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Unformatted text preview: . (2.49) to (2.51) that
R1 R 2 + R 2 R 3 + R 3 R 1 = Ra Rb Rc (Ra + Rb + Rc )
(Ra + Rb + Rc )2 Ra R b Rc
=
Ra + R b + R c c Figure 2.49
(2.52) Superposition of Y and
networks as an aid in transforming one to
the other. Dividing Eq. (2.52) by each of Eqs. (2.49) to (2.51) leads to the following
equations:
Ra = R1 R2 + R2 R3 + R3 R1
R1 (2.53) Rb = R1 R 2 + R 2 R 3 + R 3 R 1
R2 (2.54) Rc = R 1 R 2 + R 2 R 3 + R 3 R1
R3 (2.55)  v v From Eqs. (2.53) to (2.55) and Fig. 2.49, the conversion rule for Y to
is as follows:  eText Main Menu  Textbook Table of Contents  Problem Solving Workbook Contents 52 PART 1 DC Circuits Each resistor in the network is the sum of all possible products of Y resistors
taken two at a time, divided by the opposite Y resistor.
The Y and networks are said to be balanced when R1 = R2 = R3 = RY , Ra = Rb = Rc = R (2.56) Under these conditions, conversion formulas become
RY = R
3 R = 3RY or (2.57) One may wonder why RY is less than R . Well, we notice that the Yconnection is like a “series” connection while the connection is like a
“parallel” connection.
Note that in making the transformation, we do not take anything out
of the circuit or put in anything new. We are merely substituting different
but mathematically equivalent threeterminal network patterns to create
a circuit in which resistors are either in series or in parallel, allowing us
to calculate Req if necessary. EXAMPLE 2.14
Convert the network in Fig. 2.50(a) to an equivalent Y network. a b 5Ω
Rc a b 7.5 Ω
R2 R1 25 Ω Rb 10 Ω 15 Ω Ra R3 c c (b) (a) Figure 2.50 3Ω For Example 2.14: (a) original network, (b) Y equivalent network.  v v Solution:
Using Eqs. (2.49) to (2.51), we obtain  eText Main Menu  Textbook Table of Contents  Problem Solving Workbook Contents CHAPTER 2
R1 = Basic Laws 53 250
Rb Rc
25 × 10
=
=5
=
Ra + R b + R c
25 + 10 + 15
50
R2 = R c Ra
25 × 15
= 7.5
=
Ra + R b + R c
50 R3 = R a Rb
15 × 10
=
=3
Ra + R b + R c
50 The equivalent Y network is shown...
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 Spring '12
 bkav

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