The Kirchhoff’s Laws are very useful in solving electrical networks which may not be easily solved by Ohm’s Law. In this article, I will describe these laws and will show some of **Kirchhoff’s voltage law examples** to make these laws easily understandable. Kirchhoff’s Laws, two in number, are as follows:

- Kirchhoff’s Current Law
- Kirchhoff’s Voltage Law or Kirchhoff’s Loop Law

**Kirchhoff’s Current Law**: This law states that “**The algebraic sum of all the **currents** meeting at a point or a junction in an electric circuit is zero**”.

- Consider five wires carrying current I
_{1}, I_{2}, I_{3}, I_{4}, I_{5}meeting at a point O. - To take the algebraic sum, the sigh of the current is to be considered.
**If we take the flow of current towards point O as positive, then the flow of current away from point O will be negative.**

Applying Kirchhoff’s Current Law,

The algebraic sum of current at point O = zero

i.e. I_{1} + I_{2} + (-I_{3}) + (-I_{4}) + (-I_{5}) = 0

or I_{1} + I_{2} = I_{3} + I_{4} + I_{5}

i.e. Incoming currents = Outgoing currents

Hence, Kirchhoff’s Current Law can also be stated as, “**The sum of incoming currents is equal to the sum of outgoing currents at a point or junction in an electric circuit**”.

# Kirchhoff’s Voltage Law

This law relates to voltages and applied to a closed circuit or mesh, therefore, it is also known as *Kirchhoff’s Loop Law*.

This law states that “**In any closed circuit or mesh, the algebraic sum of all the EMF’s plus the algebraic sum of voltage drops is zero**”.

## Sign Conventions

Whenever Kirchhoff’s Voltage Law is to be applied to any closed circuit or a mesh algebraic sum of EMF’s and voltage drops are to be considered. Therefore, EMF’s and voltage drops must be given a proper sign.

**A rise in potential should be considered positive while fall in potential should be considered negative.**

## Signs to be given to EMF’s

While tracing any circuit, **if we go from negative terminal of a battery or source of EMF to the positive terminal, there is a rise in potential and it should be given a +ve sign.**

Whereas, **if we go from positive terminal to negative terminal, there is a fall in potential and it should be given –ve sign. **

It is worthwhile to note that **sign of EMF’s is independent of the direction of flow of current**.

## Signs to be given to Voltage Drops

When current flows through a resistance, there is a voltage drop (IR).

**If we go in the direction of flow of current, a negative sign should be given to the voltage drop** (fall in potential) as current flows from higher potential to lower potential.

However, **if we go in the opposite direction to the flow of current a positive sign should be given to the voltage drop** (rise in potential).

It is worthwhile to note that the **sign of voltage drop depends upon the direction of flow of current** through the resistance.

## Steps to solve circuits by Kirchhoff’s Voltage Law

- Mark the direction of flow of current in various branches of the circuit according to Kirchhoff’s Current Law.
- Choose as many numbers of closed circuits as the number of unknown quantities.
- Find the algebraic sum of voltage drops and EMF’s in that circuit and put their sum equal to zero.
- After solving the problem, if the calculated value of the current has a +ve sign, it indicates that the direction assumed is correct. If it has –ve sign, it means that the actual direction of flow of current is opposite to that of assumed direction.

## Kirchhoff’s Voltage Law Examples

**Example**: Apply *Kirchhoff’s voltage law* to the adjoining circuit and calculate the current in each branch of the circuit.

**Solution: **

- To solve the above circuit, mark points A, B, C, D, E, F.
- Apply Kirchhoff’s Current Law at the junction B, mark the assumed direction of flow of current in various branches i.e. Incoming currents = Outgoing currents.
- In this problem, there are two unknown quantities (I
_{1}and I_{2}). Therefore, we have to choose**two closed circuits to solve the problem**.

The current is starting from the positive plate of the 3 V cell is I_{1}. This current after flowing in 10 Ω resistance, is divided into two parts at point B.

One part I_{2} goes through 20 Ω resistance and 50 Ω resistances and the remaining part I_{1 }– I_{2} goes through 15 V Ω resistance.

Both the currents meet at point E and again current I_{1} reaches the negative plate of the cell.

**Consider the closed path ABEFA.**

If we trace the closed path in the direction ABEFA, various voltage drops and EMF’s will have the following signs:

- Voltage drop in 10 Ω = 10 I
_{1}negative (fall in potential). - Voltage drop in 15 Ω = 15(I
_{1}– I_{2}) negative (fall in potential). - EMF = 3 positive (rise in potential).

Applying Kirchhoff’s Voltage Law to mesh ABEFA, we get,

– 10 I_{1} – 15(I_{1} – I_{2}) + 3 = 0

or – 10 I_{1} – 15 I_{1} + 15 I_{2} +3 = 0

or – 25 I_{1} + 15 I_{2} + 3 = 0

or 25 I_{1} – 15 I_{2} – 3 = 0 ………. (i)

Now to get the second equation, **consider the closed path BCDEB**.

If we trace the closed path in the direction BCDEB, various voltage drops and EMF’s will have the following signs:

- Voltage drop in 20 Ω = 20 I
_{2}negative (fall in potential). - Voltage drop in 50 Ω = 50 I
_{2}negative (fall in potential). - Voltage drop in 15 Ω = 15(I
_{1}– I_{2}) positive (rise in potential). - EMF = 0 (no EMF source in this mesh).

Applying Kirchhoff’s Voltage Law to mesh BCDEB, we get,

– 20 I_{2} – 50 I_{2} + 15(I_{1} – I_{2}) + 0 = 0

or – 20 I_{2} – 50 I_{2} + 15 I_{1} – 15 I_{2}) = 0

or – 85 I_{2} + 15 I_{1} = 0

or – 17 I_{2} + 3 I_{1} = 0

or 3 I_{1} – 17 I_{2} = 0 ………..( ii)

By solving equations (i) and (ii), we get,

I_{1} = 0.134 A and

I_{2} = 0.024 A

**Example**: A circuit is shown in Figure. Calculate the current in each branch of the circuit.**Solution**:

- To solve the above circuit, mark points A, B, C, D, E, F.
- Apply Kirchhoff’s Current Law at the junction B, mark the assumed direction of flow of current in various branches i.e. Incoming currents = Outgoing currents.
- In this problem, there are two unknown quantities (I
_{1}and I_{2}). Therefore,**we have to choose two closed circuits**to solve the problem.

**Consider the closed path ABEFA**.

If we trace the closed path in the direction ABEFA, various voltage drops and EMF’s will have the following signs:

- Voltage drop in 20 Ω = 20(I
_{1}+ I_{2}) negative (fall in potential). - Voltage drop in 6 Ω = 6 I
_{2}negative (fall in potential). - EMF = 140 positive (rise in potential).

Applying Kirchhoff’s Voltage Law to mesh ABEFA, we get,

– 20(I_{1} + I_{2}) – 6 I_{2} + 140 = 0

or 20 I_{1} + 26 I_{2} = 140

or 10 I_{1} + 13 I_{2} = 70 …….. (i)

Now, to get the second equation, **consider the closed path BCDEB**.

If we trace the closed path in the direction BCDEB, various voltage drops and EMF’s will have the following signs:

- Voltage drop in 5 Ω = 5 I
_{1}negative (fall in potential). - Voltage drop in 6 Ω = 6 I
_{2}positive (rise in potential). - EMF = 90 negative (fall in potential).

Applying Kirchhoff’s Voltage Law to mesh BCDEB, we get,

– 5 I_{1} – 90 + 6 I_{2} =0

or – 5 I_{1} + 6 I_{2} = 90 ………. (ii)

By solving equations (i) and (ii), we get,

I_{2} = 10 A

I_{1} = – 6 A

Therefore, current in branch BE is 10 A, current in branch BC is 6 A (but **opposite to the assumed direction**, because I_{1} has a negative sign in the solution) and current in branch AB is 16 A.

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