Kirchhoff's circuit laws

**Kirchhoff's laws** are two equalities that deal with the current and potential difference (commonly known as voltage) in the lumped element model of electrical circuits. Widely used in electrical engineering, they can be applied in time and frequency domains and form the basis for network analysis.

## Kirchhoff's current law (KCL)

This law is also called **Kirchhoff's first law**, **Kirchhoff's point rule**, or **Kirchhoff's junction rule** (or **nodal rule**).

This law states that, for any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node; or equivalently:

> **The algebraic sum of currents in a network of conductors meeting at a point is zero**.
>
>

Recalling that current is a signed (positive or negative) quantity reflecting direction towards or away from a node, this principle can be succinctly stated as:

∑

# k
1

n

I

k

=
0

{\displaystyle \sum\_{k=1}^n {I}\_k = 0}

where *n* is the total number of branches with currents flowing towards or away from the node.

The law is based on the conservation of charge where the charge (measured in coulombs) is the product of the current (in amperes) and the time (in seconds). If the net charge in a region is constant, the KCL will hold on the boundaries of the region. This means that KCL relies on the fact that the net charge in the wires and components is constant.

### Uses

A matrix version of Kirchhoff's current law is the basis of most circuit simulation software, such as SPICE. Kirchhoff's current law is used with Ohm's law to perform nodal analysis.

KCL is applicable to any lumped network irrespective of the nature of the network; whether unilateral or bilateral, active or passive, linear or non-linear.

## Kirchhoff's voltage law (KVL)

This law is also called **Kirchhoff's second law**, **Kirchhoff's loop** (or **mesh**) **rule**, and **Kirchhoff's second rule**.

This law states that

> **The directed sum of the potential differences (voltages) around any closed loop is zero.**
>
>

Similarly to KCL, it can be stated as:

∑

# k
1

n

V
k

=
0

{\displaystyle \sum\_{k=1}^n V\_k = 0}

Here, *n* is the total number of voltages measured.

Consider some arbitrary circuit. Approximate the circuit with lumped elements, so that (time-varying) magnetic fields are contained to each component and the field in the region exterior to the circuit is negligible. Based on this assumption, the Maxwell-Faraday equation reveals that

∇
×

E

=
−

∂

B

∂
t

=

0

{\displaystyle \nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}=\mathbf{0}}

in the exterior region. If each of the components have a finite volume, then the exterior region is simply connected, and thus the electric field is conservative in that region. Therefore, for any loop in the circuit, we find that

∑

V
i

=
−
∑

∫

P

i

E

⋅

d

l

=
∮

E

⋅

d

l

=
0

{\displaystyle \sum V\_i=-\sum \int\_{\mathcal{P}\_i}\mathbf{E}\cdot\mathrm{d}\mathbf{l}=\oint\mathbf{E}\cdot\mathrm{d}\mathbf{l}=0}

where

P

i

{\textstyle \mathcal{P}\_i}

 are paths around the *exterior* of each of the components, from one terminal to another.

### Generalization

In the low-frequency limit, the voltage drop around any loop is zero. This includes imaginary loops arranged arbitrarily in space – not limited to the loops delineated by the circuit elements and conductors. In the low-frequency limit, this is a corollary of Faraday's law of induction (which is one of Maxwell's equations).

This has practical application in situations involving "static electricity".

## Example

Assume an electric network consisting of two voltage sources and three resistors.

According to the first law we have

i
1

−

i
2

−

i
3

=
0

{\displaystyle i\_1 - i\_2 - i\_3 = 0 \, }

The second law applied to the closed circuit *s*1 gives

−

R
2

i
2

E

1

−

R
1

i
1

=
0

{\displaystyle -R\_2 i\_2 + \mathcal{E}\_1 - R\_1 i\_1 = 0}

The second law applied to the closed circuit *s*2 gives

−

R
3

i
3

−

E

2

−

E

1

R
2

i
2

=
0

{\displaystyle -R\_3 i\_3 - \mathcal{E}\_2 - \mathcal{E}\_1 + R\_2 i\_2 = 0 }

Thus we get a system of linear equations in

i
1

,

i
2

,

i
3

{\displaystyle i\_1, i\_2, i\_3}

:

{

i
1

−

i
2

−

i
3

=
0

−

R
2

i
2

E

1

−

R
1

i
1

=
0

−

R
3

i
3

−

E

2

−

E

1

R
2

i
2

=
0

{\displaystyle \begin{cases}
i\_1 - i\_2 - i\_3 & = 0 \\
-R\_2 i\_2 + \mathcal{E}\_1 - R\_1 i\_1 & = 0 \\
-R\_3 i\_3 - \mathcal{E}\_2 - \mathcal{E}\_1 + R\_2 i\_2 & = 0
\end{cases}
}

Which is equivalent to

{

i
1

(
−

i
2

)
(
−

i
3

)

=
0

R
1

i
1

R
2

i
2

0

i
3

=

E

1

0

i
1

R
2

i
2

−

R
3

i
3

=

E

1

E

2

{\displaystyle \begin{cases}
i\_1 + (- i\_2) + (- i\_3) & = 0 \\
R\_1 i\_1 + R\_2 i\_2 + 0 i\_3 & = \mathcal{E}\_1 \\
0 i\_1 + R\_2 i\_2 - R\_3 i\_3 & = \mathcal{E}\_1 + \mathcal{E}\_2
\end{cases}
}

Assuming

R
1

=
100
Ω
,

R
2

=
200
Ω
,

R
3

=
300
Ω

{\displaystyle
R\_1 = 100\Omega,\ R\_2 = 200\Omega,\ R\_3 = 300\Omega
}

E

1

=
3
V
,

E

2

=
4
V

{\displaystyle
\mathcal{E}\_1 = 3\text{V}, \mathcal{E}\_2 = 4\text{V}
}

the solution is

{

i
1

=

1
1100

A

i
2

=

4
275

A

i
3

=
−

3
220

A

{\displaystyle \begin{cases}
i\_1 = \frac{1}{1100}\text{A} \\[6pt]
i\_2 = \frac{4}{275}\text{A} \\[6pt]
i\_3 = - \frac{3}{220}\text{A}
\end{cases}
}

i
3

{\displaystyle i\_3}

 has a negative sign, which means that the direction of

i
3

{\displaystyle i\_3}

 is opposite to the assumed direction i.e.

i
3

{\displaystyle i\_3}

 is directed upwards.