CHAPTER 17 Fourier Method of Waveform Analysis 420
3.7 CURRENT DIVISION
A parallel arrangement of resistors as shown in Fig. 3-6 results in acurrent divider. The ratio of the branch currenti1 to the total currentiillustrates the operation of the divider.
i¼ v R1þ v
R2þ v
R3 and i1¼ v R1 i1
i ¼ 1=R1
1=R1þ1=R2þ1=R3¼ R2R3
R1R2þR1R3þR2R3 Then
Fig. 3-5
Fig. 3-6
For a two-branch current divider we have i1
i ¼ R2 R1þR2
This may be expressed as follows: The ratio of the current in one branch of a two-branch parallel circuit to the total current is equal to the ratio of the resistance of theotherbranch resistance to the sum of the two resistances.
EXAMPLE 3.8. A current of 30.0 mA is to be divided into two branch currents of 20.0 mA and 10.0 mA by a network with an equivalent resistance equal to or greater than 10.0. Obtain the branch resistances.
20 mA 30 mA¼ R2
R1þR2
10 mA 30 mA¼ R1
R1þR2
R1R2
R1þR210:0 Solving these equations yieldsR115:0andR230:0.
Solved Problems
3.1 FindV3 and its polarity if the currentI in the circuit of Fig. 3-7 is 0.40 A.
Assume thatV3has the same polarity asV1. Applying KVL and starting from the lower left corner, V1Ið5:0Þ V2Ið20:0Þ þV3¼0
50:02:010:08:0þV3¼0 V3¼ 30:0 V Terminalbis positive with respect to terminala.
3.2 Obtain the currentsI1 andI2 for the network shown in Fig. 3-8.
aandbcomprise one node. Applying KCL,
2:0þ7:0þI1¼3:0 or I1¼ 6:0 A Also,canddcomprise a single node. Thus,
4:0þ6:0¼I2þ1:0 or I2¼9:0 A
3.3 Find the currentI for the circuit shown in Fig. 3-9.
Fig. 3-7
The branch currents within the enclosed area cannot be calculated since no values of the resistors are given. However, KCL applies to the network taken as a single node. Thus,
2:03:04:0I¼0 or I¼ 5:0 A
3.4 Find the equivalent resistance for the circuit shown in Fig. 3-10.
The two 20-resistors in parallel have an equivalent resistanceReq¼ ½ð20Þð20Þ=ð20þ20Þ ¼10. This is in series with the 10-resistor so that their sum is 20. This in turn is in parallel with the other 20- resistor so that the overall equivalent resistance is 10.
3.5 Determine the equivalent inductance of the three parallel inductances shown in Fig. 3-11.
Fig. 3-8
Fig. 3-9
Fig. 3-10
The two 20-mH inductances have an equivalent inductance of 10 mH. Since this is in parallel with the 10-mH inductance, the overall equivalent inductance is 5 mH. Alternatively,
1 Leq¼ 1
L1þ 1 L2þ 1
L3¼ 1
10 mHþ 1
20 mHþ 1
20 mH¼ 4
20 mH or Leq¼5 mH
3.6 Express the total capacitance of the three capacitors in Fig. 3-12.
ForC2andC3in parallel,Ceq¼C2þC3. Then forC1andCeq in series, CT ¼ C1Ceq
C1þCeq¼ C1ðC2þC3Þ C1þC2þC3
3.7 The circuit shown in Fig. 3-13 is a voltage divider, also called anattenuator. When it is a single resistor with an adjustable tap, it is called a potentiometer, or pot. To discover the effect of loading, which is caused by the resistanceRof the voltmeter VM, calculate the ratioVout=Vinfor (a) R¼ 1, (b) 1 M, (c) 10 k, (d) 1 k.
Vout=Vin¼ 250
2250þ250¼0:100 ðaÞ
Fig. 3-11
Fig. 3-12
Fig. 3-13
(b) The resistanceRin parallel with the 250-resistor has an equivalent resistance Req¼ 250ð106Þ
250þ106¼249:9 and Vout=Vin¼ 249:9
2250þ249:9¼0:100 Req¼ð250Þð10 000Þ
250þ10 000¼243:9 and Vout=Vin¼0:098 ðcÞ
Req¼ð250Þð1000Þ
250þ1000¼200:0 and Vout=Vin¼0:082 ðdÞ
3.8 Find all branch currents in the network shown in Fig. 3-14(a).
The equivalent resistances to the left and right of nodesaandbare ReqðleftÞ¼5þð12Þð8Þ
20 ¼9:8 ReqðrightÞ¼ð6Þð3Þ
9 ¼2:0 Now referring to the reduced network of Fig. 3-14(b),
I3¼ 2:0
11:8ð13:7Þ ¼2:32 A I4¼ 9:8
11:8ð13:7Þ ¼11:38 A Then referring to the original network,
I1¼ 8
20ð2:32Þ ¼0:93 A I2¼2:320:93¼1:39 A I5¼3
9ð11:38Þ ¼3:79 A I6¼11:383:79¼7:59 A
Supplementary Problems
3.9 Find the source voltage V and its polarity in the circuit shown in Fig. 3-15 if (a) I¼2:0 A and (b) I¼ 2:0 A. Ans. (a) 50 V,bpositive; (b) 10 V,apositive.
3.10 FindReqfor the circuit of Fig. 3-16 for (a) Rx¼ 1, (b) Rx¼0, (c) Rx¼5. Ans. (a) 36; (b) 16; (c) 20
Fig. 3-14
3.11 An inductance of 8.0 mH is in series with two inductances in parallel, one of 3.0 mH and the other 6.0 mH.
FindLeq. Ans. 10.0 mH
3.12 Show that for the three capacitances of equal value shown in Fig. 3-17Ceq¼1:5 C:
3.13 Find RH and RO for the voltage divider in Fig. 3-18 so that the current I is limited to 0.5 A when VO¼100 V. Ans: RH¼2 M;RO¼200
3.14 Using voltage division, calculateV1andV2in the network shown in Fig. 3-19. Ans. 11.4 V, 73.1 V
3.15 Obtain the source currentI and the total power delivered to the circuit in Fig. 3-20.
Ans. 6.0 A, 228 W
3.16 Show that for four resistors in parallel the current in one branch, for example the branch ofR4, is related to the total current by
I4¼IT R0 R4þR0
whereR0¼ R1R2R3 R1R2þR1R3þR2R3 Fig. 3-15
Fig. 3-16
Fig. 3-17
Fig. 3-18
Fig. 3-19
Fig. 3-20
Note: This is similar to the case of current division in a two-branch parallel circuit where the other resistor has been replaced byR0.
3.17 A power transmission line carries current from a 6000-V generator to three loads, A, B, and C. The loads are located at 4, 7, and 10 km from the generator and draw 50, 20, and 100 A, respectively. The resistance of the line is 0.1/km; see Fig. 3-21. (a) Find the voltage at loads A, B, C. (b) Find the maximum percentage voltage drop from the generator to a load.
Ans. (a) vA¼5928 V;vB¼5889 V;vC¼5859 V; (b) 2.35 percent
3.18 In the circuit of Fig. 3-22,R¼0 andi1andi2are unknown. FindiandvAC. Ans. i¼4 A;vAC¼24 V
3.19 In the circuit of Fig. 3-22,R¼1andi1¼2 A. Find,i,i2, andvAC. Ans. i¼5 A;i2¼ 16 A;vAC¼27 V
3.20 In the circuit of Fig. 3-23, is1¼vs2¼0, vs1¼9 V, is2¼12 A. For the four cases of (a) R¼0, (b) R¼6, (c) R¼9, and (d) R¼10 000, draw the simplified circuit and find iBA and vAC. Hint: A zero voltage source corresponds to a short-circuited element and a zero current source corresponds to an open-circuited element.
Ans:
ðaÞ iBA¼7;vAC¼30 ðbÞ iBA¼4:2;vAC¼21:6 ðcÞ iBA¼3:5;vAC¼19:5
ðdÞ iBA¼0:0060;vAC¼9:029 8>
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ðAll in A and V)
3.21 In the circuit of Fig. 3-23, vs1¼vs2¼0;is1¼6 A;is2¼12 A: For the four cases of (a) R¼0;ðbÞR¼6;ðcÞR¼9;andðdÞR¼10 000;draw the simplified circuit and findiBA andvAC.
Ans:
ðaÞ iBA¼6;vAC¼36 ðbÞ iBA¼3:6;vAC¼28:8 ðcÞ iBA¼3;vAC¼27 ðdÞ iBA¼0:0050;vAC18 8>
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ðAll in A and V) Fig. 3-21
Fig. 3-22
3.22 In the circuit Fig. 3-23, vs1¼0, vs2¼6 V, is1¼6 A, is2¼12 A. For the four cases of (a) R¼0, (b) R¼6, (c) R¼9, and (d) R¼10 000, draw the simplified circuit and findiBAandvAC.
Ans:
ðaÞ iBA¼5:33;vAC¼34 ðbÞ iBA¼3:2;vAC¼27:6 ðcÞ iBA¼2:66;vAC¼26
ðdÞ iBA¼0:0050;vAC¼18:0118 8>
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(All in A and V)
3.23 In the circuit of Fig. 3-24, (a) find the resistance seen by the voltage source,Rin¼v=i, as a function ofa, and (b) evaluateRinfora¼0;1;2. Ans. (a) Rin¼R=ð1aÞ; (b) R;1;R
3.24 In the circuit of Fig. 3-24, (a) find power P delivered by the voltage source as a function of a, and (b) evaluatePfora¼0;1;2. Ans. (a) P¼v2ð1aÞ=R; (b) v2=R;0;v2=R
3.25 In the circuit of Fig. 3-24, leta¼2. Connect a resistorRxin parallel with the voltage source and adjust it within the range 0Rx0:99Rsuch that the voltage source delivers minimum power. Find (a) the value ofRxand (b) the power delivered by the voltage source.
Ans. (a) Rx¼0:99R, (b) P¼v2=ð99RÞ
Fig. 3-23
Fig. 3-24
Fig. 3-25
3.26 In the circuit of Fig. 3-25,R1¼0 andb¼100. Draw the simplified circuit and findvforR¼1kand 10k. Ans. v¼1;10 V
3.27 In the circuit of Fig. 3-25,R1¼0 andR¼1k. Draw the simplified circuit and findvforb¼50;100;200.
Note thatvchanges proportionally withb. Ans. v¼0:5;1;2 V
3.28 In the circuit of Fig. 3-25, R1¼100 and R¼11k. Draw the simplified circuit and find v for b¼50;100;200. Compare with corresponding values obtained in Problem 3.27 and note that in the present casevis less sensitive to variations inb. Ans. v¼0:90;1;1:04 V
3.29 A nonlinear element is modeled by the following terminal characteristic.
i¼ 10v whenv0 0:1v whenv0
Find the element’s current if it is connected to a voltage source with (a) v¼1þsintand (b) v¼ 1þsint.
See Fig. 3-26(a). Ans. (a) i¼10ð1þsintÞ; (b) i¼0:1ð1þsintÞ
3.30 Place a 1-linear resistor between the nonlinear element of Problem 3.29 and the voltage source. See Fig.
3-26(b). Find the element’s current if the voltage source is (a) v¼1þsintand (b) v¼ 1þsint.
Ans. (a) i¼0:91ð1þsintÞ; (b) i¼0:091ð1þsintÞ Fig. 3-26
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