## Krichhoff’s Voltage Law (KVL)

Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL) are very fundamental laws in the electrical circuit. Using these laws, we can find the voltage and current in the electrical circuit.

**Statement:** The algebraic sum of all the branch voltages around any closed loop in the network or circuit is zero at all instant of time.

Let’s understand the statement through one example.

**Fig. 1 Kirchhoff’s Voltage Law in the Electrical Circuit**

Fig.1 shows the electrical circuit which consists of one voltage source and three passive circuit elements. The passive circuit elements could be a resistor, capacitor, or inductor. But in general form, here they are represented as Z_{1}, Z_{2, }and Z_{3}. Let’s say the voltages across these elements are V_{1}, V_{2,} and V_{3} respectively.

According to the Kirchhoff’s Voltage Law, if we mover around any electrical circuit (either clockwise or anti-clockwise) and add the voltages drop across each element then the algebraic sum of all the voltages will be zero.

Let’s say as shown in Fig.2, we are moving in the clockwise direction, starting from point A.

**Fig.2 Sign convention for Kirchhoff’s Voltage Law (KVL)**

While moving in the clockwise direction, the first element that we will come across is the voltage source Vs. And while we are moving, we are moving from the negative to the positive potential. So, we will use the following sign convention for the KVL.

**Sign Convention for Kirchhoff’s Voltage Law (KVL)**

Whenever we are moving from the positive (+) to the negative (-) terminal across any element or in other words, if there is a drop in the potential across the element then we can consider that voltage as the negative voltage.

Similarly, whenever we are moving from the negative (-) terminal to the positive (+) terminal across any element or in other words, if there is a rise in the potential across the element then we can consider that voltage as the positive voltage. The same is shown in Fig.3.

**Fig. 3 Sign Conventions for the Kirchhoff’s Voltage Law **

As shown in Fig.2, starting from point A, when we move in the clockwise direction then the first element is the voltage source Vs. And while moving, since there is a rise in the potential, we can consider that voltage as a positive voltage. Similarly, when we are moving from point B to point C, there is a voltage drop across the element Z_{1}. That means the voltage V_{1} can be considered as the negative voltage. Similarly, while moving across the element Z_{2} and Z_{3}, there is a drop across each element. That means the voltage across element Z_{2} and Z_{3} (V_{2} and V_{3}) will be negative.

According to the Kirchhoff’s Voltage Law, the algebraic sum of all these voltage is zero.

That means **Vs + ( – V1) + (- V _{2} ) + ( – V_{3} ) = 0**

**=> Vs – V _{1} – V_{2} – V_{3} = 0**

**=> Vs = V _{1} + V_{2} + V_{3}**

**KVL is the law of Conservation of Energy **

Kirchhoff’s Voltage Law is the low of conservation of energy. Let’s prove it. The voltage V can also be written as

In the above case, V_{S} = V_{1} + V_{2 }+ V_{3} can be written as

That means** E _{S}= E_{1 }+ E_{2} + E_{3}**

Or it can be said that the energy supplied by the voltage source is equal to the energy dissipated across three elements. That means **KVL is the law of conservation of energy**.

**Example**

**Find the current I and the voltage across 15 Ω resistor.**

**Solution: **

First, let’s denote the voltages across each element. Let’s say the voltage drop across 5 Ω, 10 Ω, and 15 **Ω** resistor are V1, V2, and V3 respectively.

Applying Kirchhoff’s voltage law,

5V – V_{1} – V_{2} – 2V – V_{3} = 0

Since, moving in the clockwise direction, there is a rise in potential only across 5V voltage source and there is a drop in the potential across remaining elements.

That means, 3V = V_{1} + V_{2} + V_{3} ——- (1)

Since all the elements are connected in series, the current flowing through each element is the same. Let’s say the current in the circuit is I.

Therefore, using Ohm’s law (V= I x R), V_{1} = 5 x I, V_{2} = 10 x I and V_{3} = 15 x I

From the above equation 1,

3V = (5 x I ) + ( 10 x I ) + (15 x I)

=> 3V = 30 x I

=>** I = 0.1 A**

And the voltage across 15 **Ω** resistor (V_{3}) = **15 x I = 15 V x 0.1A = 1.5 V**

From the above example, we can see that, using the Kirchhoff’s voltage law, it is possible to find the current and the voltage across any element in the electrical circuit.

For more information, you can refer this video tutorial on KVL,

Nice

best refresher deffinition ive seen

How can it be,

From the above equation 1,

3V = (5 x I ) + ( 10 x I ) + (15 x I)

=> 3V = 30 x I

=> I = 0.1 A