1.7 Delta-Star Transformation

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1 S Electronic ircuits D ircuits 8.7 Delta-Star Transformation Fig..(a) shows three resistors R, R and R connected in a closed delta to three terminals, and, their numerical subscripts,, and, being opposite to the terminals,, and respectively. It is possible to replace these delta-connected resistors by three star-connected resistors R a, R b, and R c in Fig..(b) If the star-connected network is said to be equivalent to the delta-connected network, the resistance between any two terminals in Fig..(b) must be the same as that between the same two terminals in Fig..(a). R R b R c R R R a Fig..(a) Fig..(b) onsider terminals and in Fig..(a), we have a circuit having a resistance R in parallel with a circuit having resistances R and R in series; hence For Fig..(b), we have Equating (.7.) and (.7.) R ( R + R ) R = (.7.) R a a b R = R + R (.7.) RR + RR + Rb = (.7.)

2 S Electronic ircuits D ircuits 9 Similarly, R b RR + RR + Rc = (.7.4) Rearrange (.7.), (.7.4) and (.7.5) to give R a RR + RR + Rc = (.7.5) RR R a = (.7.6) RR R b = (.7.7) RR R c = (.7.8).8 Star-Delta Transformation onversely, the star-connected network can be replaced with an equivalent delta-connected network. Dividing (.7.6) by (.7.7), R Similarly, dividing (.7.6) by (.7.8), = R R a R b (.7.9) R Substituting for R and R into (.7.6), = R R a R c (.7.0) R + = Rb + Rc RbRc Ra (.7.) Similarly, R = R R a c a c b (.7.) R + = Ra + Rb RaRb Rc (.7.)

3 S Electronic ircuits D ircuits 0.9 Maximum Power Transfer onsidering Fig.., a voltage source E with an internal resistance r is connected in series with an external resistance R L. Energy source I r E R L Fig.. The load current I E = (.8.) r + R L The power delivered to the load is P L E = I RL = R L (.8.) ( r + R ) To achieve maximum power transfer, differentiating (.8.) with respect to R L and setting the result equal to zero. L dpl ( r + RL) RL( r + RL) = E = 0 (.8.) dr 4 ( r + R ) L To find the value of RL to satisfy (.8.), setting the numerator of (.8.) equal to zero to obtain L R L = r (.8.4) For maximum power transfer, the load resistance is chosen to be equal to the internal resistance of the source.

4 S Electronic ircuits D ircuits dditional examples () Find the magnitude and direction of the current flowing through 7Ω resistor in the following network. Ω Ω D 8V 4Ω 7Ω 4V E pproach Kirchhoff s Law pproach F G H Using KL, the currents are assigned as in the following diagram. I Ω Ω I I D I -I I -I 8V 4Ω 7Ω 4V E F G H

5 S Electronic ircuits D ircuits pproach Mesh urrent nalysis Ω Ω D 8V I 4Ω 7Ω 4V I I E F G H pproach Superposition Theorem

6 S Electronic ircuits D ircuits () Obtain the Thevenin s equivalent circuit for the following active network. Ω a Ω 0V 0V D b y KVL, the current flowing through the resistor is I 6 Ω = = (circulate in clockwise direction) + 6 So, V is equal to 0 V = = 0V The Thevenin equivalent resistance R T is 6 R T = + = 5Ω + 6 The Thevenin s equivalent circuit is obtained as 0V 5Ω a b

7 S Electronic ircuits D ircuits 4 () Using Thevenin s Theorem, find the current in the 0Ω resistor of the following network. V Ω 4Ω 4Ω 0Ω pplying Thevenin s Theorem, simplify the networks across - and -. I I I V Ω I V I 4Ω 4Ω I The equivalent circuit is shown as follows. I V 0Ω

8 S Electronic ircuits D ircuits 5 (4) network is arranged as shown in the following figure. alculate the equivalent resistance between (a) and, and (b) and N. (a) y using Star-delta transformation, 5Ω 5Ω 4Ω N Ω R R R 0Ω 0Ω

9 S Electronic ircuits D ircuits 6 (b) y using Delta-star transformation,

10 S Electronic ircuits D ircuits 7 (5) alculate the voltage across - in the network by using delta-star transformation. 5Ω 0V (a) Network is rearranged as follows. R a 5Ω I I R c R eq R c R b 0V 0V I 0V () Transform the -5Ω- delta connection into R a -R b -R c star connection.

11 S Electronic ircuits D ircuits 8 () alculate the equivalent resistance R eq and compute the current I. () Using current division principle, compute the currents I and I. (4) alculate the voltage drop across the and resistors which on the LHS in the middle figure. ompute the voltage difference between and.

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