Prolem.14 Determine currents 1 to 3 in the circuit of Fig. P.14. 1 A 18 V Ω 3 A 1 8 Ω 1 Ω 7 Ω 4 Ω 3 Figure P.14: Circuit for Prolem.14. For the loop contining the 18-V source, Hence, 1 = 1.5 A. KCL t node gives KCL t node gives 183 8 1 = 0. 3 1 1 = 0 = 1 = 1.5 = 0.5 A. 1 3 = 0 3 = 1 = 10.5 = 1.5 A. All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
Prolem.15 Determine x in the circuit of Fig. P.15. 1 V 5 Ω x Ω 1 A Figure P.15: Circuit for Prolem.15. KVL gives: 15 x = 0. KCL gives: 1 x = 0. Solution of the two equtions yields x = 17 7 =.43 A. All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
Prolem. Find in the circuit of Fig. P.. 10 V 3 Ω Figure P.: Circuit for Prolem.. Hence, 10 3 = 0. = 10 5 = A. All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
Prolem.9 Given tht 1 = 1 A in the circuit of Fig. P.9, determine 0. 5 4 3 1 = 1 A 0 1 Ω Ω 4 Ω 8 Ω 1 Figure P.9: Circuit for Prolem.9. Since the 16-Ω nd 8-Ω resistors re connected in prllel, they hve the sme voltge cross them, nmely V = 16 1 = 16 1 = 16 V. By KCL, 0 equls the sum of the currents flowing in ll five resistors: 0 = 16 1 16 16 4 16 8 16 16 = 31 A. All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
Prolem.30 Wht should R e in the circuit of Fig. P.30 so tht = 4 Ω? 1 Ω 5 Ω Ω R Figure P.30: Circuit for Prolem.30. The prllel comintion of R nd -Ω resistor is R 1 is in series with 5-Ω resistor. Hence R is in prllel with 6-Ω resistor: nd Solving for R leds to R 1 = R R. R = R 1 5 = R R 5. R 3 = = 1R 3 = 1 ( ) R 6 R 5 6 R, R 5 ( ) R 6 R 5 R = Ω. 11 R R = 4. All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
Prolem.31 Find 0 in the circuit of Fig. P.31. 18 A 0 1 Ω 4 Ω 3 Ω Figure P.31: Circuit for Prolem.31. Comining the 3-Ω nd 6-Ω resistors in prllel gives The new circuit ecomes R = 3 6 36 = 18 9 = Ω. 18 A 0 1 Ω Current division leds to 0 = ( ) Req 18 = 6 1 18 = 6 A. 1 61 All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
Prolem.38 For the circuit in Fig. P.38, find R* t terminls (, ). 80 / -rt - P--=4-LS:qO -- \)J! v- rl ll, '4'?JL -fr-- {!3 -z-j o,l /l ) J*//3 =1 *' 4fr1 *-- 'JL lt e{l t. -J--J lrz-nll'qj ",3* /xr! //n = =/ t*/z "z All rights reserved. Do not reproduce or distriute. 0013 Ntionl Technology nd Scielce Prcss
Prolem.40 Simplify the circuit to the right of terminls (,) in Fig. P.40 to find, nd then determine the mount of power supplied y the voltge source. All resistnces re in ohms. 5 V 3 5 8 4 8 6 1 1 6 Figure P.40: Circuit for Prolem.40. 5 V 3 5 8 8 1 4 4 6 5 V 3 5 4 8 6 6 1 1 = 6 5 V 3 4 8 8 5 6 6 = 5 3 = 8 5 V 3 8 8 4 = = 3 = 5 Ω P = V = (5) = 15 W. 5 All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
Prolem.41 For the circuit in Fig. P.41, determine t () Terminls (, ) () Terminls (, c) (c) Terminls (, d) (d) Terminls (, f) d c Ω Ω e Ω Ω Ω Ω Ω f Ω Ω Ω Figure P.41: Circuit for Prolem.41. All resistnces re in ohms. () 6 6 = 1.5 () = 1.5 = 5.5 Ω. (c) (d) c d f c d f 8 6 6 = = 4 Ω. = 1.5 = 1.5 = 5.5 Ω. 4 4 4 4 = = =. All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
Prolem.4 Find for the circuit in Fig. P.4. All resistnces re in ohms. 5 10 10 10 10 Figure P.4: Circuit for Prolem.4. 5 10 5 10 0 0 5 5 5 5 = 15 Ω. All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
Prolem.43 Apply voltge nd current division to determine V 0 in the circuit of Fig. P.43 given tht V out = 0. V. 5 V 0 V 5 8 3 4 V 3 4 1 V 4 4 V 1 V 1 Figure P.43: Circuit for Prolem.43. V out = 0. V 1 = 0. 1 = 0. A = V = 1 (1) = 0.3 A 3 = 1 = 0.5 A 4 = V 4 4 = V 3 V = 4 3 4 4 = 0.65 A 5 = 3 4 = 1.15 A V 0 = V 4 V 5 = 4 4 8 5 = 11.8 V. All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
Prolem.44 Apply source trnsformtions nd resistnce reductions to simplify the circuit to the left of nodes (,) in Fig. P.44 into single voltge source nd resistor. Then, determine. 3 A 10 Ω 5 A Ω 1 Ω 4 Ω Figure P.44: Circuit of Prolem.44. All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
3 A 10 Ω 5 A Ω 1 Ω 30 V Ω 10 Ω 10 V 1 Ω 1 Ω 40 V 1 Ω 40 1 1 Ω 1 Ω 40 1 0 V 4 Ω Fig. P.44 () = 0 64 = A. All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
Prolem.45 Fig. P.45. Determine the open-circuit voltge V oc cross terminls (,) in 5 Ω 30 V 3 Ω A V oc 6 A 5 Ω 3 Ω A V oc 8 A R = 3 5 = 15 Ω 3 5 8 V oc 15 Ω 8 15 V V oc Fig. P.45 () Hence, V oc = 15 V. All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
Prolem.46 Use circuit trnsformtions to determine in the circuit of Fig. P.46. 3 A 4 Ω 3 Ω A 4 Ω 30 V Ω 1 V 10 A 4 Ω 4 Ω 8 V 3 Ω Ω 6 3 9 = Ω 8 Ω 0 V 10 A Ω 0 V 8 Ω Ω 0 V Ω Fig. P.46 () = 0 0 8 = 0. All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press
Prolem.47 Determine currents 1 to 4 in the circuit of Fig. P.47. 1 Ω 1 3 3 Ω 4 1 V Figure P.47: Circuit of Prolems.47 nd.48. 1 = 1 1 = 1 A, = 1 6 = A, 3 = 1 3 = 4 A, 4 = 1 6 = A. All rights reserved. Do not reproduce or distriute. c 013 Ntionl Technology nd Science Press