Star Delta Transformation (Solved Problems)

Solved Examples on Star/Delta Transformation 

Q1). Determine the resistance between the terminals A&B and hence find the current through the voltage source. Refer figure 16.1

Answer:

See figure 16.1(a)

The resistors in between point 1, 2&3 are about to replace by a star connected system. Otherwise is difficult to find the total resistance.

So we have to use the delta to star transformation equations.

R₁ = R₁₂R₃₁ / (R₁₂+R₂₃+R₃₁)

R₁ = (60*40)/ (60+40+100)

R₁ = 12Ω

R₂ = R₂₃R₁₂ / (R₁₂+R₂₃+R₃₁)

R₁ = (100*60)/ 200

R₁ = 30Ω

R₃ = R₃₁R₂₃ / (R₁₂+R₂₃+R₃₁)

R₃ = (100*40)/ 200

R₃ = 20Ω

So we can redraw the network as shown in figure 16.2

Now we can easily find the total resistance between A&B terminals

R­_total = [(80+20)//(88+12)] + 30

R_total = 50 + 30

R_total = 80Ω

Applying ohm’s law to the total resistance,

I = V/R

I = 160v/80Ω

I = 2A

Q2) Find the total resistance between A&B terminals for the network shown in figure 16.3

Answer:

See figure 16.3(a)

We are about to replace the delta system by star system in between point 1, 2 &3

So we have to use the delta to star transformation equations.

R₁ = R₁₂R₃₁ / (R₁₂+R₂₃+R₃₁)

R₁ = (3*6)/ (3+6+9)

R₁ = 1Ω

R₂ = R₂₃R₁₂ / (R₁₂+R₂₃+R₃₁)

R₂ = (9*3)/18

R₂ = 1.5Ω

R₃ = R₃₁R₂₃ / (R₁₂+R₂₃+R₃₁)

R₃ = (6*9)/18

R₃ = 3Ω

So now we can replace the system as shown in figure 16.4

Now we can easily find the total resistance between A&B terminals

R_AB = (7Ω+3Ω) + (8.5Ω+1.5Ω) + 1Ω

R_AB = 6Ω

Q3). Find the total resistance between A&B terminals (R_AB) shown in figure 16.5

Answer:

You must understand that you have to use star/delta transformation for this problem. Unlike other problems, in this case it is not pointed out which system of resistance you must replace. So you yourself have to point it out.

This is very important. Though the tutorial problems guide you to find the replaceable systems, in practical level you will have to guide yourself manually. This means you must know how to choose the correct system to apply delta/star transformation.

See figure 16.6

See the circled systems in the figure. You have to replace these systems with delta systems. If you see it carefully, you’ll see that both systems are same (one is upside down of the other). So you don’t need to find two different sets of delta systems. See figure 16.7

This figure shows you the star to delta transformation. As the required equation for transformation are given in my previous post, I’ve directly put the values for the delta system shown in the above figure. Steps for this calculation are shown below.

R₁₂ = R₁ + R₂ + (R₁R₂/R₃)

R₁₂ = 3 + 2 + (3*2)/2

R₁₂ = 8Ω

R₂₃ = R₂ + R₃ + (R₂R₃/R₁)

R₂₃ = 2 + 2 + (2*2)/3

R₂₃ = 16/3Ω

R₃₁ = R₃ + R1 + (R₃R₁/R₂)

R₁₃ = 3 + 2 + (3*2)/2

R₁₃ = 8Ω

So we can redraw the network as shown in figure 16.8

Now we can easily find the total resistance between A&B terminals. For your better understanding I’ve simplified the network. See figure 16.9

So now it is simple.

R_AB = { [ (7+5)//8//8 ] + 5 } //8//4

R_AB = (3 + 5) // 8 // 4

R_AB = 4//4

R_AB = 2Ω

Pabindu lakshitha

B.Sc. (Engineering Undergraduate)

Star, Delta circuits

Star, Delta circuits

Delta/Star transformation

When solving networks with considerable number of branches, sometimes we experiences a great difficulty due to a large number of unknown variable have to be find. Such complicated networks can be simplified by successively replacing delta meshes by equivalent star systems and vice versa.

Consider we have three resistances connected in delta fashion between terminals 1, 2 and 3 as shown in figure 15.1(a). These three resistances can be replaced by three star connected (or ‘Y’ connected) resistances as shown in figure 15.1(b).

If we can arrange these three resistances in such a way that both delta and star systems will show the same resistance between any pair of terminals we can replace any of these arrangement instead of the other one.  This means these two arrangements are electrically equivalent.

How to convert a delta connection to star connection?

In the delta connection, there are two parallel paths between terminals 1 & 2. One is having a resistance of R₁₂ and the other is having a resistance of (R₂₃+R₃₁).

So the resistance between terminals 1&2 = R₁₂ // (R₂₃+R₃₁)

In the star connection, the resistance between the same terminals = R₁ + R₂

For electrically equivalent arrangements these two values should be equal, so

R₁ + R₂ = R₁₂X (R₂₃+R₃₁) / (R₁₂+R₂₃+R₃₁) --------------------- (A)

Similarly for terminals 2 & 3 and terminal 3 & 1, we get

R₂ + R₃ = R₂₃X (R₃₁+R₁₂) / (R₁₂+R₂₃+R₃₁) --------------------- (B)

R₃ + R₁ = R₃₁X (R₁₂+R₂₃) / (R₁₂+R₂₃+R₃₁) ---------------------(C)

By solving these three equations,

R₁ = R₁₂R₃₁ / (R₁₂+R₂₃+R₃₁)

R₂ = R₂₃R₁₂ / (R₁₂+R₂₃+R₃₁)

R₃ = R₃₁R₂₃ / (R₁₂+R₂₃+R₃₁)

Star/Delta transformation

This can be easily done by using equations (A), (B) and (C)

1.       Multiply A&B , B&C and A&C

2.       Add them together and simplify them.

Then we get,

R₁₂ = R₁ + R₂ + (R₁R₂/R₃)

R₂₃ = R₂ + R₃ + (R₂R₃/R₁)

R₃₁ = R₃ + R1 + (R₃R₁/R₂)

*You do not need to hardly remember these equations. You must be intelligent enough to understand the pattern of these equations. We will discuss some solved problems in my next post.

My next post will be a “solved problems” post regarding everything we’ve learnt so far in network analysis

Pabindu lakshitha

B.Sc (Engineering Undergraduate)  

Norton's theorem solved problems(3)

Q4) In order to find the V₁ & R₁ values in the circuit shown in Figure14.12, power dissipation at the variable resistor R_load is measured, for the different values of R_load. The maximum power dissipated at R_load is measured as 81mW when the resistance R is set to 9kΩ. Find V₁ & R₁.

Answer:

You have to solve this problem from the end to beginning. This means we must go backward in the problem.

As the problem says, the maximum power output occurs when R_LOAD = 9kΩ. Therefore according to the maximum power transfer theorem, internal resistance as seen from A-B terminals is also equal to 9kΩ. So this resistance is therefore equal to the Norton’s resistance of A-B terminals.

As the maximum power dissipation at R_LOAD = 9kΩ is equal to 81mW, we can find the current flowing through the load resistance R_LOAD .

P = I²R

81mW = I² X 9KΩ

I² = 9 X 10^-6

I = 3mA

So now we can draw the Norton’s equivalent circuit as shown is figure 14.12.a to determine the Norton’s current.

We can simply understand the Norton’s current is equal to 6mA, as the two resistance are equal and parallel connected.

Now we know the Norton’s current.

So now see figure 14.13.

As the Norton’s resistance is equal to 9kΩ, we can obtain the value of R₁ now.

(R₁//12kΩ) + 5kΩ = 9kΩ

R₁//12KΩ = 4KΩ

R₁ = 6KΩ

Now, we are about to draw the figure that we draw when we wants to find the Norton’s current. Of course we know the Norton’s value. But here we draw it to find V₁. See figure 14.14

So as we know the value of I_N, we can find the voltage across 5kΩ which is equal to the voltage at point c (V).

Ohm’s law to 5kΩ resistor,

V = IR

V = 6mA x 5kΩ

V = 30V

Now by applying nodal equation on point C,

V/12kΩ  + (V-V₁)/6kΩ + 6mA = 0

30/12kΩ + (30-V₁)/6kΩ + 6mA = 0

30/12kΩ + 30/6kΩ + 6mA = V₁/6kΩ

V₁ = 81V

If you have any problem please leave a comment.

Pabindu lakshitha

B.Sc (Engineering Undergraduate)

Norton's theorem solved problems(2)

solved problems on Norton's theorem (2)

Q3) Using Norton’s theorem calculates the current flowing through 12Ω resistor. You may refer figure 14.8

Hope you tried individually to solve this problem as I mentioned in my previous post. I’ll show you how to solve this problem here.

First we remove the 12Ω resistor and short circuit the two terminals to find the Norton’s current as shown in figure 14.9

There are many simple ways to find I_N. I’ll use the nodal equation method. Therefore I grounded the point D.

Applying nodal equation to point D,

V/6 + V/4 + (V-36)/3 = 0

V/6 + V/4 + V/3 = 36/3

3V/4 = 12

V = 16V

Now apply Ohm’s law to 4Ω resistor,

I = V/R

I_N = 16/4

I_N = 4A

Now we are about to find the Norton’s resistance R_N . see figure  14.10

It is very easy.

R_N = (6//3) + 4

R_N = 6Ω

 Now we can redraw the circuit as shown in figure 14.11

I₁ + I₂ = 4A ------- (1)

As the two resistors are connected parallel, the voltage across them should be equal, therefore,

6I₁ = 12I₂

I₁ = 2I₂ ----------- (2)

By (1) & (2)

I₂ = 4/3A

Q4) In order to find the V₁ & R₁ values in the circuit shown in Figure14.12, power dissipation at the variable resistor R_load is measured, for the different values of R_load. The maximum power dissipated at R_load is measured as 81mW when the resistance R is set to 9kΩ. Find V₁ & R₁.

See my next post for answers.

Please try to solve this problem at least three times.

pabindu lakshitha

B.Sc(Engineering Undergraduate) 

NORTON'S THEOREM (Solved problems)

Solved problems on Norton’s theorem

Q1) Find the Norton’s equivalent circuit across A-B terminals for the circuit shown in figure 14.1

Answer:

First we remove the 10Ω resistor and short circuit the terminals A&B. see figure 14.2

The current flowing through the short circuited terminals is called the Norton’s curren I_N.

To find the I_N we apply nodal equation for point C

(V – 30)/5 + V/10 + V/3 = 0

V = 180/19

Ohms law to 3Ω resistance

I = V/R

I_N = (180/19) / 3

I_N = 60/19

This is the Norton’s current.

Now we are about to find the Norton’s resistance. Note that this is equal to the thevenin’s resistance also.

See figure 14.3

R_N = (10//5) + 3

R_N = 19/3Ω

Q2) Find the Norton’s equivalent circuit across A-B terminals for the circuit shown in figure 14.4

Answer:

First we short circuit the two terminals A&B and calculate the current flowing

See figure 14.5

When we short circuit A-B terminals it results in shorting out 12Ω resistor as shown in figure 14.6

Now we can simply find the current I_N by applying Ohms law to 4Ω resistor,

I  = V/R

I_N = 24/4

I_N = 6A

Then we find the R_N . See figure 14.7

R_N = 12//4

R_N = 3Ω

Q3) Using Norton’s theorem calculates the current flowing through 12Ω resistor. You may refer figure 14.8

Wait for my next post to see the solution. Till then, try this yourself. It is very similar to Q1.

NORTON'S THEOREM

Norton’s theorem

Theorem:

Any two terminals of linear active bilateral network can be replaced by an equivalent current source in parallel to a resistance. The current source being the short circuit current through the load terminals and the resistance being internal resistance of the source network looking through the open circuit load terminals.

In thevenin’s theorem we simplified our circuit with a voltage source in series with a resistance. In Norton’s theorem we use a current source in parallel to a resistance instead of a voltage source in series to a resistance.

This theorem reduces a normally complicated network to a simple parallel circuit consisting of

( 

   a)    An ideal current source I_N Of infinite internal resistance and,

(  

   b)   A resistance R_N  in parallel with it.

I_N is the current which would flow through a short circuit placed across terminals A&B. R_N is equal to the thevenin’s resistance R^th . So R_N is the resistance looking back from A-B terminals.

How to Nortonise a given circuit:

Consider we have to nortonise the network in figure 13.1 as seen from A-B terminals

First we short circuit the given two terminals (A&B) and find the current flowing through. 

See figure 13.2, this current is called the Norton’s current (I_N).

Then again we open circuit the two terminals. And find the resistance as we do in the thevenin’s theorem. Note that for a given circuit, Norton’s resistance is equal to the thevenin’s resistance.

After that we can redraw the figure as shown in figure 13.3

Note that we connect the Norton’s resistance in parallel with Norton’s current source.

 We will discuss some solved problems on Norton's in my next post.

pabindu lakshitha
B.Sc(Engineering Undergraduate)

THEVENIN'S THEOREM SOLVED PROBLEMS (2)

SOLVED PROBLEMS (2)

Q4) In order to determine the unknown voltage V₀ and the resistance R in the circuit shown in figure 12.1 , power dissipation at the variable resistor R_L is measured, for the different values of R. the maximum power dissipation at the variable resistor R_L is measured as 147mW when the resistance R_L is set to 3kΩ. Determine the values of V₀ & R.

It is reccomended to workout previous problems before this. And you should try to solve this problem without looking at the answers given below. Don’t even look at  single word in the answers if you didn’t try to solve it atleast three times.

Answer:

It says that the maximum power dissipation at R_L occurs at 3kΩ. So you may understand that the thevenin resistance R^TH is also 3kΩ. This is one of the most important practical usages of thevenin’s theorem. This is the combination of both maximum power transfer theorem and thevenin’s theorem.

If you still can’t understand this method, consider the tehvenin equivalent circuit with the load connected. Now this circuits it exactly equivalent to the original circuit given. So instead of the given circuit we can use the thevenin equivalent circuit with load connected. Now, you may see that there are only two resistances available (R_LOAD and R^TH). According to maximum power transfer theorem, to get the maximum power output these two resistance values should be equal.

This means,

At maximum power dissipation,

R_LOAD = R^TH

So as it given in the problem, at maximum power dissipation R_load = 3kΩ

Therefore

R^TH = 3KΩ

Now we can find the resistance ‘R’

From figure 12.2

[ (R+2) // 3 ] + 1 = 3 (all values in kΩ)

[ (R+2)*3 / R+5 ] + 1 = 3

3R + 6 = 2R + 10

R = 4Ω

Now we have find R.

We have given that the maximum power dissipation in R_LOAD is 147mW at R_LOAD = 3.

Apply,

 p = V²/R     to  R_LOAD

147mW = V²/ 3kΩ

V = 21v

This is the voltage across load when the maximum power dissipates, See figure 12.2(a).

So

V^TH/2 = 21V

V^TH = 42V

Now see figure 12.3

Note that V_C = V^TH = 42V

We have grounded the point B.

Apply nodal equation to point C

(V_C – V)/2kΩ  +  (V_C – 0)/3kΩ – 12mA – 13mA = 0

(42 – V) /2kΩ + 42/3kΩ -25mA = 0

(42 – V) x 10^-3/2 = 11x10^-3

42 – V = 22

V = 20v

But this is not what we need. We need to find V₀

V = V₀ + V₁  (see figure 12.3)

V₀ + V₁ = 20 ---------------------------(1)

Applying nodal equation to point B

(0 – V₁)/ 4kΩ  + (0 - V_C)/ 3kΩ  + 13mA = 0

-V₁/4 + (-42)/3 +13 = 0

-V₁/4 -1 = 0

V₁  = -4v -------------(2)

substitute (2) on (1)

V₀  = 24v

 LEAVE YOUR PROBLEM AS A COMMENT.

pabindu lakshitha
B.Sc(engineering undergraduate)

Maximum power transfer theorem

Maximum power transfer theorem:

This is a very important theorem. Most usually for analyzing electronic and communication networks where main consideration is to transfer maximum power to the load irrespective of the efficiency, this theorem is highly important.

Theorem:

A resistive load connected to a dc network receives maximum power when the load resistance is equal to the internal resistance of the source network as seen from load terminals.

As this theorem is very easy to understand, there is no need of a big clarification. You can simply understand the theorem. If you understood it, then you will understand the important of thevenin’s theorem. After determine the thevenin’s voltage, to transfer the maximum power to the load, we have to add the load resistance equal to the thevenin’s theorem.

So here onwards we will use this theorem with thevenin’s theorem. 

SOLVED PROBLEMS ON THEVENIN'S THEOREM (1)

SOLVED PROBLEMS

Q1) Determine the thevenin’s equivalent  circuit between the terminals A&B For the circuit shown in figure 11.1 .

Answer:

First we are about to find the thevenin’s resistance. To find the thevenin’s resistance we remove the resistance R_L and open circuit the AB terminals. Then we remove the voltage source and short circuit it. See figure 11.2

We can easily find the thevenin’s resistance now.

R^th = (5//10 + 3)

Now we have to find the thevenin’’s voltage. For this we remove the load resistance R_L . See figure 11.3

Note that there is no current flow through 3ohms resistance as it is open ended from point A. Hence in nodal equation we don’t have to add the current through this resistance.

Applying nodal equation method to point c,

(V-30)/5  + (V-0)/10  = 0

V = 20V

From figure 11.3 you can see that,

V_A = V_C

V_A = V^TH

Therefore V = V^TH

V^TH = 20V

Q2) Find the equivalent e.m.f of the network when viewed from terminals A&B and the equivalent resistance of the network when looked from terminals A&B and the current through load resistance R_L . Refer figure 11.4

Answer:

Current in the network after load resistance has been removed (figure 11.5)

I = 24/(12+3+1)

I = 1.5A.

Applying Ohm’s law on 12  Ohm’s resistance,

Voltage across A and B = V^TH = 12 Ohms x 1.5A

Now we are about to find the R^TH

It is very easy. Just remove and short circuit the voltage source and find the equivalent resistance as seen from A&B terminals.

NOTE: YOU SHOULD LEAVE THE INTERNAL RESISTANCE OF THE VOLTAGE SOURCE THOUGH YOU REMOVE IT. SEE FIGURE 11.4.(a)

R^th  = (3+1)//12 = 3 Ohms

Hence we have find the thevenins equivalent circuit. So we can easily find the current through load resistance. See figure 11.5.(a)

I_LOAD = 18 / (3+15)

I_LOAD = 1A

Q3) determine the thevenin’s equivalent circuit between the terminals A&B. you may refer figure 

11.6

Answer:

This network contains a current source. So it will be important to you to understand how to deal with  a current source in thevenin’s theorem.

Figure 11.7 shows the circuit to find R^TH. Note that two terminals between the current source has been open circuited while the voltage source do the opposite.

R^TH =  5 + { [ (10//10 ) + 2 ] // 3 }

R^TH = 7.1 Ω

From figure 11.8

Voltage at point C = V_c = V^th  (as there is no current through 5Ω resistor)

We have grounded the point B to make our calculation easier.

Nodal equation to point C

(V_C – 0)/3 + (V_C-V_D)/ 2 = 0

 5V_C – 3V_D = 0 -------------------------------------------(1)

Nodal equation to point D

(V_D - V_C)/2 + V_D/10 + (V_D -5)/10 – 1 = 0

 7V_D – 5V_C = 15 ---------------------------------------(2)

By solving 1&2

V_c  = 2.25V

V_C = V^TH

V^TH  = 2.25V

I’ll add more examples in my next post.

PLEASE LEAVE YOUR PROBLEM AS A COMMENT. OR MAIL TO: pabindu@gmail.com

Pabindu lakshitha

B.Sc (Engineering Undergraduate)