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A parallel arrangement of resistors as shown in Fig. 3-6 results in a current divider. The ratio of the branch current i1 to the total current i illustrates the operation of the divider.
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Fig. 3-6
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v v v v and i1 R1 R2 R3 R1 i1 1=R1 R2 R3 i 1=R1 1=R2 1=R3 R1 R2 R1 R3 R2 R3 i
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CIRCUIT LAWS
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For a two-branch current divider we have i1 R2 i R1 R2 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 the other branch resistance to the sum of the two resistances.
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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
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. Obtain the branch resistances. 20 mA R2 30 mA R1 R2 10 mA R1 30 mA R1 R2 R1 R2 ! 10:0
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R1 R2
Solving these equations yields R1 ! 15:0
and R2 ! 30:0
Solved Problems
3.1 Find V3 and its polarity if the current I in the circuit of Fig. 3-7 is 0.40 A.
Fig. 3-7 Assume that V3 has the same polarity as V1 . Applying KVL and starting from the lower left corner, V1 I 5:0 V2 I 20:0 V3 0 50:0 2:0 10:0 8:0 V3 0 V3 30:0 V Terminal b is positive with respect to terminal a.
Obtain the currents I1 and I2 for the network shown in Fig. 3-8.
a and b comprise one node. Applying KCL, or I1 6:0 A 2:0 7:0 I1 3:0 Also, c and d comprise a single node. Thus, or I2 9:0 A
4:0 6:0 I2 1:0
Find the current I for the circuit shown in Fig. 3-9.
CIRCUIT LAWS
[CHAP. 3
Fig. 3-8
Fig. 3-9 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, 2:0 3:0 4:0 I 0 or I 5:0 A
Find the equivalent resistance for the circuit shown in Fig. 3-10.
Fig. 3-10 The two 20-
resistors in parallel have an equivalent resistance Req 20 20 = 20 20 10
. 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
Determine the equivalent inductance of the three parallel inductances shown in Fig. 3-11.
CHAP. 3]
CIRCUIT LAWS
Fig. 3-11 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 1 1 1 1 1 1 4 Leq L1 L2 L3 10 mH 20 mH 20 mH 20 mH or Leq 5 mH
Express the total capacitance of the three capacitors in Fig. 3-12.
Fig. 3-12 For C2 and C3 in parallel, Ceq C2 C3 . CT Then for C1 and Ceq in series,
C1 Ceq C C C3 1 2 C1 Ceq C1 C2 C3
The circuit shown in Fig. 3-13 is a voltage divider, also called an attenuator. When it is a single resistor with an adjustable tap, it is called a potentiometer, or pot. To discover the e ect of loading, which is caused by the resistance R of the voltmeter VM, calculate the ratio Vout =Vin for (a) R 1, (b) 1 M
, (c) 10 k
, (d) 1 k
a Vout =Vin 250 0:100 2250 250
Fig. 3-13
CIRCUIT LAWS
[CHAP. 3
(b) The resistance R in parallel with the 250-
resistor has an equivalent resistance Req c d 250 106 249:9
and 250 106 250 10 000 Req 243:9
250 10 000 250 1000 200:0
Req 250 1000 Vout =Vin and and 249:9 0:100 2250 249:9
Vout =Vin 0:098 Vout =Vin 0:082
Find all branch currents in the network shown in Fig. 3-14(a).
Fig. 3-14 The equivalent resistances to the left and right of nodes a and b are Req left 5 Req right 12 8 9:8
20 6 3 2:0
Now referring to the reduced network of Fig. 3-14(b), I3 2:0 13:7 2:32 A 11:8 9:8 13:7 11:38 A I4 11:8
Then referring to the original network, I1 8 2:32 0:93 A 20 3 I5 11:38 3:79 A 9 I2 2:32 0:93 1:39 A I6 11:38 3:79 7:59 A
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