Kirchhoff Current Law and Kirchhoff Voltage Law

Kirchhoff Laws

There are some simple relationship between currents and voltages of different branches of an electrical circuit. These relationship are determined by some basic laws which are known as Kirchhoff Laws or more specifically Kirchhoff Current and Voltage laws. These laws are very helpful in determining the equivalent resistance or impedance (in case of AC) of a complex network and the currents flowing in the various branches of the network. These laws are first derived by Guatov Robert Kirchhoff and hence these laws are also referred as Kirchhoff Laws.

Kirchhoff Current Law

In an electrical circuit the electric current flows rationally as electrical quantity. As the flow of current is considered as flow of quantity, at any point in the circuit the total current enters is exactly equal to the total current leaves the point. The point may be considered any where in the circuit.
Suppose the point is on the conductor through which the current is flowing, then the same current crosses the point which can alternatively said that the current enters at the point and same will leave the point. As we said the point may be any where on the circuit, so it can also be a junction point in the circuit. So total quantity of current enters at the junction point must be exactly equal to total quantity of current leave the junction. This is very basic thing about flowing of electric current and fortunately Kirchhoff Current law says the same. The law is also known as Kirchhoff First Law and this law stated that at any junction point in the electrical circuit, the summation of all the branch currents is zero. If we consider all the currents enter in the junction are considered as positive current then convention of all the branch currents leaving the junction are negative. Now if we add all these positive and negative signed currents obviously we will get result of zero.
The mathematical form of Kirchhoff Current Law is as follows,
We have a junction where n number of beaches meet together.

Let’s I₁, I₂, I₃, …………………. I_m are the current of branches 1, 2, 3, ……m and
I_{m + 1}, I_{m + 2}, I_{m + 3}, …………………. I_n are the current of branches m + 1, m + 2, m + 3, ……n respectively.


The currents in branches 1, 2, 3 ….m are entering to the junction
whereas currents in branches m + 1, m + 2, m + 3 ….n are leaving from the junction.

So the currents in the branches 1, 2, 3 ….m may be considered as positive as per general convention
and similarly the currents in the branches m + 1, m + 2, m + 3 ….n may be considered as negative.


Hence all the branch currents in respect of the said junction are

+ I₁, + I₂, + I₃,…………….+ I_m, − I_{m + 1}, − I_{m + 2}, − I_{m + 3}, ……………… and − I_n.


Now, the summation of all currents at the junction is
I₁ + I₂ + I₃ + …………….+ I_m − I_{m + 1} − I_{m + 2} − I_{m + 3}………………− I_n.


This is equal to zero according to Kirchhoff Current Law.
⇒ I₁ + I₂ + I₃ + …………….+ I_m − I_{m + 1} − I_{m + 2} − I_{m + 3}………………− I_n = 0


The mathematical form of Kirchhoff First Law is ∑ I = 0 at any junction of electrical network

Kirchhoff Voltage Law

Kirchhoff Voltage Law

This law deals with the voltage drops at various branches in an electrical circuits. Think about one point on an closed loop in an electrical circuit. If some one goes to any other point on the same loop, he or she will find that the potential at that second point may be different from first point. If he or she continues to go to some different point in the loop, he or she may find some different potential at that new location. If he or she goes on further along that closed loop ultimately he or she reaches the initial point from where the journey was started. That means he or she comes back to the same potential point after crossing through different voltage levels. It can be alternatively said that net voltage gain and net voltage drops along a closed loop are equal. That is what Kirchhoff Voltage law states. This law is alternatively known as Kirchhoff Second Law.
If we consider a closed loop, conventionally if we consider all the voltage gains along the loop are positive then all the voltage drops along the loop should be considered as negative. The summation of all these voltages in a closed loop is equal to zero. Suppose n numbers of back to back connected elements form a closed loop. Among these circuit element m number elements are voltage source and n – m number of elements drop voltage such as resistors.

The voltages of sources are V₁, V₂, V₃,………………. V_m
and voltage drops across the resistors respectively, V_{m + 1},
V_{m + 2}, V_{m + 3},………………… V_n.
As it said that the voltage gain conventionally considered as positive, and voltage drops are considered as negative, the voltages along the closed loop are
+ V₁, + V₂, + V₃,………………. + V_m, − V_{m + 1}, − V_{m + 2}, − V_{m + 3},…………………− V_n.
Now according to Kirchhoff Voltage law the summation of all these voltages results to zero.
That means, V₁ + V₂ + V₃ + ………………. + V_m − V_{m + 1} − V_{m + 2} − V_{m + 3} + …………………− V_n = 0
So accordingly Kirchhoff Second Law, ∑V = 0

Application of Kirchhoff laws to circuits

The current distribution in various branches of a circuit can easily be found out by applying Kirchhoff Current law at different nodes or junction points in the circuit. After that Kirchhoff Voltage law is applied each possible loops in the circuit and generate algebraic equation for every loop. By solving these all equations, one can easily find out different unknown currents, voltages and resistances in the circuits.

Some popular conventions we generally use during applying KVL

1) The resistive drops in a loop due to current flowing in clockwise direction must be taken as positive drops.

2) The resistive drops in a loop due to current flowing in anti-clockwise direction must be taken as negative drops.

3) The battery emf causing current to flow in clockwise direction in a loop is considered as positive.

4) The battery battery emf causing current to flow in anti-clockwise direction is referred as negative.

Electrical Power Factor | Calculation & Power Factor Improvement

In general power is the capacity to do work. In electrical domain, electrical power is the amount of electrical energy that can be transferred to some other form (heat, light etc) per unit time. Mathematically it is the product of voltage drop across the element and current flowing through it.
Considering first the DC circuits, having only DC voltage sources, the inductors and capacitors behave as short circuit and open circuit respectively in steady state. Hence the entire circuit behaves as resistive circuit and the entire electrical power is dissipated in the form of heat. Here the voltage and current are in same phase and the total electrical power is given by
Electrical Power = Voltage across the element* Current through the element.

Its unit is Watt = Joule/sec.
Now coming to AC circuits, here both inductor and capacitor offer certain amount of impedance given by X_L = 2*π*f*L and X_C = 1/(2*π*f*C). The inductor stores electrical energy in the form of magnetic energy and capacitor stores electrical energy in the form of electrostatic energy. Neither of them dissipates it. Further there is a phase shift of 90-°between voltage and current. Hence when we consider the entire circuit consisting of resistor, inductor and capacitor, there exists some phase difference between the source voltage and current. The cosine of this phase difference is called electrical power factor.
This factor (0 < cosφ < 1 ) represents the fraction of total power that is used to do the useful work.
The other fraction of electrical power is stored in the form of magnetic energy or electrostatic energy in inductor and capacitor respectively.
The total power in this case is

Total Electrical Power = Voltage across the element*Current through the element

This is called Apparent power and its unit is VA (Volt Amp) and denoted by ‘S’
A fraction of this total electrical power which actually does our useful work is called as active power. It is denoted as ‘P’

P = Active power = Total Electrical Power * cosφ

Its unit is watt.
The other fraction of power is called reactive power. This does no useful work, but it is required for the active work to be done. It is denoted by ‘Q’ and mathematically is given by

Q = Reactive power = Total Electrical Power*sinφ

Its unit is Var. (Volt amp reactive)

This reactive power oscillates between source and load every twice in every cycle
To help understand this better all these power are represented in the form of triangle.

Power Factor Triangle

Mathematically S² = P² + Q² and Electrical Power Factor is Active power / Apparent power.

Power Factor Improvement

The term power factor comes into picture in AC circuits only. Mathematically it is cosine of the phase difference between source voltage and current. It refers to the fraction of total power (apparent power) which is utilized to do the useful work called active power.

cosφ = Active power / Apparent power.

Need for power factor improvement

• Real power is given by P = V*I* cosφ. To transfer a given amount of power at certain voltage, the electrical current is inversely proportional to cosφ. Hence higher the pf lower will be the current flowing. A small current flow requires less cross sectional area of conductor and thus it saves conductor and money.
• From above relation we saw having poor power factor increases the current flowing in conductor and thus copper loss increases. Further large voltage drop occurs in alternator, electrical transformer and transmission & distribution lines which gives very poor voltage regulation.
• Further the KVA rating of machines is also reduced by having higher power factor as

KVA = KW / cosφ, hence the size and cost of machine also reduced.
Hence electrical power factor should be maintained close to unity.

Methods of power factor improvement

• Capacitors: Improving power factor means reducing the phase difference between voltage and current. Since majority of loads are of inductive nature, they require some amount of reactive power for them to function. This reactive power is provided by the capacitor or bank of capacitors installed parallel to the load. They act as a source of local reactive power and thus less reactive power flows through the line. Basically they reduces the phase difference between the voltage and current.
• Synchronous condenser: They are 3 phase synchronous motor with no load attached to its shaft. The synchronous motor has the characteristics of operating under any power factor leading, lagging or unity depending upon the excitation. For inductive loads, synchronous condenser is connected towards load side and is overexcited. This makes it behave like a capacitor. It draws the lagging current from the supply or supplies the reactive power.
• Phase advancer: This is an ac exciter mainly used to improve pf of induction motor. They are mounted on shaft of the motor and is connected in the rotor circuit of the motor. It improves the power factor by providing the exciting ampere turns to produce required flux at slip frequency. Further if ampere turns are increased, it can be made to operate at leading power factor.

Power factor calculation

In power factor calculation, we measure the source voltage and current drawn using a voltmeter and ammeter respectively. A wattmeter is used to get the active power.

Now we know P = V*I*cosφ watt

From this cosφ = P/ (V*I) or Watt meter reading /(voltmeter reading * ammeter reading). Hence we can get the electrical power factor.

Now we can calculate the reactive power Q= V*I* sinφ VAR

This reactive power can now be supplied from the capacitor installed in parallel with load in local.

Value of capacitor is calculated as per following formula:

Q = V² /X_c ⇒ C = Q/(2*π*f*V²) farad

Resistances in Series and Resistances in Parallel

More than one electrical resistances can be connected either in series or in parallel in addition to that, more than two resistances can also be connected in combination of series and parallel both. Here we will discuss mainly series and parallel combination

Resistances in Series

suppose you have, three resistors, R₁, R₂ and R₃ and you connect them end to end as shown in the figure below, then it would be referred as resistances in series. In case of series connection, the equivalent resistance of the combination, is sum of these three electric resistances.
That means, resistance between point A and D is the figure below, is equal to the sum of three individual resistances. The current enters in to the point A of the combination, will also leave from point D as there is no other parallel path provided in the circuit. Now say this current is I. So this current I will pass through the resistance R₁, R₂ and R₃. Applying Ohm’s law, it can be found that voltage drops across the resistances will be V₁ = IR₁, V₂ = IR₂ and V₃ = IR₃. Now, if total voltage applied across the combination of resistances in series, is V.
Then obviously V = IR₁ + IR₂ + IR₃ ………….(1)

Since, sum of voltage drops across the individual resistance is nothing but the equal to applied voltage across the combination.
Now, if we consider the total combination of resistances as a single resistor of electric resistance value R, then according to Ohm’s law,
V = IR ………….(2)

Now, comparing equation (1) and (2), we get

IR = IR₁ + IR₂ + IR₃

IR = I(R₁ + R₂ + R₃)

R = R₁ + R₂ + R₃
So the above proof, shows that equivalent resistance of a combination of resistances in series is equal to the sum of individual resistance. If there were n number of resistances instead of three resistances, the equivalent resistance will be R = R₁ + R₂ + R₃ + ………………..+R_n

Resistances in Parallel

Let’s three resistors of resistance value R₁, R₂ and R₃ are connected in such a manner, that right side terminal of each resistor are connected together as shown in the figure below, and also left side terminal of each resistor are also connected together.

This combination is called resistances in parallel. If potential difference is applied across this combination, then it will draw a current I(say).
As this current will get, three parallel paths through these three resistances, the current will be divided in to three parts. Say currents I₁, I₁ and I₁ pass through resistor R₁, R₂ and R₃ respectively.
Where total source current

I = I₁ + I₂ + I₃
Now, as from the figure it is clear that, each of the resistances in parallel, is connected across the same voltage source, the voltage drops across each resistor is same, and it is same as supply voltage V (say).
Hence, according to Ohm’s law,
V = I₁ R₁ + I₂ R₂ + I₃ R₃

Now, if we consider the equivalent resistance of the combination is R
Then, V = IR ⇒ I = V ⁄ R

Now putting the values of I, I₁, I₂ and I₃ in equation (1) we get,

The above expression represents equivalent resistance of resistances in parallel. If there were n number of resistances connected in parallel, instead of three resistances, the expression of equivalent resistance would be
(1/R₁ + 1/R₂ + 1/R₃ + ………. + 1/R_n) ^{– 1}