delta to star conversion formula | what is delta to star resistance conversion formula problems with solution questions answers

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
R₁ = R₁₂R₃₁/R₁₂ + R₂₃ + R₃₁, R₂ = R₁₂R₂₃/R₁₂ + R₂₃ + R₃₁
R₃ = R₂₃R₃₁/R₁₂ + R₂₃ + R₃₁
Star to Delta (or T to) Transformation
Superimposed delta and star networks
R₁₂ = R₃, R₂₃ = R₁, R₃₁ = R₃
R₁ R₂ + R₂ R₃ + R₃ R₁
Voltage and Current Division
Voltage Division in Series Circuit
V_R1 = [R₁/R₁ + R₂]V,   V_R2 = [R₂/R₁ + R₂] V
This circuit is a voltage divider circuit.
Current Division in Parallel Circuit
I₁ = [R₂/R₁ + R₂]I
I₂ = [R₁/R₁ + R₂]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

1. superposition theorem
2. thevenin’s theorem
3. norton’s theorem
4. maximum power transfer theorem
5. reciprocity theorem
6. compensation theorem
7. millman’s theorem
8. tellegen’s theorem
9. 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

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

This theorem is not applicable to

1. Non-linear network
2. 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 I₁, V + V_R1 + V_C   and in loop  I₂, 0 = V_L – V_C
By KCL at node 1, I = – I₁ + I₂ + I₃
Let V_C (t) be the total response due to sources V (t) and I (t) from the three equations,
R₁ d/dt (CV_C (t)) + V_C (t) + R₁/L |V_C (P) dp
= V (t) + R₁ I (t) ………………….(1)

2. (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 = – I₁ + I₂ + I₃
Combining these equations,
R₁ d/dt [CV_C‘(t)] + V’_C (t) + R₁/L |V’_C(P) dp = V (t)  ………….(2)

3. (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),

R₁d/dt (CV”_C(t)) + V”_C(T) + R₁/L |V’_C (P) dp = R₁ I (t) …………..(3)
solving eqs. (i), (ii), and (iii) for V_c (t), V_‘C (T) and V²_C (t), respectively, we can verify the superposition principle by showing
V_C (t) = V_C‘ (t) + V_C‘ (t) ……………..(4)

2. Thevenin’s Theorem

V_OC = voltage (open circuit) between aa’ (when I = 0) Z_EQ = equivalent impedance between aa’ terminals when

1. all independent voltage sources are short circuited.
2. 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

3. 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

4. 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_S²R_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_S² + X_S²
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_TH²/4R_L

5. 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’

6. 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

7. 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

8. 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

1. 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

2. 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

3. 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

4. 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

5. 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 <0⁰ and Z_TH <0_TH

(a) Z_L = Z_TH
(b) 0_L = – 0_TH
(c) 0_L = 0_TH
(d) All of these

6. 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

7. 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

8. 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

9. 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

10. A delta-connected network with its wyes equivalent is shown in figure. the resistances R₁,R₂ and R₃ (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

11. 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

12. 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

13. A simple equivalent circuit of the two-terminal network shown in figure, is
14. Find the value R_TH in the figure given below

(a) ₀₀
(b) 0
(c) 3/125
(d) 125/3

15. 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)

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

(a) 4
(b) 3
(c) 2
(d) 1

17. Find the value of R_EQ in the figure given below

(a) 18
(b) 72/13
(c) 36/13
(d) 9

18. 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

19. 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

20. 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

1. (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

2. (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

3. (c)

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

4. (d)

Reciprocity theorem is applicable for linear and bilateral networks.

5. (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.

6. (d)

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

7. (c)

According to maximum power transfer theorem,
Z_P = Z_G

8. (b)

Input voltage = ramp voltage = 100t
maximum output voltage of RC differentiator circuit
= RC dv/dt
= 5 x 10³ x 4 x 10^-6 x 100 = 20 x 10^-1
= 2 x 10^-1 x 10¹ = 2 v

9. (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

10. (d)

R₁ = R_AB x R_AC/R_AB + R_BC + R_CA = 5 x 30/50 = 3
R₂ = R_ab x R_BC/R_AB = 5 x 15/50 = 1.5
R₃ = R_CA x R_CB/R_AB = 15 x 30/50 = 9
R_AB = R_CB + R_BC + R_CA

11. (b)

R_TH = (3||6) + 2
R_TH = 4
V_TH = (6) (6)/3 + 6
V_TH = 4 V

12. (a)

R_EQ = R_N = (2||4) + 2
R_N = 10/3
V₁ = (15/2)/(1/2) + (1/2 + (1/4) = 6 V
I_SC = I_N = V₁/2 = 3A

13. (b)

After replacing all the sources by their internal impledance, equivalent resistance is R and open circuit voltage is V₁.

14. (d)

Using source transformation method,
V_X = 100I₁ + 200I₁ + 50(I₁ + 1)
V_X = 100I₁ – V_X
V_X = 50I₁
50I₁ = 300I₁ + 50I₁ + 50 = I₁  – 1/6
V_TEST – 50 (1-1/6) = 125/3 V
R_TH = V_TEST/I = (125/3) /1=(125/3)

15. (d)

Z_QR = [(5 x 10) + (10 x j10) + (j x 10 x 5)]/5
Z_QR = (10 + J30)

16. (b)

V_S = 4 x 3i/4         V_S/I = 3

17. (d)

Changing the delta network to Y-network
R_EQ = 18||{14 + 10||(6 + 2/3)}
R_EQ = 18||(14 + 4)
R_EQ = 9

18. (c)

R_TH = R_EQ = (7||5) + (6||9) = 6.52

19. (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

20. (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