A.C Fundamentals (Solved Examples 2)

1) Find the R.M.S of,

V (t) =   Sin 2πt +  Sin 6πt

Answer

First of all, We have to find the Period of this function.

We know that,

T = 2π/w

Here we have two sinusoidal functions (V1 & v2).

Figure 22.1 Show the Two sine functions individually. 

  ----- Sin 2πt

  -----Sin 6πt

Figure 22.1 Sin 2πt & Sin k2πt Functions individually, 

So now it is clear that sin2πt is the Primary wave.

T = 2π/w

Here w = 2π ,

Therefore,

T = 1

Now, 

Now,

If you have any problems Leave a comment. or Click Here to Contact on Face Book

Pabindu Lakshitha

B.Sc(Engineer) 

Electrical and Information Engineering. 

A.C Fundamentals (Solved Examples)

Problem 1

aa)      Prove that the R.M.S of,

V = k + A Sin wt   is,

Where, v₁ = A Sin wt   and k is a D.C Component (constant).

bb)      Find the R.M.S Voltage of

V = 10 + 5 Sin (300t + 60⁰ )  

_____________________________________________________________________

Answers

aa)      

Equation for the R.M.S is,

 

Since,

And

Cos 2π – Cos 0  =  1 – 1  = 0

bb)      This is a sin wave with a D.C Component. So We can use the equation we derived above.

Pabindu Lakshitha

B.Sc(Engineer) 

Electrical and Information Engineering. 

Click Here to Contact on Face Book

Fundamentals of Alternative Currents

In this chapter, methods of finding R.M.S & Average values of Alternating Voltages and currents are discussed with examples.

Note: Generation of Alternative currents and Alternative voltages will not be discussed here.

Equations for alternative currents and voltages

An Alternative voltage can be represented by,

E(t) = E_m Sin wt

Where ‘E(t)’ is the Voltage at any instant, ‘E_m’ is the Peak voltage.

Similarly an Alternative current can be represented by,

I(t) = I_m Sin wt

Where ‘I(t)’ is the Current at any instant, ‘I_m’ is the Peak Current.

Root Mean Squared (R.M.S) Value

The average value of a Sine Wave (For a cycle) is zero. Hence, Method of R.M.S is used.

Equation for the R.M.S voltage is,

Or in terms of phase,

Where, ‘v’ is the instantaneous voltage.

For a Sine Wave form,

It can be proved that the R.M.S of a Sine wave is,

. 

where v_{m is the peak (or the maximum) value.} 

You can use the above equation anywhere (Only for Sine waves) without proving. 

Average value

Since the average value for a full cycle of a sine wave-form is zero, we calculate the average value for a half cycle.

Or in terms of phase,

Where, ‘v’ is the instantaneous voltage.

All the other terms have their usual meanings.

This is the End of the quick revision. See you soon in the next article with solved examples. 

Pabindu Lakshitha

B.Sc(Engineer) 

Electrical and Information Engineering. 

Click Here to Contact on Face Book

Wien Bridge

Wien Bridge

You can analyze the circuit shown in figure 4 easily if you followed my previous posts. But here we are about to talk something special on this circuit.

Consider the current goes through the resistor ‘r’ is zero. This means that the voltage difference between ‘r’ is zero. (If you are not familiar with these things, please refer my previous posts regarding ohms low, KCL & star delta transformation)

So that,

V_BD = 0

 Therefore we can see that the voltage between the resistor R₄ and R₂ must be equal.

By voltage dividing method (or you can simply use the ohms low on each branch)

V_DC = (V x R₄ ) / (R₃ + R₄) ---(1)

V_BC = (V x R₂)  / (R₁ + R₂) ----(2)

But we know that V_DC = 0

So that,

V_DC = V_BC

R₄ / (R₃ + R₄) =  R₂ / (R₁ + R₂)

(R₃ + R₄) /R₄ =  (R₁ + R₂) / R₂

1 + (R₃/R₄) = 1 + (R₁/R₂)

(R₃/R₄) = (R₁/R₂)

Or

(R₁/R₃) = (R₂/R₄)

So If the above equations are true for any bridge circuit, we can say that the voltage difference between D & C is zero and therefore there is no use of the resistor ‘r’. Then we can redraw the circuit as shown in figure 17.2.

 This kind of circuits are called Wien bridges. This is an important point. There are so many uses of this Wien bridge method.

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)