Kirchhoff's Current Law (KCL)

I. Charge (current flow) conservation law

(the Kirchhoff’s Current law)

Pipe 1

Pipe 2

Pip e 3

Total volume of water per second flowing through pipe 1 = total volume of water per second flowing through pipe 2 + total volume of water per second flowing through pipe 3

Total current (charge per second) entering the node through the wire 1 =

total current leaving the node through the wire 2

+

I

I

I

I. Charge (current flow) conservation law

(the Kirchhoff’s Current law)

Kirchhoff's Current Law (KCL)

"The algebraic sum

of all currents entering and leaving a node

must equal zero"

Established in 1847 by Gustav R. Kirchhoff

Σ (

Σ (

Σ (

KCL Example 1

R

R

The rest of the circuit

V

KCL Example 2

• KCL can be applied to any single node of the network.

Problem 1

180

180

I_C1 = 4 mA; I_C2 = 3 mA; I_C3 = 2 mA

Find the current I

in mA

Circuits with multiple sources

V

V

-V

V

-In circuits with more than one source, the current directions are not obvious up front.

V

V

+

-+

-The actual current directions depend on the potential profile in the circuit.

ϕ₁ = 8 V; ϕ₂ = 4.5 V;

12V _6V

Suppose the potentials are known. Then the current directions are as shown. (Of course, knowing the potentials requires solving the circuit!)

For different potential distribution, the current directions could be different:

ϕ₁ = 7 V; ϕ₂ = 9 V;

6V _12V

Suppose the potentials are known. Then the current directions are as shown. (Of course, knowing the potentials requires solving the circuit!)

R = 1 k

V₁₂ = ϕ₁ – ϕ₂

ϕ₁ = 7 V

If ϕ₁ > ϕ₂, the current 5 mA flows from the node #1 to the node #2

I 12 1 2 12

V

I

R

R

ϕ

−

ϕ

=

=

The actual current direction

R = 1 k

V₂₁ = ϕ₂ – ϕ₁

ϕ₁ = 7 V

If ϕ₁ < ϕ₂, the actual current 5 mA flows from node #2 to node #1

+5 mA 21 2 1 21

12

7

5

1

V

V

V

I

mA

R

R

k

ϕ

−

ϕ

−

=

=

=

=

We can also say that, the current defined as flowing from node#1 to node# 2 is negative in this case.

V₁₂ = ϕ₁ – ϕ₂ 12 1 2 12

7

12

5

0

1

V

V

V

I

mA

R

R

k

ϕ

−

ϕ

−

=

=

=

= −

<

The actual current direction

Σ (

Σ (

Σ (

Σ (Entering) = Σ (

Σ (

Σ (

Σ (Leaving)

Σ (

Σ (

Σ (

Σ (Entering) - Σ (

Σ (

Σ (

Σ (Leaving) =0

General form of KCL

Assigning positive

signs to the currents entering the node and

negative

signs to the currents leaving the node, the KCL can be

re-formulated as:

Σ (

Σ (

Σ (

Problem 2

120

120

Find the current I

in A

I 1 I 2 I 3 I 4 I 1 = 1 A I 2 = 3 A I 3 = 0.5 A Timed response

Problem 2

120

120

Find the current I

in A

I 1 I 2 I 3 I 4 I 1 = 4 A I 2 = 3 A I 3 = 0.5 A Timed response

The defining characteristic of a parallel circuit is that all components are connected between the same two wires (ideal conductors).

In a parallel circuit, the voltages across all the components are the same, no matter how many components are connected.

There could be many paths for currents to flow.

Simple parallel circuits

The voltage drops are equal across all the components in the circuit. Why?

V₁₂ = V₂₃ = V₃₄ =0 (voltage drops across the wires = 0)

φ

φ

φ

φ

=

φ

φ

φ

φ

=

φ

φ

φ

φ

=

φ

φ

φ

φ

= E;

φ

φ

φ

φ

=

φ

φ

φ

φ

=

φ

φ

φ

φ

=

φ

φ

φ

φ

= 0 ;

From these: V

= V

= V

= E;

Currents in the parallel circuits

E =

Using the Ohm’s law:

I

= V

/R

= E/R

I

= V

/R

= E/R

I

= V

/R

= E/R

What is the

total current

Now apply the KCL,

SUM (Currents) = 0

I

– I

– I

– I

= 0;

I

= I

+ I

+ I

= E/R

+ E/R

+ E/R

= E×(1/R

+ 1/R

+ 1/R

)

E =

I_T I₁ I₂ I₃

Currents in the parallel circuits

I

= V

/R

= E/R

= 9V/10kΩ = 0.9 mA

I

= V

/R

= E/R

= 9V/2kΩ = 4.5 mA

I

= V

/R

= E/R

= 9V/1kΩ = 9 mA

I

= 0.9 + 4.5+ 9 = 14.4 mA

Equivalent resistance for parallel circuits

I

= I

+ I

+ I

;

I

= E×(1/R

+ 1/R

+ 1/R

)

E =

I_T I₁ I₂ I₃

Let us replace the part of network containing R

, R

and R

with a

single resistor R

. Then I

= E/R

(the Ohm’s law)

1/R

= 1/R

+ 1/R

+1/R

R

If some resistors in the network or a part of it, are

Equivalent resistance for parallel circuits

Another formulation of the parallel connection rule:

the equivalent conductance = sum (all the parallel conductances)

E =

I_T I₁ I₂ I₃

1/R

= 1/R

+ 1/R

+1/R

Note: G = 1 / R;

G

= G

+ G

+ G

When the circuit contains only two parallel resistors:

The equivalent resistance

1/R

= 1/R

+ 1/R

1

1

1

R

R

R

R

R

R

R

R

R

R

R

R

+

=

+

=

+

=

Current division in a parallel circuit

R

E

I

=

R

E

I

=

R

R

I

I

=

G

G

I

I

=