Chapter 3, Problem 9(8). Find V x in the network shown in Fig. 3.78. Figure 3.78 Chapter 3, Solution 9(8). Consider the circuit below. 2 Ω 2 Ω -j 8 30 o I j 4 j 4 I 2 -j2v For loop, 8 30 = (2 j4)i ji 2 () For loop 2, ((j4 2 j)i 2 ji (j2) = 0 or I = (3 j2)i 2 2 (2) Substituting (2) into (), 8 30 (2 j4)2 = (4 j7)i 2 I 2 = (0.928 j2)/(4 j7) =.037 2.2 V x = 2I 2 = 2.074 2.2
Chapter 3, Problem 2(4). Find I and I 2 in the circuit of Fig. 3.90. Calculate the power absorbed by the 4-Ω resistor. Figure 3.90 Chapter 3, Solution 2(4). For mesh, 36 30 = (7 j6)i (2 j)i 2 () For mesh 2, 0 = (6 j3 j4)i 2 2I ji = (2 j)i (6 j)i 2 (2) Placing () and (2) into matrix form, 36 30 0 = 7 j6 2 j 2 j I 6 j I 2 = 48 j35 = 59.4 36., = (6 j)36 30 = 29 20.54 2 = (2 j)36 30 = 80.5 56.56, I = / = 3.69 5.56, I 2 = 2 / =.355 20.46 Power absorbed fy the 4-ohm resistor, = 0.5(I 2 ) 2 4 = 2(.355) 2 = 3.672 watts
Chapter 3, Problem 22(5). Find current I o in the circuit of Fig. 3.9. Figure 3.9 Chapter 3, Solution 22(5). With more complex mutually coupled circuits, it may be easier to show the effects of the coupling as sources in terms of currents that enter or leave the dot side of the coil. Figure 3.9 then becomes, -j50 I o I 3 j20i c I a j40 j0i b j30i c j30i b j60 I x j20i a 50 0 V j80 I I 2 00 Ω I b j0i a
2 Note the following, I a = I I 3 I b = I 2 I I c = I 3 I 2 and I o = I 3 Now all we need to do is to write the mesh equations and to solve for I o. Loop #, -50 j20(i 3 I 2 ) j 40(I I 3 ) j0(i 2 I ) j30(i 3 I 2 ) j80(i I 2 ) j0(i I 3 ) = 0 j00i j60i 2 j40i 3 = 50 Multiplying everything by (/j0) yields 0I 6I 2 4I 3 = - j5 () Loop # 2, j0(i I 3 ) j80(i 2 I ) j30(i 3 I 2 ) j30(i 2 I ) j60(i 2 I 3 ) j20(i I 3 ) 00I 2 = 0 Loop # 3, -j60i (00 j80)i 2 j20i 3 = 0 (2) -j50i 3 j20(i I 3 ) j60(i 3 I 2 ) j30(i 2 I ) j0(i 2 I ) j40(i 3 I ) j20(i 3 I 2 ) = 0 -j40i j20i 2 j0i 3 = 0 Multiplying by (/j0) yields, -4I 2I 2 I 3 = 0 (3) Multiplying (2) by (/j20) yields -3I (4 j5)i 2 I 3 = 0 (4) Multiplying (3) by (/4) yields -I 0.5I 2 0.25I 3 = 0 (5) Multiplying (4) by (-/3) yields I ((4/3) j(5/3))i 2 (/3)I 3 = -j0.5 (7) Multiplying [(6)(5)] by 2 yields (-22 j20)i 2 7I 3 = 0 (8) Multiplying [(5)(7)] by 20 yields -22I 2 3I 3 = -j0 (9) (8) leads to I 2 = -7I 3 /(-22 j20) = 0.2355 42.3 o = (0.748j0.5849)I 3 (0) (9) leads to I 3 = (j0 22I 2 )/3, substituting () into this equation produces, I 3 = j3.333 (-.2273 j.623)i 3 or I 3 = I o =.3040 63 o amp.
Chapter 3, Problem 26(9). Find I o in the circuit of Fig. 3.95. Switch the dot on the winding on the right and calculate I o again. Fig. 3.95. Chapter 3, Solution (26)9. M = k L L 2 ωm = k ω LωL 2 = 0.6 20 x40 = 7 The frequency-domain equivalent circuit is shown below. 50 Ω j30 j7 I o 200 60 I j2 j4 I2 0 Ω For mesh, 200 60 = (50 j30 j20)i j7i 2 = (50 j0)i j7i 2 () For mesh 2, 0 = (0 j40)i 2 j7i (2) In matrix form, 200 60 = 0 50 j0 j7 j7 I 0 j40 I = 900 j00, = 2000 60 ( j4) = 8246.2 36, 2 = 3400 30 I 2 = 2 / = 3.755 36.34 I o = I 2 = 3.755 36.34 A Switching the dot on the winding on the right only reverses the direction of I o. This can be seen by looking at the resulting value of 2 which now becomes 3400 50. Thus, I o = 3.755 43.66 A 2
Chapter 3, Problem 43(30). Obtain V and V 2 in the ideal transformer circuit of Fig. 3.07. Chapter 3, Solution 43(30). Transform the two current sources to voltage sources, as shown below. 0 Ω : 4 2 Ω 20 V v v 2 I I 2 2V Using mesh analysis, 20 0I v = 0 20 = v 0I () 2 2I 2 v 2 = 0 or 2 = v 2 2I 2 (2) At the transformer terminal, v 2 = nv = 4v (3) I = ni 2 = 4I 2 (4) Substituting (3) and (4) into () and (2), we get, 20 = v 40I 2 (5) 2 = 4v 2I 2 (6) Solving (5) and (6) gives v = 4.86 V and v 2 = 4v = 6.744 V
Chapter 3, Problem 44(3). In the ideal transformer circuit of Fig. 3.08, find i (t) and i 2 (t). Chapter 3, Solution 44(3). We can apply the superposition theorem. Let i = i i and i 2 = i 2 i 2 where the single prime (i, i 2 ) is due to the DC source and the double prime (i, i 2 ) is due to the AC source. Since we are looking for the steady-state values of i and i 2, i =V 0 /R i 2 = 0. (No induction on the secondary at DC) For the AC source, consider the circuit below. R : n i v v 2 i 2 V m 0 V 2 /V = n, I 2 /I = /n But V 2 = V m, V = V m /n I = - V /R= V m /(Rn) I 2 = I /n = V m /(Rn 2 ) Hence, i (t) = V 0 /R (V m /Rn)cosωt A, (DC and AC together) i 2 (t) = (V m /(n 2 R))cosωt A (Only AC)
Chapter 3, Problem 56(40). Find the power absorbed by the 0-Ω resistor in the ideal transformer circuit of Fig. 3.20. Chapter 3, Solution 56(40). We apply mesh analysis to the circuit as shown below. 2 Ω : 2 v v 2 46V I I 2 0 Ω 5 Ω For mesh, 46 = 7I 5I 2 v () For mesh 2, v 2 = 5I 2 5I (2) At the terminals of the transformer, v 2 = nv = 2v (3) Substituting (3) and (4) into () and (2), Combining (5) and (6), 46 =.5I 2 or I 2 = 4 I = ni 2 = 2I 2 (4) 46 = 9I 2 v (5) v = 2.5I 2 (6) P 0 = 0.5I 2 2 (0) = 80 watts.
Chapter 3, Problem 6(45). For the circuit in Fig. 3.25 below, find I, I 2, and V o. Chapter 3, Solution 6(45). We reflect the 60-ohm load to the middle circuit. Z R = Z L /n 2 = 60/(4/3) 2 = 90 ohms, where n = 4/3 2 Ω I : 5 I o 4 Ω I o 24 0 v v o 60 Ω 90 Ω 4 60 90 = 4 36 = 50 ohms We reflect this to the primary side. Z R = Z L /(n ) 2 = 50/5 2 = 2 ohms when n = 5 I = 24/(2 2) = 6A 24= 2I v or v = 24 2I = 2 V v o = nv = 60 V, I o = I /n = 6/5 =.2 I o = [60/(60 90)]I o = 0.48A I 2 = I o /n = 0.48/(4/3) = 0.36 A