Question

- Review Learning Goal: To reduce series-parallel combinations of inductors to an equivalent inductance Inductors in series and parallel combine like resistors in series and parallel. It is possible to use Kirchhoff's current law to find the current through the equivalent inductance. - Part A - Determine the equivalent inductance to the right of terminals band e, including L4 For the network in Figure 1(Figure 1), determine the equivalent inductance to the right of terminals bande, including L4. Assume that L1 = 2.0 mH, L2 = 6.0 mH L3 = 10.0 mH, L4 = 4.0 mH. L5 = 5.0 MH, LE = 8.0 mH, L7 = 6.0 mHand Lg = 6.0 MH. Express the equivalent inductance, Lp, in millihenries to three significant figures. View Available Hint(s) fempBles Symbole Slope/Displacement Vector Ado rede reslet keyboard shortcuts hele mH Submit Part B - Determine the equivalent inductance to the right of terminals a and d For the network in Figure 1(Figure 1), determine the equivalent inductance to the right of terminals a and d. Assume that L1 = 2.0 mH, L2 = 6.0 mH, L3 = 10.0 mH, L = 1.0 mH, L = 5.0 mH, LG = 8.0 mH, L = 6.0 mH, and Lg = 6.0 mH. Express your answer in millihenries to three significant figures. View Available Hint(s) templates Symbols Slope/Displacement Vectorvando redo redet keyboard shortcuts help Lada mH Submit Figure 1 of 2 - Part C-Determine the current, e1, through the equivalent inductance to the right of terminals a and d For the network in Figure 2(Figure 2), determine the current, 81, through the equivalent inductance to the right of terminals a and d. Assume that is = -10.0 mA, i4 = 9.0 mA, and 15 = 1.0 mA Express your answer in milliamperes to three significant figures. View Available Hint(s) Templates Symbols Slope/Displacement Vector do redo reset keyboard shortcuts help 31 #1 - mA Submit Figure 1

Answer 1

Solution 9) b وا a. mm į . lu is 13 mm LS mums al e 4 Lo Network to the right of terminals 6 and e. L2=6my b L5 Given that 4=2mH, L2=60H, 13 = lomH, L = 4.0mH, 25=50H, L6 8 mH, Ly = 60H, Lq = 6mH. Leq 24 umy, ما 2 m smy =PMH La = 6mH 6mH Smye 3 (6+8) my (1485 14+5 Lag - [lan18+1]111]ma 1114 184x4 or her = b) Now, equivalent resistance to the right of terminals 736 e = b 220 184 9 +4 u ml Il my 184 = 3.345 mH 9 ss e a and d 2my Leq=164 Leg (calculated lomy above bove) 604 2m4 m le 184 .+6. 55 Lei 3 184 +6 ) my = 514 55 d. my

519 χιο 55 4,83mH 574 55 +10. 2mH Leg 4.8300H l'eq = (2+4,830 mm L'eq 6.830mh b a ☺ m Ginen that z = -loma 1429 mH is alma. -is Now kee at node C izzig tis. lIto z IMA L7 KCL at node d. iz tiz zo iz --i3 =-(-10) iz = 10mA as Lg and 4 are inseries i 18 = 16 Now kcc at node e izzlyti5 tip i8 = iq-14-15 is = 10-9-1 is OMA iGzip= OmA. At last kee at node b. i = iz tlu + 12 i, = (-10) +91 OmA - The current is, through the equivalent inductance to the right of the terminal a and a 15 om A