Fundamentals of Electric Circuits

A major advantage of analyzing circuits using Kirchhoff's laws like we did in Chapter 3 is that we can analyze a circuit without modifying the original configuration but the major disadvantage is that it will take some time to solve it because of the long and complex circuitry.

Linear Property

Linear property is the linear relationship between cause and effect of an element. This property gives linear and nonlinear circuit definition. The property can be applied in various circuit elements. The homogeneity (scaling) property and the additivity property are both the combination of linearity property.


The homogeneity property is that if the input is multiplied by a constant k then the output is also multiplied by the constant k. Input is called excitation and output is called response here. As an example if we consider ohm’s law. Here the law relates the input i to the output v.


Mathematically, v= iR


If we multiply the input current i by a constant k then the output voltage also increases correspondingly by the constant k. The equation stands, kiR = kv


The additivity property is that the response to a sum of inputs is the sum of the responses to each input applied separately.


Using voltage-current relationship of a resistor if


v1 = i1R and v2 = i2R


Applying (i1 + i2) gives


V = (i1 + i2)R = i1R+ i2R = v1 + v2


We can say that a resistor is a linear element. Because of the voltage-current relationship satisfies both the additivity and the homogeneity properties.


We can tell a circuit is linear if the circuit both the additive and the homogeneous. A linear circuit always consists of linear elements, linear independent and dependent sources.


What is linear circuit?


A circuit is linear if the output is linearly related with its input.


The relation between power and voltage is nonlinear. So this theorem cannot be applied in power.


See a circuit in figure 1. The box is linear circuit. We cannot see any independent source inside the linear circuit.




                                                      Figure 1


The linear circuit is excited by another outer voltage source vs. Here the voltage source vs acts as input. The circuit ends with a load resistance R. we can take the current I through R as the output.


Suppose vs = 5V and i = 1A. According to linearity property if the voltage is multiplied by 2 then the voltage vs = 10V and then the current also will be multiplied by 2 hence i = 2A.


The power relation is nonlinear. For example, if the current i1 flows through the resistor R, the power p1 = i12R and when current i2 flows through the resistor R then power p2 = i22R.


If the current (i1 + i2) flows through R resistor the power absorbed


P3 = R(i1 + i2)2 = Ri12 + Ri22 + 2Ri1i2 ≠ p1 + p2


So the power relation is nonlinear. Circuit solution method superposition is based on linearity property.

Superposition

The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or current through) that element due to each independent source acting alone.




To apply this principle for analysis, we must follow these procedures:


1. Turn off all independent sources except one. Find the output (voltage or current) due to that source.


2. Repeat Step 1 for each independent source.


3. Add the contribution of each source to find the total output.



Also:


*Superposition is based on circuit linearity


*Must analyze as many circuits as there are independent sources


*Dependent sources are never turned off


*As with examples, is usually more work than combining resistors, the node voltage analysis, or mesh current analysis


*Is an important idea


*If you want to consider a range of values for an independent source, is sometimes easier than these methods

We continue with our analysis of linear circuits by introducing two new passive and linear elements: the capacitor and the inductor.
All the methods developed so far for the analysis of linear resistive circuits are applicable to circuits that contain capacitors and inductors.
Unlike the resistor which dissipates energy, ideal capacitors and inductors store energy rather than dissipating it.



Capacitor

A capacitor is formed from two conducting plates separated by a thin
insulating layer.
If a current i flows, positive change, q, will
accumulate on the upper plate. To preserve
charge neutrality, a balancing negative charge
will be present on the lower plate.












Types of Capacitor

Capacitor symbol represents the two separated
plates. Capacitor types are distinguished by
the material used as the insulator.

Polystyrene: Two sheets of foil separated by a
thin plastic film and rolled up to save space.
Values: 10 pF to 1 nF.

Ceramic: Alternate layers of metal and ceramic
(a few μm thick). Values: 1 nF to 1 μF.

Electrolytic: Two sheets of aluminium foil
separated by paper soaked in conducting
electrolyte. The insulator is a thin oxide layer
on one of the foils. Values: 1 μF to 10mF.
Electrolytic capacitors are polarised: the foil with the oxide layer must
always be at a positive voltage relative to the other (else explosion).
Negative terminal indicated by a curved plate in symbol.








Inductors

Inductors are formed from coils of wire, often
around a steel or ferrite core.

In this chapter we will tackle about some techniques that is commonly applied in circuit design and analysis. These techniques includes combining resistors in series or parallel, voltage division, current division and delta-to-wye and wye-to-delta transformations.






To make a current flow through a resistance there must be a voltage across that resistance. Ohm's Law shows the relationship between the voltage (V), current (I) and resistance (R). It can be written in three ways:



V = I × R                 or                 I =    V                           or                             R = V
                                                              R                                                                    I


where:

V = voltage in volts (V)

I = current in amps (A)

R = resistance in ohms (Ω)

or:

V = voltage in volts (V)

I = current in milliamps (mA)

R = resistance in kilohms (kΩ)


For most electronic circuits the amp is too large and the ohm is too small, so we often measure current in milliamps (mA) and resistance in kilohms (kΩ). 1 mA = 0.001 A and 1 kΩ = 1000Ω.

The Ohm's Law equations work if you use V, A and Ω, or if you use V, mA and kΩ. You must not mix these sets of units in the equations so you may need to convert between mA and A or kΩ and Ω.




A node is an uninterrupted segment of wire of any shape. Unless otherwise stated, the wire is considered to be perfect and have no resistance. Therefore, all points on the node have the same voltage. Physically speaking, a node may connect components inches away. However, electrically speaking all the components are connected to a single point. In a schematic representation, node1 extends from the top terminal of the 120V source to the left terminal of the 33 ohm resistor. Node2 includes all the wire segments between R1, R2 and R4. No matter how long the lines extend between components, no potential difference exists between them as long as they are on the same node. A branch is any element connected between two nodes. R4 and R3 are examples of a branch. A loop is any sequence of elements connected between nodes that starts and ends at the same node. You only include a node once in a loop. There may be loops within loops as loop1 and loop2 are within loop3.






The voltage across each element and the current through each element in an electric circuit are governed by two general results which are summarized in Kirchhoff's two laws. Since Kirchhoff's laws are derived from general physical properties of electricity, they are applicable to all kinds of electric circuits.




             Kirchhoff's Current Law


As a direct consequence of the conservation of charge, namely charge can neither be created nor destroyed, the node, being of negligible physical size, holds no charge. For instance, referring to figure 1.6, the sum of , and must equal zero.





Figure 1.6: Kirchhoff's current law



Formally, KCL states that the algebraic sum of the currents in all the branches that converge in a common node is equal to zero. In mathematical form, for n branches converging into a node, KCL states that  where is the current flowing in the kth branch and its direction is assumed to be pointing towards the node.



                 Kirchhoff's Voltage Law

When a charged particle is moved from a point to another, the work done is , where V is the voltage across the two points and is the amount of charge on the particle. Consider a particular case where the two points are actually the same point in the circuit. In this case, the work done is zero. By the same argument, if a unit charge is moved around a closed path such as the square path shown in figure 1.7, the work done is zero, i.e.,



Figure 1.7: Kirchhoff's voltage law

Formally, KVL states that the algebraic sum of the voltages between successive nodes in a closed path in a circuit is equal to zero. In mathematical form, for a closed path with successive nodes , KVL states that
Remarks -- Any circuit that has a solution must satisfy Kirchhoff's laws. From the properties of independent sources, we can immediately conclude that a circuit cannot be solved if there exists a loop that is formed exclusively of independent voltage sources. Thus, short-circuiting an independent voltage source, as remarked earlier, is a particular case where KVL is violated. Similarly, a circuit cannot be solved if there exists a node to which only independent current sources are connected. Also, open-circuiting an independent current source is a particular case where KCL is violated.



The Voltage Divider


For a single resistor with a voltage across it, Ohm's Law is quite an impressive tool. Put a few resistors together in series, however, and things get even more interesting! The total resistance of resistors in series is simply the sum of their individual resistances. By combining this concept with Ohm's Law we can design one of the most commonly used circuits in a guitar amplifier: the voltage divider.
Problem


Consider the power supply of a Mesa Boogie Bass 400 amplifier. It uses two 150k resistors in series to equalize the voltage across the filter capacitors and to act as bleed resistors that slowly discharge the capacitors when the amplifier is turned off. (If only all amplifiers were designed this way - it would be a much safer world!) The total voltage across both resistors is 534 volts. What is the voltage across each individual resistor?



Solution


The total resistance across both resistors is


150k + 150k = 300k


According to Ohm's Law, the current passing through the resistors is


534/300k = 1.78mA


Thus 1.78mA passes through each of the resistors individually. By applying Ohm's Law once more, we get the voltage across one resistor:


(1.78mA)(150k) = 267V




A voltage divider is often used to attenuate a signal by a prescribed amount. We often see them in guitar amplifier input circuits, for example, where they attenuate the signal to give the guitar player a low-gain input jack. They are also used in the second stage of a paraphase inverter to equalize the amplitudes of the two output phases. Other guitar circuits may not appear to be voltage dividers at first, but we frequently discover that at certain frequencies the components effectively create one. The voltage divider thus represents an essential tool of guitar amplifier design.
Problem


Consider the second stage paraphase inverter of the Gibson GA-20T amplifier. An input signal is applied across 220k and 4.7k resistors in series. The output voltage is the voltage across the 4.7k resistor. Given that the input signal is 1 volt, what is the output voltage?



Solution


The total resistance of both resistors in series is 224.7k. This means that according to Ohm's Law the current is


1/224.7k = 0.00445mA


This current passes through the 4.7k resistor, so according to Ohm's Law, the voltage across it is


(0.00445mA)(4.7k) = 0.021


So the input is 1 volt and the output is 0.021 volts. The output is thus only 2.1 percent of the input. This is quite a bit of attenuation and not a random value picked by the engineers in Kalamazoo. The resistor values were chosen to reduce the input voltage by exactly this amount.


Notice that we computed the output voltage in two separate steps. First we computed the current that passes through both resistors. Then we used this current to determine the output voltage. We really weren't interested in the current. We just needed it as an intermediate step. Let's avoid the middleman and combine the two steps into one equation to simplify the calculation. Looking at our procedures in reverse, the output voltage is


(0.0445mA)(4.7k) = 0.21


and the current is


1/224.7k = 0.0445mA


We combine these into a single equation by noting that the 0.0445mA in the first formula is equal to 1/224.7k in the second. Thus


(1/224.7k)(4.7k) = 0.021


So the output voltage is equal to the input voltage divided by the total resistance in series, times the resistance across the output. This works for any combination of two resistors R1 and R2, as shown in this circuit:





The output voltage is equal to the input voltage times the output resistor R2 divided by the total resistance R1+R2. This is the fundamental Principle of a voltage divider.





Consider the circuit shown in figure where two resistors are connected parallel. Supplying voltage for current resistor. From Ohm’s law we get


v = i1R1

= i2R2

or i1 = V/R1, i2 = V/R2


Applying KCL,

i = i1 + i2


Substituting and adding the value in equation (i) then we get,


i = V/R1+ V/R2

= V (1/R1 + 1/R2)

= V/Req


Here Req is the equivalent resistance.


1/Req = (1/R1) + (1/R2)

Or 1/Req = (R1 + R2)/ R1 R2

Req = R1 R2/ (R1 + R2)


Using current divider rule,


I1= R2i/ (R1 + R2)


I2= R1i/ (R1 + R2)







Most of the time we can solve the circuit using our idea of series and parallel connection and ohm’s law or KCL or KVL. But sometimes we encounter some circuit where the formations of elements are neither in series nor in parallel. For example look at the figure of bridge network below:





Now look at the following two formations which look like Wye and Delta. These formations usually found in three phase connections, filters, or in matching networks.









(a) Y netwrk (b) Delta Network





(a) Delta Network





(b) Pie Network



When we analyze a circuit we may find a certain formation to be helpful. And we might want to change one formation into other. Suppose it might helpful for us to work with Wye formation in three phase rather in Delta formation. So we have to know how to transform a Delta network into Wye network.





Delta to Wye conversion:


Look at the figure below:








Now for Delta to Wye conversion:


R1 = (Rb . Rc) / (Ra + Rb + Rc)


R2 = (Rc . Ra) / (Ra + Rb + Rc)


R3 = (Ra . Rb) / (Ra + Rb + Rc)


We don’t have to memorize these formulas rather we can easily remember these.Each resistor of the Wye network is the product of the two adjacent resistors of Delta network, divided by the sum of all three resistors of the Delta network.


Wye to Delta Conversion:


Look at the figure again:





Now for Wye to Delta conversion:


Ra = (R1.R2 + R2.R3 + R3.R1) / R1


Rb = (R1.R2 + R2.R3 + R3.R1) / R2


Rc = (R1.R2 + R2.R3 + R3.R1) / R3


We don’t have to memorize this also. The easy way to remember is, each resistor in the Delta Network is the sum of all possible product formation of Wye network’s resistors taken two at a time, divided by the opposite Wye resistor.