is equal to the current in the open circuit. Consequently, i2 = 0 A . The controlling current of the CCVS is expressed in terms of the mesh currents as i a = i1 − i 2 = i1 − 0 = i1 Apply KVL to mesh 1 to get 3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 3 i1 − 2 ( i1 − 0 ) + 6 ( i1 − 0 ) − 10 = 0 ⇒ i1 = 10 = 1.43 A 7 Apply KVL to mesh 2 to get 5 i 2 + voc − 6 ( i1 − i 2 ) = 0 ⇒ voc = 6 ( i1 ) = 6 (1.43) = 8.58 V Figure (c) Calculating the open circuit voltage, voc, using mesh equations. As a check, notice that R t isc = ( 7.57 ) (1.13) = 8.55 ≈ voc (checked using LNAP 8/16/02) 5-30 P5.6-5 To determine the value of the short circuit current, Isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 4.6-5a after adding the short circuit and labeling the short circuit current. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = −24 V . The voltage at node 3 is equal to the voltage across a short, v3 = 0 . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the voltage across a short, i.e. v3 = 0 . Apply KCL at node 2 to get v1 − v 2 3 = v 2 − v3 6 ⇒ 2 v1 + v 3 = 3 v 2 ⇒ − 48 = 3 v a ⇒ v a = −16 V Apply KCL at node 3 to get v2 − v3 6 + 4 v 2 = isc 3 ⇒ 9 v a = isc 6 ⇒ isc = 9 ( −16 ) = −24 A 6 Figure (a) Calculating the short circuit current, Isc, using mesh equations. To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as v R th = T iT Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. 5-31 In Figure (b), node voltage v1 is equal to the across a short circuit, i.e. v1 = 0 . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the voltage across the current source, i.e. v3 = vT . Apply KCL at node 2 to get v1 − v 2 3 = v 2 − v3 6 ⇒ 2 v1 + v 3 = 3 v 2 ⇒ vT = 3 v a Apply KCL at node 3 to get v2 − v3 6 + 4 v 2 + iT = 0 ⇒ 9 v 2 − v3 + 6 iT = 0 3 ⇒ 9 v a − vT + 6 iT = 0 ⇒ 3 v T − vT + 6 iT = 0 ⇒ 2 vT = −6 iT Finally, Rt = vT = −3 Ω iT Figure (b) Calculating the Thevenin resistance, R th = vT , using mesh equations. iT To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure P 4.6-5a after adding the open circuit and labeling the open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = −24 V . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the open circuit voltage, i.e. v3 = voc . Apply KCL at node 2 to get 5-32 v1 − v 2 3 = v 2 − v3 6 ⇒ 2 v1 + v 3 = 3 v 2 ⇒ − 48 + v oc = 3 v a Apply KCL at node 3 to get v2 − v3 6 + 4 v 2 = 0 ⇒ 9 v 2 − v 3 = 0 ⇒ 9 v a = v oc 3 Combining these equations gives 3 ( −48 + voc ) = 9 v a = voc ⇒ voc = 72 V Figure (c) Calculating the open circuit voltage, voc, using node equations. As a check, notice that R th I sc = ( −3) ( −24 ) = 72 = Voc (checked using LNAP 8/16/02) Section 5-7: Maximum Power Transfer P5.7-1 a) For maximum power transfer, set RL equal to the Thevenin resistance: R L = R t = 100 + 1 = 101 Ω b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin equivalent circuit: 5-33 The voltage across RL is vL = Then pmax 101 (100 ) = 50 V 101 + 101 2 v 502 =L= = 24.75 W R L 101 P5.7-2 Reduce the circuit using source transformations: Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 Ω and (b) the value of that maximum power is P = i 2 ( R) = (0.03)2 (60) = 54 mW max R 5-34 P5.7-3 RL v L = vS RS + R L 2 2 vL vS RL ∴ pL = = R L ( RS + R L )2 By inspection, pL is max when you reduce RS to get the smallest denominator. ∴ set RS = 0 P5.7-4 Find Rt by finding isc and voc: The current in the 3 Ω resistor is zero because of the short circuit. Consequently, isc = 10 ix. Apply KCL at the top-left node to get ix + 0.9 = 10 ix ⇒ ix = 0.9 = 0.1 A 9 so Next isc = 10 ix = 1A Apply KCL at the top-left node to get 5-35 ix + 0.9 = 10 ix ⇒ ix = 0.9 = 0.1 A 9 Apply Ohm’s law to the 3 Ω resistor to get voc = 3 (10 ix ) = 30 ( 0.1) = 3 V For maximum power transfer to RL: R L = Rt = voc 3 = =3Ω isc 1 The maximum power delivered to RL is given by 2 pmax = voc 32 3 = =W 4 R t 4 ( 3) 4 P5.7-5 The required value of R is R = Rt = 8 + ( 20 + 120 ) (10 + 50 ) = 50 Ω ( 20 + 120 ) + (10 + 50 ) 30 170 voc = ( 20 ) 10 − 170 + 30 ( 20 ) 50 170 + 30 170(20)(10) − 30(20)(50) 4000 = = = 20 V 200 200 The maximum power is given by 2 v 202 pmax = oc = =2W 4 R t 4 ( 50 ) 5-36 PSpice Problems SP5-1 a = 0.3333 b = 0.3333 c =33.33 V/A (a) (b) vo = 0.3333 v1 + 0.3333 v2 + 33.33 i 3 7 = 0.3333 (10 ) + 0.3333 ( 8 ) + 33.33 i 3 18 3 = 3 = 30 mA ⇒ i3 = 100 100 3 7− 5-37 SP5-2 Before the source transformation: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 V_V2 -3.000E-02 -4.000E-02 After the source transformation: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V2 -4.000E-02 5-38 SP5-3 voc = −2 V VOLTAGE SOURCE CURRENTS NAME CURRENT V_V3 V_V4 -7.500E-01 7.500E-01 isc = 0.75 A Rt = −2.66 Ω 5-39 SP5-4 voc = 8.571 V VOLTAGE SOURCE CURRENTS NAME CURRENT V_V5 -2.075E+00 V_V6 1.132E+00 X_H1.VH_H1 9.434E-01 isc = 1.132 A Rt = 7.571 Ω 5-40 Verification Problems VP5-1 Use the data in the first two lines of the table to determine voc and Rt: voc Rt + 0 voc = 39.9 V ⇒ voc R t = 410 Ω 0.0438 = R t + 500 0.0972 = Now check the third line of the table. When R= 5000 Ω: v 39.9 i = oc = = 7.37 mA R t + R 410 + 5000 which disagree with the data in the table. The data is not consistent. VP5-2 Use the data in the table to determine voc and isc: voc = 12 V (line 1 of the table) isc = 3 mA so Rt = (line 3 of the table) voc = 4 kΩ isc Next, check line 2 of the table. When R = 10 kΩ: v 12 i = oc = = 0.857 mA 3 R t + R 10 (10 ) + 5 (103 ) To cause i = 1 mA requires which agrees with the data in the table. v 12 0.001 = i = oc = ⇒ R = 8000 Ω R t + R 10 (103 ) + R I agree with my lab partner’s claim that R = 8000 causes i = 1 mA. 5-41 VP5-3 60 60 voc 11 i= = = 11 = 54.55 mA R t + R 6 110 + 40 60 + 40 () 11 The measurement supports the prelab calculation. Design Problems DP5-1 The equation of representing the straight line in Figure DP 5-1b is v = − R t i + voc . That is, the slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the 0−5 open circuit voltage. Therefore: R t = − = 625 Ω and voc = 5 V. 0.008 − 0 Try R1 = R 2 = 1 kΩ . (R1 || R2 must be smaller than Rt = 625 Ω.) Then 5= R2 1 vs = vs 2 R1 + R 2 ⇒ vs = 10 V and R1 R2 = R3 + 500 ⇒ R3 = 125 Ω R1 + R2 Now vs, R1, R2 and R3 have all been specified so the design is complete. 625 = R 3 + DP5-2 The equation of representing the straight line in Figure DP 5-2b is v = − R t i + voc . That is, the slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the 0 − ( −3 ) = 500 Ω and voc = −3 V. open circuit voltage. Therefore: R t = − −0.006 − 0 From the circuit we calculate R 3 ( R1 + R 2 ) R1 R 3 and voc = − is Rt = R1 + R 2 + R 3 R1 + R 2 + R 3 so R 3 ( R1 + R 2 ) R1 R 3 500 Ω = is and −3 V = − R1 + R 2 + R 3 R1 + R 2 + R 3 5-42 Try R 3 = 1kΩ and R1 + R 2 = 1kΩ . Then R t = 500 Ω and −3 = − 1000 R1 is = − R1 i s ⇒ 6 = R1 i s 2000 2 This equation can be satisfied by taking R1 = 600 Ω and is = 10 mA. Finally, R2 = 1 kΩ - 400 Ω = 400 Ω. Now is, R1, R2 and R3 have all been specified so the design is complete. DP5-3 The slope of the graph is positive so the Thevenin resistance is negative. This would require R1 R 2 R3 + < 0 , which is not possible since R1, R2 and R3 will all be non-negative. R1 + R 2 Is it not possible to specify values of vs, R1, R2 and R3 that cause the current i and the voltage v in Figure DP 5-3a to satisfy the relationship described by the graph in Figure DP 5-3b. DP5-4 The equation of representing the straight line in Figure DP 5-4b is v = − R t i + voc . That is, the slope of the line is equal to the Thevenin impedance and the "v - intercept" is equal to the open −5 − 0 circuit voltage. Therefore: R t = − = −625 Ω and voc = −5 V. 0 − 0.008 The open circuit voltage, voc, the short circuit current, isc, and the Thevenin resistance, Rt, of this circuit are given by R 2 ( d + 1) voc = vs R1 + ( d + 1) R 2 , ( d + 1) v isc = s R1 and R1 R 2 Rt = R1 + ( d + 1) R 2 Let R1 = R2 = 1 kΩ. Then −625 Ω = R t = and −5 = ( d + 1) vs d +2 1000 1000 ⇒ d= − 2 = −3.6 A/A d +2 −625 ⇒ vs = −3.6 + 2 ( − 5 ) = −3.077 V −3.6 + 1 Now vs, R1, R2 and d have all been specified so the design is complete. 5-43 ...
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This note was uploaded on 06/10/2011 for the course PHYSICS physics taught by Professor Physics during the Spring '11 term at HKU.

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