7 CHPTER PROLEMS.1 Determine the voltages and V in the networ in Fig..1 using voltage division. 1V Ω Ω Ω Ω V Fig..1. Find the currents 1 and 0 in the circuit in Fig.. using current division. Ω Ω 1 Ω 1Ω 9m 0 Fig... Find the resistance of the networ in Fig.. at the terminals. 8Ω 10Ω Ω 1Ω Ω Ω 1Ω 18Ω Ω Ω Ω Fig... Find the resistance of the networ shown in Fig.. at the terminals. Ω Ω 1Ω 18Ω Ω Fig.. 1Ω 1Ω
8.5 Find all the currents and voltages in the networ in Fig..5. Ω 10Ω 6 1 Ω Ω 8V Ω V1 V Ω V Ω 5 Fig..5.6 n the networ in Fig..6, the current in the Ω resistor is m. Find the input voltage V S. Ω 1Ω 9Ω V S Ω Ω m Ω Ω Fig..6
9.1 We recall that if the circuit is of the form CHPTER SOLUTONS R V 1 1 R V 0 Fig. S.1(a) Then using voltage division V R R 1 R 0 V1 That is the voltage divides between the two resistors in direct proportion to their resistances. With this in mind, we can draw the original networ in the form Ω 1V Ω Fig. S.1(b) Ω Ω V The series combination of the Ω and Ω resistors and their parallel combination with the Ω resistor yields the networ in Fig. S.1(c). 1V Ω Ω Fig. S.1(c) Now voltage division can be sequentially applied. From Fig. S.1(c). Then from the networ in Fig. S.1(b) 1 6V
10 V V V 1. f we combine the and 1 ohm resistors, the networ is reduced to that shown in Fig. S.(a). Ω 1 9m Fig. S.(a) Ω Ω The current emanating from the source will split between the two parallel paths, one of which is the Ω resistor and the other is the series combination of the and Ω resistors. pplying current division 1 9 m ( ) Using KCL or current division we can also show that the current in the Ω resistor is 6m. The original circuit in Fig. S. (b) indicates that 1 will now be split between the two parallel paths defined by the and 1Ω resistors. 1 m Ω 6m Ω 9m Ω 1Ω 0 Fig. S.(b) pplying current division again 1 18 1m 0 1 0 Liewise the current in the Ω resistor can be found by KCL or current division to be m. Note that KCL is satisfied at every node.
11. To provide some reference points, the circuit is labeled as shown in Fig. S.(a). 8 ' 10 " 18 1 ' " Fig. S.(a) Starting at the opposite end of the networ from the terminals, we begin looing for resistors that can be combined, e.g. resistors that are in series or parallel. Note that none of the resistors in the middle of the networ can be combined in anyway. However, at the righthand edge of the networ, we see that the and 1 ohm resistors are in parallel and their combination is in series with the Ω resistor. This combination of 1 is in parallel with the Ω resistor reducing the networ to that shown in Fig. S.(b). 8 ' 10 " 18 1 ' " Fig. S.(b) 1 ( 1 ) Repeating this process, we see that the Ω resistor is in series with the 10Ω resistor and that combination is in parallel with the1ω resistor. This equivalent Ω resistor ( 10) 1 is in series with the Ω resistor and that combination is in parallel with the 18Ω resistor that ( ) 18 Ω and thus the networ is reduced to that shown in Fig. S.(c). 8 ' ' Fig. S.(c)
1 t this point we see that the two Ω resistors are in series and their combination in parallel with the Ω resistor. This combination ( ) Ω which is in series with 8Ω resistors yielding total resistance R 8 11Ω.. n examination of the networ indicates that there are no series or parallel combinations of resistors in this networ. However, if we redraw the networ in the form shown in Fig. S.(a), we find that the networs have two deltas bac to bac. 1 18 1 1 Fig. S.(a) f we apply the Y transformation to either delta, the networ can be reduced to a circuit in which the various resistors are either in series or parallel. Employing the Y transformation to the upper delta, we find the new elements using the following equations as illustrated in Fig. S.(b) 18 R R 1 R 1 Fig. S.(b) R 1 R R ( ) ( 18) Ω 1 18 ( ) ( 1) Ω 1 18 ( 1) ( 18) 6 Ω 1 18 The networ is now reduced to that shown in Fig. S.(c).
1 1 Fig. S.(c) 1 Now the total resistance, R is equal to the parallel combination of ( 1) and ( 1) in series with the remaining resistors i.e. R (1 18) 16.875Ω f we had applied the Y transformation to the lower delta, we would obtain the networ in Fig. S.(d). 18 Fig. S.(d) n this case, the total resistance R is R ( ) (18 ) 16.875Ω which is, of course, the same as our earlier result..5 Our approach to this problem will be to first find the total resistance seen by the source, use it to find 1 and then apply Ohm s law, KCL, KVL, current division and voltage division to determine the remaining unnown quantities. Starting at the opposite end of the networ from the source, the and Ω resistors are in series and that combination is in parallel with the Ω resistor yielding the networ in Fig. S.5(a).
1 10 8V 1 V Fig. S.5(a) Proceeding, the and 10 ohm resistors are in series and their combination is in parallel with both the and ohm resistors. The combination (10 ) Ω. Therefore, this further reduction of the networ is as shown in Fig. S.5(b). 8 1 Fig. S.5(b) Now 1 and can be easily obtained. nd by Ohm s law 8 1 1m or using voltage division 1 V 8 V once is nown, and can be obtained using Ohm s law V1 V1 6m m can be obtained using KCL at node. s shown on the circuit diagram. 1
15 1 6 m The voltage V is then V 10 10 V ( ) or using voltage division V 10 1 6 V V 1 Knowing V, 5 can be derived using Ohm s law and also 5 V m 6 V m current division can also be used to find 5 and 6. and 5 m
16 6 m Finally V can be obtained using KVL or voltage division and V V 6 8 V V V 8 V.6 The networ is labeled with all currents and voltages in Fig. S.6. V V S V 5 V 1 Fig. S.6 Given the m current in the Ω resistor, the voltage ( ) 1V Now nowing, 1 and can be obtained using Ohm s law as pplying KCL at node 1 V1 1 m 1 V1 1 1m 9 1 9
17 Then using Ohm s law 1 6m KVL can then be used to obtain V i.e. V (1) 6V Then V V 6 1 18V V 9m nd 5 6 9 15m using Ohm s law V () 5 0V and finally V S V V 8V