Chapter 7: Alternating Current

AC VOLTAGE APPLIED TO A RESISTOR

 ​​Alternating Currents are used almost as a standard by electricity distribution companies. In India, 50 Hz Alternating Current is used for domestic and industrial power supply. Many of our devices are in fact nothing but resistances. These resistances cause some voltage drop but since the voltage this time is alternating, these voltage drops are dealt with differently

AC VOLTAGE APPLIED TO AN INDUCTOR

 

AC voltage is applied to the inductor. In order to find out the equation, we will consider the circuit as shown in the figure below. we have an inductor and an AC voltage V, which is represented by the symbol ~. The voltage produces a potential difference across its terminals that varies using a sinusoidal equation. The difference that is, the AC voltage thus can be given as,


From the equation, we deduce that vis used to signify the amplitude for the oscillating potential to denotes the differences. The angular frequency is given by ω. The current can be calculated by using the Kirchhoff’s loop rule. The equation which forms is as under,

Using the above equation in the given circuit,

The value of current as given by,

Therefore, the integration constant is zero.

Here, the amplitude of the current is given by

The quantity ωL can be said to be equivalent to the resistance of this device and is termed as the inductive resistance. We denote the inductive resistance of the device as XL.

Thus, we can say that the amplitude of current in this circuit is given as

AC Voltage Applied to a Capacitor
An AC voltage source is connected to a capacitor. The expression for the voltage from the voltage source is given by v = vmsin(ωt). A capacitor is an electrical device that stores electrical energy. It is a passive electronic component with two terminals. The effect of the capacitor is known as capacitance. A capacitor when connected to a voltage source draws current from the source so as to charge itself. Once the capacitor is charged, the potential at its plates becomes equal to the potential at the battery. At this point, the current stops flowing into the capacitor. This is called the charging of the capacitor. 
At a particular time “t”, denotes the charge on the capacitor by “q”. The instantaneous voltage across the capacitor is given by, 

Using the kirchhoff’s rule, 

Since the current is continuously changing, to find the current. Derivative of the charge is required, 

Differentiating the given equation, 

 

 

i = vmωC cos(ωt) 

Rearranging the above equation,

i = imsin(ωt + π/2) 

Here, im = vmωC. It is the amplitude of the oscillating current. It can also be re-written as, 


This equation when compared to the ohm’s law gives 1/ωC as resistance. It is called capacitive reactance and it is denoted by XC

Now, the amplitude of the current becomes, 

im = 

AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT
we will follow the analytical analysis of the circuit.
Analytical solution
As i

, we can write

Hence, writing the voltage equation in terms of the charge q through the circuit, we can write,

The above equation can be considered analogous to the equation of a forced, damped oscillator. In order to solve the equation, we assume a solution given by,


So,

And

Substituting these values in the voltage equation, we can write,

Here, we have substituted the value of Xc and XL by Xc = 1/ωC and XL = ω L.

As we know,

hence, substituting this value in the above equation, we get,

Now, let

So we can say,

Now, comparing the two sides of the equation, we can write,

And,

Hence, the equation for current in the circuit can be given as,

LC OSCILLATION
The Difference between the Direct and Alternating current is that the direct current (DC), travels only in one direction while the alternating current (AC) is an electric current that alternates direction on occasion and alters its amplitude continuously over time.In LC Oscillations, the charge equation is as follows:
q = qm cos(ωt)
We derive the present equation by differentiating this equation:
i = dq/dt
i = –qm ωsin(ωt)
The formula for calculating the energy stored in a capacitor is:
U= q2/2C
Substituting the equation for a given time interval t;
U= qm2/2C × (ωt)
The formula for calculating the energy stored in an inductor is:
U= 1/2 L i2
Substituting the capacitor’s equation for the same amount of time;
U= 1/2 L qmω(ωt)
Since the angular frequency, ω=1/ √LC
 U= qm2/2C × (ωt)
As a result, the LC Oscillations’ total energy will be;
U = U+ UC
U = qm2/2C × (ωt)+qm2/2C × (ωt)
U = qm/ 2C

Transformer 
There are usually two coils primary coil and secondary coil on the transformer core. The core laminations are joined in the form of strips. The two coils have high mutual inductance. When an alternating current pass through the primary coil it creates a varying magnetic flux. As per faraday’s law of electromagnetic induction, this change in magnetic flux induces an emf (electromotive force) in the secondary coil which is linked to the core having a primary coil. This is mutual induction.

Overall, a transformer carries the below operations:

  1. Transfer of electrical energy from circuit to another
  2. Transfer of electrical power through electromagnetic induction
  3. Electric power transfer without any change in frequency
  4. Two circuits are linked with mutual induction

1. Core

The core acts as a support to the winding in the transformer. It also provides a low reluctance path to the flow of magnetic flux. The winding is wound on the core as shown in the picture. It is made up of a laminated soft iron core in order to reduce the losses in a transformer. The factors such as operating voltage, current, power etc decide core composition. The core diameter is directly proportional to copper losses and inversely proportional to iron losses.

2. Windings

Windings are the set of copper wires wound over the transformer core. Copper wires are used due to:

  • The high conductivity of copper minimizes the loss in a transformer because when the conductivity increases, resistance to current flow decreases.
  • The high ductility of copper is the property of metals that allows it to be made into very thin wires.

There are mainly two types of windings. Primary windings and secondary windings.

  • Primary winding: The set of turns of windings to which supply current is fed.
  • Secondary winding: The set of turns of winding from which output is taken.

The primary and secondary windings are insulated from each other using insulation coating agents.

3. Insulation Agents

Insulation is necessary for transformers to separate windings from each other and to avoid short circuit. This facilitates mutual induction. Insulation agents have an influence on the durability and the stability of a transformer.

Following are used as an insulation medium in a transformer:

  • Insulating oil
  • Insulating tape
  • Insulating paper
  • Wood-based lamination

Ideal Transformer

The ideal transformer has no losses. There is no magnetic leakage flux, ohmic resistance in its windings and no iron loss in the core.

Vs = (Ns/ Np) Vp

Power at the input end is same as the power at the output end.

Therefore Pintput = Poutput

IpVp =  IsVs