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Kirchhoff’s Voltage Law

Kirchhoff’s Voltage Law

Kirchhoff’s Voltage Law (KVL) is Kirchhoff’s second law that deals with the conservation of energy around a closed circuit path.

Gustav Kirchhoff’s Voltage Law is the second of his fundamental laws we can use for circuit analysis. His voltage law states that for a closed loop series path the algebraic sum of all the voltages around any closed loop in a circuit is equal to zero. This is because a circuit loop is a closed conducting path so no energy is lost.

In other words the algebraic sum of ALL the potential differences around the loop must be equal to zero as: ΣV = 0. Note here that the term “algebraic sum” means to take into account the polarities and signs of the sources and voltage drops around the loop.

This idea by Kirchhoff is commonly known as the Conservation of Energy, as moving around a closed loop, or circuit, you will end up back to where you started in the circuit and therefore back to the same initial potential with no loss of voltage around the loop. Hence any voltage drops around the loop must be equal to any voltage sources met along the way.

So when applying Kirchhoff’s voltage law to a specific circuit element, it is important that we pay special attention to the algebraic signs, (+ and ) of the voltage drops across elements and the emf’s of sources otherwise our calculations may be wrong.

But before we look more closely at Kirchhoff’s voltage law (KVL) lets first understand the voltage drop across a single element such as a resistor.

A Single Circuit Element

a single circuit element

For this simple example we will assume that the current, I is in the same direction as the flow of positive charge, that is conventional current flow.

Here the flow of current through the resistor is from point A to point B, that is from positive terminal to a negative terminal. Thus as we are travelling in the same direction as current flow, there will be a fall in potential across the resistive element giving rise to a -IR voltage drop across it.

If the flow of current was in the opposite direction from point B to point A, then there would be a rise in potential across the resistive element as we are moving from a potential to a + potential giving us a +I*R voltage drop.

Thus to apply Kirchhoff’s voltage law correctly to a circuit, we must first understand the direction of the polarity and as we can see, the sign of the voltage drop across the resistive element will depend on the direction of the current flowing through it. As a general rule, you will loose potential in the same direction of current across an element and gain potential as you move in the direction of an emf source.

The direction of current flow around a closed circuit can be assumed to be either clockwise or anticlockwise and either one can be chosen. If the direction chosen is different from the actual direction of current flow, the result will still be correct and valid but will result in the algebraic answer having a minus sign.

To understand this idea a little more, lets look at a single circuit loop to see if Kirchhoff’s Voltage Law holds true.

A Single Circuit Loop

kirchhoffs voltage law in a single loop

 

Kirchhoff’s voltage law states that the algebraic sum of the potential differences in any loop must be equal to zero as: ΣV = 0. Since the two resistors, R1 and R2 are wired together in a series connection, they are both part of the same loop so the same current must flow through each resistor.

Thus the voltage drop across resistor, R1 = I*R1 and the voltage drop across resistor, R2 = I*R2 giving by KVL:

kirchhoffs voltage law, kvl

 

We can see that applying Kirchhoff’s Voltage Law to this single closed loop produces the formula for the equivalent or total resistance in the series circuit and we can expand on this to find the values of the voltage drops around the loop.

kirchhoffs voltage drops

Kirchhoff’s Voltage Law Example No1

Three resistor of values: 10 ohms, 20 ohms and 30 ohms, respectively are connected in series across an ideal 12 volt DC battery supply. Calculate: a) the total resistance, b) the circuit current, c) the current through each resistor, d) the voltage drop across each resistor, e) verify that Kirchhoff’s voltage law, KVL holds true.

a) Total Resistance (RT)

RT = R1 + R2 + R3  =  10Ω + 20Ω + 30Ω = 60Ω

Then the total circuit resistance RT is equal to 60Ω

b) Circuit Current (I)

kirchhoffs voltage law circuit current

Thus the total circuit current I is equal to 0.2 amperes or 200mA

c) Current Through Each Resistor

The resistors are wired together in series, they are all part of the same loop and therefore each experience the same amount of current. Thus:

IR1 = IR2 = IR3 = ISERIES  =  0.2 amperes

d) Voltage Drop Across Each Resistor

VR1 = I x R1 = 0.2 x 10  =  2 volts

VR2 = I x R2 = 0.2 x 20  =  4 volts

VR3 = I x R3 = 0.2 x 30  =  6 volts

e) Verify Kirchhoff’s Voltage Law

kirchhoffs voltage law

Thus Kirchhoff’s voltage law holds true as the individual voltage drops around the closed loop add up to the total.

Kirchhoff’s Circuit Loop

kirchhoff's voltage law circuit loop

 

We have seen here that Kirchhoff’s voltage law, KVL is Kirchhoff’s second law and states that the algebraic sum of all the voltage drops, as you go around a closed circuit from some fixed point and return back to the same point, and taking polarity into account, is always zero. That is ΣV = 0

The theory behind Kirchhoff’s second law is also known as the law of conservation of voltage, and this is particularly useful for us when dealing with series circuits, as series circuits also act as voltage dividers and the voltage divider circuit is an important application of many series circuits.

298 Comments

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  • Sayed Omran

    You great

  • John Thompson

    In your circuit diagrams it is a mistake to fail to show batteries having an internal resistance. This gives a false view of the true situation. Your basic work on currents, potential differences and emf details is sound and good to follow. Thanks.

    • Wayne Storr

      No, it is not. Batteries are taken as being ideal DC voltage sources, unless otherwise stated, for circuit analysis with zero internal resistance for ease of calculation and undestanding, (the same as for ideal zero-tolerance resistances). This assumption also implies an infinite current amount to and from any connected load. Typically, real or non-ideal batteries have negligibly small values of internal cell resistance (less than 0.5Ω) which would be added in series with the external load.

      Then for Example No1 to which you refer. Assuming an internal battery resistance of about 0.3Ω’s. The total series circuit resistance would therefore be 60.3Ω, giving a closed-loop circuit current of 199mA, less than 1mA difference, and an internal voltage drop in the battery of less than 60mV. Clearly, the higher the load resistance value, the less effect a batteries internal resistance will have on any voltage or current calculations.

      You can learn and understand more about the effects of internal resistance by reading our tutorial about Connecting Batteries Together.

  • CJK

    I had both theories, electron flow in the NAVY (EM Nuclear) and positive hole flow as an Electrical Engineer. It took awhile to get use to positive flow after the NAVY. It is the standard in civilian engineering as started by Benjamin Franklin. By the time it was figured out that current is due to electron flow it was to late. So hard to change ie going metric still using lb, oz, mile, feet, inches….. go Newton.

  • Peter Mek

    I want more notes

  • PissedoffVet

    Are you aware that you show current flowing from positive to negative in the circuits and that is exactly WRONG!
    Current in the battery flows from positive to negative but in the circuit it flows negative to positive.

    • Wayne Storr

      No, electron flow is from a negative electrode to a positive electrode as a result of a potential difference between them. It is standard practice to say that current (measured in Amperes) flows from positive to negative. Its basic electrical theory.

  • Chikumbutso

    I understand it well

  • Mokati Mookholi

    I want to be a part of this study

  • IhsanMalik

    It’s so well explain.
    Thank you.

  • Tarkeshwar yadav

    Electrica basic MCQ question

  • Kenneth munene

    Cool

  • Simphiwe

    its helpfull i now understand more it enyoyable. i love to be part of this platform

  • Lense

    How we can determine the voltage by two loops

  • Nadia Candelaria

    Where can I find the information to use this as a source for one of my reports?

  • Lamin Kargbo

    I love ❤ to be part of this educational forum..

  • Aakash

    Impossible

  • biranchihembram

    Kvl

  • Touch Music Empire

    nice presentation, it was helpful

  • Kuppili Harish

    I. Am harish

  • Kuppili Harish

    Hii