Kirchhoff's Current Law (KCL) - ee.sc.eduee.sc.edu/personal/faculty/simin/ELCT102/09 KCL &...

Kirchhoff's Current Law (KCL)

I. Charge (current flow) conservation law

(the Kirchhoff’s Current law)

Pipe 1

Pipe 2

Pipe 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 +

total current leaving the node through the wire 3

I1

I2

I3

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 (Σ (Entering Currents) = Σ (Σ (Σ (Σ (Leaving Currents)

KCL Example 1

I0

=10 mA

R1

R2

I1= 4 mA

I2 =?

The rest of the

circuit

V0

Entering current: I0

Leaving currents: I1, I2

I0 = I1 + I2;

I2 = I0 – I1;

I2 =10 mA – 4 mA = 6 mA

KCL Example 2

Network fragment

I1

I2

I3

I4

I0

I1= 2 mA

I2 = 5 mA

I0 = ?

Considering node A:

I0 = I1+I2 = 7 mA

A

I3= 0.5 mA

I4 = ?

Considering node B:

I4 = I1- I3 = 2 mA – 0.5 mA

= 1.5 mA

B

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

• KCL is valid for any circuit component: diode, resistor, transistor etc.

Problem 1

0

of

40

180180

R1 R2 R3 R4

T1 T2 T3

I0IC1 IC2 IC3 I4

I0 = 20 mA

IC1 = 4 mA; IC2 = 3 mA; IC3 = 2 mA Find the current I4 in mA

Timed response

Circuits with multiple sources

VB1 VB2

+

-

+

-

VB1 VB2

+

-

+

-

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

VB1 VB2

+

-

+

-

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 1 212

VI

R R

ϕ ϕ−= =

1

ϕ2 = 2 V

2

The actual current direction

depends on the potential difference across the component

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 mA21 2 1

2112 7

51

V V VI mA

R R k

ϕ ϕ− −= = = =

1

ϕ2 = 12 V

2

We can also say that, the current defined as flowing from node#1 to node# 2

is negative in this case.

V12 = ϕ1 – ϕ212 1 2

127 12

5 01

V V VI mA

R R k

ϕ ϕ− −= = = = − <

- 5 mA

The actual current direction

depends on the potential difference across the component

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 (Σ (All currents at the node) = 0000

Problem 2

0

of

40

120120

Find the current I4 in A

I1

I2

I3

I4

I1

= 1 A

I2

= 3 A

I3

= 0.5 A

Timed response

Problem 2

0

of

40

120120

Find the current I4 in A

I1

I2

I3

I4

I1

= 4 A

I2

= 3 A

I3

= 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).

Parallel Circuits

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)

φφφφ1 = φφφφ2 = φφφφ3 = φφφφ4 = E;

Similarly,

φφφφ5555 = φφφφ6 = φφφφ7 = φφφφ8 = 0 ;

From these: V27 = V36= V45 = E;

E =

Currents in the parallel circuits

E =

Using the Ohm’s law:

I1 = V27/R1 = E/R1

I2 = V36/R2 = E/R2

I3 = V45/R3 = E/R3

What is the total current in the circuit?

Now apply the KCL, SUM (Currents) = 0

IT – I1 – I2 – I3 = 0;

IT = I1 + I2 + I3 = E/R1+ E/R2+ E/R3 = E×(1/R1+ 1/R2+ 1/R3)

E =

IT I1 I2 I3

Currents in the parallel circuits

Currents in the parallel circuits

I1 = V27/R1 = E/R1 = 9V/10kΩ = 0.9 mA

I2 = V36/R2 = E/R2 = 9V/2kΩ = 4.5 mA

I3 = V45/R3 = E/R3 = 9V/1kΩ = 9 mA

IT = 0.9 + 4.5+ 9 = 14.4 mA

E =

IT I1 I2 I3

Equivalent resistance for parallel circuits

IT = I1 + I2 + I3;

IT = E×(1/R1+ 1/R2+ 1/R3)

E =

IT I1 I2 I3

Let us replace the part of network containing R1, R2 and R3 with a

single resistor RT. Then IT = E/REQ (the Ohm’s law)

1/REQP = 1/R1 + 1/R2+1/R3

REQ

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

connected in parallel, then the equivalent resistance is:

Equivalent resistance for parallel circuits

Another formulation of the parallel connection rule:

the equivalent conductance = sum (all the parallel conductances)

E =

IT I1 I2 I3

1/REQP = 1/R1 + 1/R2+1/R3

Note: G = 1 / R;

GT = G1 + G2 + G3

When the circuit contains only two parallel resistors:

The equivalent resistance

1/REQ = 1/R1 + 1/R2

21

21

21

21

21

111

RR

RRR

RR

RR

RRR

EQ

EQ

+=

+=+=

Current division in a parallel circuit

11

R

EI =

E

22

R

EI =

1

2

2

1

R

R

I

I=

2

1

2

1

G

G

I

I=