Chapter 5 Solution P5.2- 2, 3, 6 P5.3-3, 5, 8, 15 P5.4-3, 6, 8, 16 P5.5-2, 4, 6, 11 P5.6-2, 4, 9 P 5.2-2 Consider the circuit of Figure P 5.2-2. Find ia by simplifying the circuit (using source transformations) to a single-loop circuit so that you need to write only one KVL equation to find ia. Figure P 5.2-2 Solution: 16 Finally, apply KVL: −+10 3ii + 4 − = 0 ∴= i 2.19 A aa3 a (checked using LNAP 8/15/02) P 5.2-3 Find vo using source transformations if i = 5/2 A in the circuit shown in Figure P 5.2-3. Hint: Reduce the circuit to a single mesh that contains the voltage source labeled vo. Answer: vo = 28 V Figure P 5.2-3 Solution: 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: Finally, apply KVL to loop −+6 iv (9 + 19) − 36 −o = 0 iv=5/ 2 ⇒=−+o 42 28 (5/ 2) = 28 V (checked using LNAP 8/15/02) P 5.2-6 Use source transformations to find the value of the voltage va in Figure P 5.2-6. Answer: va = 7 V Figure P 5.2-6 Solution: 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: Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources with an equivalent current source: 50( 100) 100 Finally, v = (0.21) =( 0.21) = 7 V a 50+ 100 3 (checked using LNAP 8/15/02) P5.3-3 The circuit shown in Figure P5.3-3 has two inputs, vs and is, and one output io. The output is related to the inputs by the equation ioss= ai + bv Given the following two facts: The output is io = 0.45 A when the inputs are is = 0.25 A and vs = 15 V. and The output is io = 0.30 A when the inputs are is = 0.50 A and vs = 0 V. Determine the values of the constants a and b and the values of the resistances are R 1 and R 2. Answers: a = 0.6 A/A, b = 0.02 A/V, R 1 = 30 Ω and R 2 = 20 Ω. Figure P5.3-3 Solution: From the 1st fact: 0.45=ab( 0.25) + ( 15) From the 2nd fact: 0.30 0.30=ab( 0.50) +( 0) ⇒= a =0.60 0.50 0.45− ( 0.60)( 0.25) Substituting gives 0.45=( 0.60)( 0.25) +bb( 15) ⇒= =0.02 15 Next, consider the circuit: R1 aiso= io1 = i = is vs =0 + RR12 R1 so 0.60 = ⇒=2RR12 3 RR12+ vs and bvso= io2 = i = is =0 RR12+ 11 so 0.02 = ⇒RR12 += =Ω50 RR12+ 0.02 Solving these equations gives R 1 = 30 Ω and R 2 = 20 Ω. Figure P5.3-5 P5.3-5 Determine v(t), the voltage across the vertical resistor in the circuit in Figure P5.3-5. Solution; We’ll use superposition. Let v1(t) the be the part of v(t) due to the voltage source acting alone. Similarly, let v2(t) the be the part of v(t) due to the voltage source acting alone. We can use these circuits to calculate v1(t) and v2(t). Notice that v1(t) is the voltage across parallel resistors. Using equivalent resistance, we calculate 40||10 = 8 Ω. Next, using voltage division we calculate 8 vt( ) =(12) = 2 V 1 8+ 40 Similarly v2(t) is the voltage across parallel resistors Using equivalent resistance we first determine 40||40 = 20 Ω and then calculate 20 vt( ) = (12cos( 5t)) = 8cos( 5 t) V 2 10+ 20 Using superposition vt( ) =+=+ v12( t) v( t) 2 8cos( 5t) V P 5.3-8 Use superposition to find the value of the current ix in Figure P 5.3-8. Answer: i = 3.5 mA Figure P5.3-8 Solution: 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(iii111) ++−= 3( ) 3( ) 80 82 i = = A 1 12 3 Consider 2 A source only (short the 8 V source) Let i2 be the part of ix due to the 2 A current source. Apply KVL to the supermesh: 6(ii22) + 3( ++ 23) i 2 = 0 −61 i = = − A 2 12 2 21 1 Finally, i=+=−= ii A x 12326 P 5.3-15 The circuit shown in Figure P 5.3-15 has three inputs: v1, i2, and v3. The output of the circuit is the current io. The output of the circuit is related to the inputs by i1 = avo + bv2 + ci3 where a, b, and c are constants. Determine the values of a, b, and c. Figure P 5.3-15 Solution: 10 v1 10 20 ii=− +− o2 10+ 40 20 ++ 12( 40� 10) 10+ 40 20++ 12( 40� 10) 20+ 12 v3 +− 40++( 20 12) 10++ 40�( 20 12) 111 io=− vi 12 +− +− v 3 200 10 62.5 So ab=−=−=−0.05, 0.1 and c0.016 (checked: LNAP 6/19/04) P 5.4-3 The circuit shown in Figure P 5.4-3b is the Thévenin equivalent circuit of the circuit shown in Figure P 5.4-3a. Find the value of the open-circuit voltage, voc, and Thévenin resistance, Rt. Answer: voc = 2 V and Rt = 4 Ω Figure P 5.4-3 Solution: The circuit from Figure P5.4-3a can be reduced to its Thevenin equivalent circuit in five steps: (a) (c) (b) (d) (e) Comparing (e) to Figure P5.4-3b shows that the Thevenin resistance is Rt = 4 Ω and the open circuit voltage, voc = 2 V. (checked using LNAP 8/15/02) P 5.4-6 Find the Thévenin equivalent circuit for the circuit shown in Figure P 5.4-6. Figure P 5.4-6 Solution: Find voc: 2 vv− v Apply KCL at the top, middle node: aa= a ++3 0 ⇒v = 18 V 36 a The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V Find isc: 2 vv− v v Apply KCL at the top, middle node: aa= a ++3 a ⇒v =−18 V 363 a v −18 Apply Ohm’s law to the right-hand 3 Ω resistor : i =a = = −6 V sc 33 voc 18 Finally: Rt = = =−Ω3 i sc −6 (checked using LNAP 8/15/02) P 5.4-8 A resistor, R, was connected to a circuit box as shown in Figure P 5.4-8. The voltage, v, was measured. The resistance was changed, and the voltage was measured again. The results are shown in the table. Determine the Thévenin equivalent of the circuit within the box and predict the voltage, v, when R = 8 kΩ. Figure P 5.4-8 Solution: From the given data: 2000 6 = v + oc Rt 2000 voc =1.2 V ⇒ 4000 R =−Ω1600 2 = v t + oc Rt 4000 When R = 8000 Ω, R vv= 8000 RR+ oc v = (1.2) = 1.5 V t −+1600 8000 P 5.4-16 An ideal voltmeter is modeled as an open circuit. A more realistic model of a voltmeter is a large resistance. Figure P 5.4- 16a shows a circuit with a voltmeter that measures the voltage vm. In Figure P 5.4-16b the voltmeter is replaced by the model of an ideal voltmeter, an open circuit. The voltmeter measures vmi, the ideal value of vm. As Rm → ∞, the voltmeter becomes an ideal voltmeter and vm → vmi. When Rm < ∞, the voltmeter is not ideal and vm > vmi. The difference between vm and vmi is a measurement error caused by the fact that the voltmeter is not ideal. (a) Determine the value of vmi. (b) Express the measurement error that occurs when Rm = 1000 Ω as a percentage of vmi. (c) Determine the minimum value of Rm required to ensure that the measurement error is smaller than 2 percent of vmi. Figure P 5.4-16 Solution: Replace the circuit by its Thevenin equivalent circuit: R v = m 5 m + Rm 50 (a) vvmi =limm = 5 V Rm →∞ (b) When Rvmm=1000 Ω=, 4.763 V so 5− 4.762 % error = ×=100 4.76% 5 R 55− m + Rm 50 Rm (c) 0.02 ≥ ⇒ ≥0.98 ⇒ Rm ≥Ω2450 5 Rm + 50 (checked: LNAP 6/16/04) P 5.5-2 Two black boxes are shown in Figure P 5.5-2. Box A contains the Thévenin equivalent of some linear circuit, and box B contains the Norton equivalent of the same circuit. With access to just the outsides of the boxes and their terminals, how can you determine which is which, using only one Figure P 5.5-2 shorting wire? Solution: 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.