what is delta to star resistance conversion formula problems with solution questions answers , delta to star conversion formula meaning and definition

**Network Laws and Theorems**

**Network laws and theorems **As we know, a network is just a combination of various components such as resistive etc., interconnected in all sorts of manner. most of these network cannot be solved merely by appliynig laws of series and parallel circuits. of course kirchhoff’s laws can always be used. but often it makes the solution quite long and laborious. hence, various network theorems have been developed which provide very short and time-saving methods to sovle these complicated circuits.

**Delta to Star (or to T) Transformation**

R_{1} = R_{12}R_{31}/R_{12} + R_{23} + R_{31}, R_{2} = R_{12}R_{23}/R_{12} + R_{23} + R_{31}

R_{3} = R_{23}R_{31}/R_{12} + R_{23} + R_{31}

**Star to Delta (or T to) Transformation**

Superimposed delta and star networks

R_{12} = R_{3}, R_{23} = R_{1}, R_{31} = R_{3}

R_{1} R_{2} + R_{2} R_{3} + R_{3} R_{1}

**Voltage and Current Division**

**Voltage Division in Series Circuit**

V_{R1} = [R_{1}/R_{1} + R_{2}]V, V_{R2} = [R_{2}/R_{1} + R_{2}] V

This circuit is a voltage divider circuit.

**Current Division in Parallel Circuit**

I_{1} = [R_{2}/R_{1} + R_{2}]I

I_{2} = [R_{1}/R_{1} + R_{2}]I

Thus, the current in any branch is equal to the ration of opposite branch resistance to the total resistance value, multiplied by the total current in the circuit.

**Network Theorems**

- superposition theorem
- thevenin’s theorem
- norton’s theorem
- maximum power transfer theorem
- reciprocity theorem
- compensation theorem
- millman’s theorem
- tellegen’s theorem
**Superposition theorem**

The response in any element of a linear bilateral RLC network containing more than one independent voltage and current sources is the sum of responses produced by the sources each acting alone when

- All other independent voltage sources are short circuited.
- All other independent current sources are open circuited.
- All dependent voltage and current sources remain same, as they are

This theorem is not applicable to

- Non-linear network
- non-linear parameters such as power

**Example 1. **The network is shown in figure given below. Determine V_{C} (t) the voltage across capacitor C,by the superposition principle and verify.

**Sol.** 1. Consider the network in fig. (a) :

by KVL in loop I_{1}, V + V_{R1} + V_{C} and in loop I_{2}, 0 = V_{L} – V_{C}

By KCL at node 1, I = – I_{1} + I_{2} + I_{3}

Let V_{C} (t) be the total response due to sources V (t) and I (t) from the three equations,

R_{1} d/dt (CV_{C} (t)) + V_{C} (t) + R_{1}/L |V_{C} (P) dp

= V (t) + R_{1} I (t) ………………….(1)

- (a) Remove I (t), i.e., I (t) = 0, by an open circuit. find the voltage V’
_{C}(t) across capacitor C due to input V (t) alone. Applying KVL and KCL to the network in fig. (b),

V = V_{R1} + V’_{C}

0 = V_{L} – V_{C}

0 = – I_{1} + I_{2} + I_{3}

Combining these equations,

R_{1} d/dt [CV_{C}‘(t)] + V’_{C} (t) + R_{1}/L |V’_{C}(P) dp = V (t) ………….(2)

- (b) Remove V (t) by short-circuiting, i.e., V (t) = 0. find the voltage V”
_{C}(t) across capacitor C due to input I (t) alone. applying KVL and KCL to the network in fig. (c),

R_{1}d/dt (CV”_{C}(t)) + V”_{C}(T) + R_{1}/L |V’_{C} (P) dp = R_{1} I (t) …………..(3)

solving eqs. (i), (ii), and (iii) for V_{c} (t), V_{‘C} (T) and V^{2}_{C} (t), respectively, we can verify the superposition principle by showing

V_{C} (t) = V_{C}‘ (t) + V_{C}‘ (t) ……………..(4)

**Thevenin’s Theorem**

V_{OC} = voltage (open circuit) between aa’ (when I = 0) Z_{EQ} = equivalent impedance between aa’ terminals when

- all independent voltage sources are short circuited.
- all independent current sources are open circuited.

A linear active RLC network which contains sources can independent or dependent voltage and current sources can be replaced by single voltage source V_{OC} in series with equivalent impedance Z_{EQ}

**Norton’s Theorem**

I_{SC} = short circuit current between aa’ (when V = 0)

Z_{EQ} = V_{OC}/I_{SE}

A linear active RLC network which contains one or more independent or dependent voltage and current sources can be replaced by a single current source I_{SC} in parallel with the impedance Z_{EQ}.

This theorem is not valid for

(i) unilateral elements

(ii) non-linear elements

The transformation between the thevenin’s and the norton’s models can be represented in terms of the source transformation.

**Example 2.** To find (i) thevenin’s and norton’s models

**Sol.** 10 A current is divided into three different branches.

10 = 4 + I + I’

I’ = 0

I = 6 A

V_{OC} = I x 10

V_{OC} = 6 x 10 = 60 V

R_{EQ} = 10

I_{SC} = 6 A

R_{EQ} = V_{OC}/I_{SC}

R_{EQ} = 60/10

R_{EQ} = 10

**Maximum Power Transfer Theorem**

This theorem states that, the maximum power is absorbed by one network from another joined to it at two terminals, when the impedance of one is the conjugate of other.

Z_{S} = R_{S} + JX_{S}, Z_{L} = R_{L }+ JX_{L}

**Conditions for maximum power transfer from source to load**

I = V_{S}/Z_{L} + Z_{S} = V_{S}/ (R_{L} + JX_{L}) + (R_{S} + JX_{S})

I = V_{S}/(R_{L} + R_{S}) + J(X_{L} + X_{S})

P_{L} = [I]^{2} R_{L} = V_{S}^{2}R_{L}/(R_{L} = R_{S})^{2} + (X_{L} + X_{S})^{2}

For maximum power,

[P_{L}]/R_{L} = 0 R_{L} = R_{S}

[P_{L}]/X_{L} = 0 X_{L} = – X_{S}

Z_{L} = R_{L} + JX_{L} = R_{S} – JX_{S} = Z*_{S}

Z_{S} = R_{S} + JX_{S}

R_{L} = R_{S}, X_{L} = – X_{S}

Z_{L} = Z_{S}*

So, if source impedance is inductive then load impedance must be capacitive and vice-versa.

. the theorem is applicable for the resistive or reactive circuits.

. the theorem is applicable for the DC voltage source as well as AC voltage source.

. the theorem is not applicable for unilateral or non-linear networks.

**Case 1**

R_{L} = R_{S}

The maximum power transfer takes place only by 50% efficiency.

For an ideal voltage source R_{S} = 0. Then the maximum power transfer takes place with 100% efficiency.

**Case 2**

R_{L} = R_{S}^{2} + X_{S}^{2}

**Case 3**

Z_{S} = R_{S} + JX_{S}

Z_{L} = R_{L} + JX_{L}

Z_{L} = Z_{S}

R_{L} = R_{S}

X_{L} = – X_{S}

**Steps for solving the network related to maximum power transfer theorem**

(i) Remove the load resistance and find the thevenin’s resistance R_{TH} of the source networks.

(ii) From maximum power transfer theorem, this resistance R_{TH} equals load resistance, i.e., R_{TH} = R_{L} . for maximum power transfer.

(iii) Find the thevenin’s voltage V_{TH} across the open-circuited load terminal.

(iv) Maximum power transfer is given by

P_{L, (MAX)} = V_{TH}^{2}/4R_{L}

**Reciprocity Theorem**

In any linear network containing bilateral linear impedances and single source, the ratio of a voltage V introduced in one mesh to the current I in any second mesh is the same as the ratio obtained if the positions of V and I are interchanged.

Following points should be noted :

(i) Any reciprocal network should not have any dependent source.

(ii) All the elements should be time invariants.

(iii) Reciprocal networks do not have initial conditions.

If the networks are reciprocal then from reciprocity theorem

V/I = V’/I’

**Compensation Theorem**

In a linear time invariant network when the impedance Z of an uncoupled branch carrying a current I is changed by Z, the currents in all the branches would change and can be obtained by assuming that an ideal voltage source of V_{C} = IZ has been connected in series with (Z + Z) when all other sources in the network are replaced by their internal resistances.

Consider the network N in fig. 11 (a) having branch impedance Z. let the current through Z be I and its voltage be V.

let Z be the change in Z, the new current I’ is

I’ = V_{OC}/Z + Z + Z_{TH}

I = I’ – I = V_{OC}/Z + Z + Z_{TH} – V_{OC}/ Z + Z_{TH}

I = – I Z/ Z + Z + Z_{TH }= – V_{C}/Z + Z + Z_{TH}

V_{C} + I Z

**Millman’s Theorem**

Millman’s theorem states that if several voltage source in series with admittances are connected.

**Example 6.** Using millman’s theorem. find the current through load R_{L} in the circuit shown in figure below. also find the voltage drop across R_{L}?

**Sol.** let V be the equivalent voltage source and R is the resistance to be placed in series with voltage source of the millman’s equivalent circuit.

– 4 x 1/4 – 2 x 1/4 + 10 x 1/4

v = 1/4 + 1/4 + 1/4 = 4/3 V

R = 1/1/4 + 1/4 + 1/4 = 4/3

4/3

I = 4/3 + 10 = 0.12 A

V_{L} = 0.12 x 10 = 1.2 V

**Tellegen’s Theorem**

In any network, the sum of instantaneous power absorbed by various elements is always equal to zero.

Therefore, the tatal power given out by an active elements (sources) is always equal to total power absorbed by various elements in the branches of the network.

b = number of branches in the network

This theorem is vaild irrespective of

(i) Shape of the network

(ii) Elements contain in the network

(iii) Value of each element contains in the network. The theorem is applicable to DC or AC voltage or current sources so long as the KVL and KCL equations are applicable.

**Intro Exercise-3**

- Which one of the following is applicable to any network linear or non-linear, active or passive, time varying or invariant as long as kirchhoff’s laws are not violated?

(a) tellegen’s theorem

(b) reciprocity theorem

(c) maximum power transfer theorem

(d) superposition theorem

- Which of the following truly represents the thevenin’s equivalent circuit when a voltage source to a load current of 1 A?

(a) V_{TH} = 24 V, R_{TH} = 0.6

(b) V_{TH} = 24 V, R_{TH} = 24

(c) V_{TH} = 23.4 V, R_{TH} = 0.6

(d) V_{TH} = 23.4 V, R_{TH} = 23.4

- For which type of the following thevenin’s and norton equivalents cannot be developed?

(a) DC independent sources

(b) AC independent sources

(c) Independent and dependent

(d) Independent and dependent sources with all controlling current and voltage contained within the network whose equivalent is being developed

- Which of the following is essential for the reciprocity theorem to be applicable ?

(a) linearity

(b) bilateralism

(c) no initial history

(d) all of these

- Which of the following is not a condition for maximum power transfer across a load Z
_{L}<0_{L}in an AC thevenin equivalent circuit of voltage V_{TH}<0^{0}and Z_{TH}<0_{TH}

(a) Z_{L} = Z_{TH}

(b) 0_{L} = – 0_{TH}

(c) 0_{L} = 0_{TH}

(d) All of these

- In which of the following, it is not desired to attain the condition of maximum power transfer?

(a) Electronic circuits

(b) Communicational circuits

(c) Computer circuits

(d) Electric circuits

- A generator of internal impedance [Z
_{C}] delivers maximum power to a load impedance Z_{P}only if

(a) Z_{P} < Z_{C}

(b) Z_{P} > Z_{G}

(c) Z_{P} = Z_{C}

(d) Z_{P} = 2Z_{C}

- A ramp voltage V(t) = 100t V, is applied to an RC differencing circuit with R = 5 K and C = uf. the maximum output voltage is

(a) 0.2 v

(b) 2.0 v

(c) 10.0 v

(d) 50.0 v

- The value of the resistance, R connected across the terminals A and B (from figure) which will absorb the maximum power is

(a) 4.00 k

(b) 4.11 k

(c) 8.00 k

(d) 9.00 k

- A delta-connected network with its wyes equivalent is shown in figure. the resistances R
_{1},R_{2}and R_{3}(in ohm) are respectively)

(a) 1.5, 3 and 9

(b) 3, 9 and 1.5

(c) 9,3 and 1.5

(d) 3, 1.5 and 9

- Find V
_{TH}, R_{TH}in the figure is given below.

(a) 2V, 4

(b) 4 V , 4

(c) 4 V, 5

(d) 2 V, 5

- Find I
_{N}and R_{N}in the figure given below.

(a) 3 A, 10/3

(b) 10 A, 4

(c) 1.5 A, 6

(d) 1.5 A, 4

- A simple equivalent circuit of the two-terminal network shown in figure, is
- Find the value R
_{TH}in the figure given below

(a)_{ 00}

(b) 0

(c) 3/125

(d) 125/3

- In the delta equivalent of the given star connected Circuit Z
_{QR}is equal to

(a) 40

(b) (20 + j10)

(c) (5 + j10/3)

(d) (10 + j30)

- In the circuit shown in figure the effective resistance faced by the voltage source is

(a) 4

(b) 3

(c) 2

(d) 1

- Find the value of R
_{EQ}in the figure given below

(a) 18

(b) 72/13

(c) 36/13

(d) 9

- In the lattice network, the value of R
_{L}for the maximum power transfer to it, is

(a) 6.67

(b) 9

(c) 6.52

(d) 8

- A battery has a short-circuit current of 30 A and an open-circuit voltage of 24 V. if the battery is connected to an electric bulb of resistance 2, the power dissipated by the bulb is

(a) 80 W

(b) 1800 W

(c) 147.378

(d) 228 W

- Measurement made on terminal ab of a circuit of figure yields the current voltage characteristics shown in figure, the thevenin resistance is

(a) 300

(b) – 300

(c) 100

(d) -100

**Answers with Solutions**

- (a)

Tellegen’s theorem is applicable for linear or non-linear, active or passive time varying or invariant networks.

_{k = 1} V_{K} (t) I_{K} (t) = 0

- (a)

I_{L} = 1 A

Voltage drop across voltage source = 0.6 V

R_{S} = V/I_{L} = 0.6/1

R_{S} = 0.6

Now, to draw thevenin equivalent circuit,

R_{TH} = R_{S} when calculate R_{EQ} across ab when voltage source is short-circuited and V_{TH} = 24 V

- (c)

When we use independent and dependent source in a circuit, we cannot make thevenin and norton’s equivalent circuits.

- (d)

Reciprocity theorem is applicable for linear and bilateral networks.

- (c)

For maximum power transfer,

Z_{L} = Z*_{TH}

Z_{TH} = R + JX

0_{TH} = tan^{-1}(X/R)

Z_{L} = Z*_{TH} = R – JX

0_{L} = – tan^{-1}(X/R)

0_{L} = – 0_{TH}

and Z_{L} = Z_{TH} when network is purely resistive. so, option (c) is not condition for maximum power transfer across a load.

- (d)

It is not desired to attain the condition of maximum power transfer for electric circuits.

- (c)

According to maximum power transfer theorem,

Z_{P} = Z_{G}

- (b)

Input voltage = ramp voltage = 100t

maximum output voltage of RC differentiator circuit

= RC dv/dt

= 5 x 10^{3} x 4 x 10^{-6} x 100 = 20 x 10^{-1}

= 2 x 10^{-1} x 10^{1} = 2 v

- (a)

We know that,

According to maximum power transfer theorm,

R = resistance across AB = R_{TH}

Thevenin equivalent resistance,

R_{AB} = 3 x 6/9 + 4 x 4 /8 = 2 + 2 = 4 k

R_{TH} = R_{AB} = 4 K

- (d)

R_{1} = R_{AB} x R_{AC}/R_{AB} + R_{BC} + R_{CA} = 5 x 30/50 = 3

R_{2} = R_{ab }x R_{BC}/R_{AB} = 5 x 15/50 = 1.5

R_{3} = R_{CA} x R_{CB}/R_{AB} = 15 x 30/50 = 9

R_{AB} = R_{CB} + R_{BC} + R_{CA}

- (b)

R_{TH} = (3||6) + 2

R_{TH} = 4

V_{TH} = (6) (6)/3 + 6

V_{TH} = 4 V

- (a)

R_{EQ} = R_{N} = (2||4) + 2

R_{N} = 10/3

V_{1} = (15/2)/(1/2) + (1/2 + (1/4) = 6 V

I_{SC} = I_{N} = V_{1}/2 = 3A

- (b)

After replacing all the sources by their internal impledance, equivalent resistance is R and open circuit voltage is V_{1}.

- (d)

Using source transformation method,

V_{X} = 100I_{1} + 200I_{1} + 50(I_{1} + 1)

V_{X} = 100I_{1} – V_{X}

V_{X} = 50I_{1}

50I_{1} = 300I_{1} + 50I_{1} + 50 = I_{1 } – 1/6

V_{TEST }– 50 (1-1/6) = 125/3 V

R_{TH} = V_{TEST}/I = (125/3) /1=(125/3)

- (d)

Z_{QR} = [(5 x 10) + (10 x j10) + (j x 10 x 5)]/5

Z_{QR} = (10 + J30)

- (b)

V_{S} = 4 x 3i/4 V_{S}/I = 3

- (d)

Changing the delta network to Y-network

R_{EQ} = 18||{14 + 10||(6 + 2/3)}

R_{EQ} = 18||(14 + 4)

R_{EQ} = 9

- (c)

R_{TH} = R_{EQ} = (7||5) + (6||9) = 6.52

- (c)

R_{N} = R_{EQ} = V_{OC}/I_{SE} = 24/30 = 8/10 = 0.8

P = (V_{OC})^{2}/(R_{EQ} + R_{L})^{2} x R_{L}

P = (24)^{2}/(0.8 + 2)^{2}

P = 146.378 W

- (d)

At V = 0 I_{SC} = 30 mA

At I = 0, V_{OC }= – 3 V

R_{TH} = V_{OC}/I_{SC} = -3/30

R_{TH }= – 100