Kirchhoff's Voltage Law (KVL)

Kirchhoff's voltage law

In this article, we explain Kirchhoff's voltage law (KVL).

KVL is a way of analyzing a circuit.

KVL states that sum of voltages around a loop in a circuit is equal to 0.

This is shown according to the formula, ΣV= 0.

KVL is essentially a principle of the conservation of energy.

There's a power source in a circuit and there are elements such as resistors, capacitors, inductors, and other type of loads.

What KVL really says, in simple terms, is that the sum of the power source along with the voltage drops across all the elements in the circuit is equal to 0.

In this way, the voltage of the power source and the voltage drops across each of the elements of the circuits (resistors, loads, etc) must equal 0. In other words, the voltage drops across the loads cannot be greater than or less than the voltage rise (many times the power source). The voltage drops across the loads is equal to (with opposite polarity) the voltage rise, in which case it adds to zero.

So if we have a circuit with a 5V power source and a 1KΩ resistor and 5KΩ resistor, the power source and the voltage drops across the resistor is equal to 0. The voltage drops across the resistors will be equal to the voltage rise (they will have opposite polarities), giving a sum of 0.

KVL is a form of mesh analysis. We analyze each mesh, or loop, of a circuit and we're able to calculate unknown current and voltage values. We can also check a circuit with KVL to see if the conservation of energy has been observed, in which case it shows that the circuit has sound values.

KVL Example

So, as a first example, we'll use the most basic circuit with a single loop. Later, we'll show more advanced circuits.

KVL example

So this is a basic circuit consisting of a single power source and a single resistor (a 1KΩ resistor).

KVL is a method of analyzing a circuit where you literally loop through a loop in a circuit, either clockwise or counterclockwise and do all the math based on the elements.

So, in the image above, the loop is going clockwise. So we'll analyze the circuit in a clockwise. But you can also do it counterclockwise and get the same results.

So, going clockwise, through the above loop, we get the result, -5V + (1KΩ)(I)= 0.

Because in KVL, we add up all the voltages, we have use ohm's law to find the voltage drop across the 1KΩ resistor. Since voltage= current * resistance (V=IR), the voltage across the 1KΩ resistor is equal to, V=IR=(1KΩ)(I). Now we just have to solve for I.

So we have, -5V + (1KΩ)(I)= 0.

Adding +5V to both sides gives us, (1KΩ)(I) =5V.

Dividing both sides by 1KΩ gives us, I= 5V/(1KΩ)= 5mA.

So we've solved for the current using KVL.

Using ohm's law, we know that the voltage across the resistor is 5V, since V=IR=(5mA)(1KΩ)=5V.

Just to recap, going clockwise, you can see the negative polarity is first from the power supply, which is why we get -5V. Because the positive polarity of the battery is on the left side of the 1KΩ resistor, the resistor has a polarity from + to - (going from left to right). Checking our circuit with the now known voltage value, we get -5V + 5V = 0.

Therefore, the circuit has proven to observe the conservation of energy principle and the values are sound.

Now we'll do another example where we add another resistor, still a single loop.

The principle is exactly the same.

KVL example

Again, we go through the circuit clockwise.

Because the circuit is in series, the same current flows through both resistors.

So, doing KVL, we get -5V + (1KΩ)(I) + (4KΩ)(I) = 0.

Since both the 1KΩ and 4KΩ resistors have the same variable (I), we can simply add them. This gives us now, -5V + (5KΩ)I= 0.

Adding +5V to both sides gives us, (5KΩ)I= 5V.

Dividing both sides by 5 gives us, I= 1mA.

So the circuit flowing through the circuit is 1mA.

This gives a voltage of 1V across the 1KΩ resistor and 4V across the 4KΩ resistor.

Doing KVL gives us, -5V + 1V + 4V= 0.

Therefore, the circuit observes the principle of conservation.

KVL Example with Multiple Loops

Even though KVL can be applied to any circuit, it is most often used for multi-loop circuits. The above examples could also easily be solved using ohm's law without using KVL. However, KVL is more crucial when it comes to solving multi-loop circuits.

So the circuit we analyze is shown below.

KVL multi-loop circuit example

So using KVL, we analyze each loop mesh in the circuit independent.

When you are doing multi-loop analysis, you must be consistent with the direction you use. Wtih all loops, you must analyze them in the same direction.

For this example (and throughout this whole web page), we analyze the circuits in a clockwise direction.

Using KVL, we can find the currents flowing through all parts of the circuit.

The first equation we must have is the flow of currents going through the node of the circuit. Being that the first current (I1) enters the node and the second and third currents (I2 and I3) leave the node, this gives us, -I1 + I2 + I3= 0. This is assuming that entering is negative and leaving is positive. However, you can do it the other way, but you must be consistent.

So doing KVL on the first loop gives us, -15V + (5KΩ)(I1) + (10KΩ)(I2) = 0. Changing this around gives us, (5KΩ)(I1) + (10KΩ)(I2) = 15.

Doing KVL on the second loop gives us, (10KΩ)(-I2) + (3KΩ)(I3) + (7KΩ)((I3))= 0. Reducing this equation gives us, (-10KΩ)(I2) + (10KΩ)(I3) = 0.

We now have 3 equations and can now use the linear system method of linear algebra to get the values for the currents, I1, I2, and I3.

This gives us the equations shown below.

KVL example simultaneous equations calculation

You can solve this, by hand, by rearranging equations, or you can use a systems of equations calculator. We can now plug this into a 3x3 simultaneous equation solver, which gives us the answers, I1 = 1.5mA, I2= 0.75mA, and I3= 0.75mA.

Now, you can find all the voltages across each of the resistors in the circuit using KVL>

The voltage across the 5KΩ is 7.5V, since V= (5KΩ)(1.5mA).

The voltage across the 10KΩ is 7.5V, since V= (10KΩ)(0.75mA).

The voltage across the 3KΩ resistor is 2.25V, since V=(3KΩ)(0.75mA).

The voltage across the 7KΩ resistor is 5.25V, since V=(7KΩ)(0.75mA).

We've use KVL to find all the currents and voltages throughout the circuit.

Now we can use KVL to check our answers to see if there is conservation of energy in the circuit.

Using KVL on the first loop (clockwise) gives us, -15V + 7.5V + 7.5V =0.

On the second loop (clockwise) , this gives us, -7.5V + 2.25V + 5.25V=0.

So our circuit has sound values.

Similarly, without calculating the voltages, we can use KVL with just the known currents.

So, again, we're going through the loops clockwise. This gives us -15V +(5KΩ)(1.5mA)+ (10KΩ)(0.75mA)=0.

If we go through the other loop, this gives us, -(10KΩ)(0.75mA) + (3KΩ)(0.75mA)+ (7KΩ)(0.75mA)= 0.

So KVL confirms the energy of conservation within each of the loops for this circuit.

So KVL is a great tool for finding unknown currents and a great tool for checking to see if the results are sound by seeing if there is conservation of energy in the circuit.

Related Resources

Superposition Theorem

Thevenin's Theorem

Norton's Theorem

What is Voltage?

What is an Ideal Voltage Source?

What is a Constant Voltage Source?

What is Open Circuit Voltage?

What is Bias Voltage?

What is Negative Voltage?

RMS Voltage and Current- Explained

What is Peak Voltage?

What is Peak-to-Peak Voltage (VPP)?

How to Increase Voltage in a Circuit

How to Reduce Voltage with Resistors

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