Kirchhoff's Current Law (KCL)
I. Charge (current flow) conservation law
(the Kirchhoff’s Current law)
2
e
p
Pi
Pipe 1
Pip
e3
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
I. Charge (current flow) conservation law
(the Kirchhoff’s Current law)
I2
I1
I3
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
+
Kirchhoff's Current Law (KCL)
"The algebraic sum
of all currents entering and leaving a node
must equal zero"
Σ (Entering
Currents) = Σ (Leaving
Currents)
(
(
Established in 1847 by Gustav R. Kirchhoff
KCL Example 1
I0 =10 mA
I2 =?
R2
I1= 4 mA
The rest of the
circuit
V0
R1
Entering current: I0
Leaving currents: I1, I2
I0 = I1 + I2;
I2 = I0 – I1;
I2 =10 mA – 4 mA = 6 mA
KCL Example 2
A
I0
B
I1
I3
I2
I1= 2 mA
Network fragment
I4
I3= 0.5 mA
I2 = 5 mA
Considering node A:
I0 = ?
I0 = I1+I2 = 7 mA
Considering node B:
I4 = ?
I4 = I1- I3 = 2 mA – 0.5 mA
= 1.5 mA
• KCL can be applied to any single node of the network.
• KCL is valid for any circuit component: diode, resistor, transistor etc.
Problem 1
R1
R2
R3
IC1
I0
R4
IC2
T1
T2
I0 = 20 mA
IC1 = 4 mA; IC2 = 3 mA; IC3 = 2 mA
IC3
I4
T3
Find the current I4 in mA
0
of
40
180
Timed response
Circuits with multiple sources
+
+
VB1
VB2
-
-
In circuits with more than one source, the current directions are not obvious up front.
+
VB1
+
VB2
-
-
+
+
VB1
-
VB2
-
The actual current directions depend on the potential profile in the circuit.
ϕ1 = 8 V;
12V
ϕ2 = 4.5 V;
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;
6V
ϕ2 = 9 V;
12V
Suppose the potentials are known. Then the current directions are as shown.
(Of course, knowing the potentials requires solving the circuit!)
The actual current direction
depends on the potential difference across the component
ϕ1 = 7 V
1
2
R=1k
V12 = ϕ1 – ϕ2
ϕ2 = 2 V
I
V
ϕ −ϕ
I12 = 12 = 1 2
R
R
If ϕ1 > ϕ2, the current 5 mA flows from the node #1 to the node #2
The actual current direction
depends on the potential difference across the component
ϕ1 = 7 V
1
ϕ2 = 12 V
2
R=1k
V21 = ϕ2 – ϕ1
V21 ϕ2 − ϕ1 12V − 7V
I 21 =
=
=
= 5mA
R
R
1k
+5 mA
If ϕ1 < ϕ2, the actual current 5 mA flows from node #2 to node #1
We can also say that, the current defined as flowing from node#1 to node# 2
is negative in this case.
V12 = ϕ1 – ϕ2
V12 ϕ1 − ϕ2 7V −12V
I12 =
=
=
= − 5mA < 0
1k
R
R
- 5 mA
General form of KCL
Σ (Entering)
= Σ (Leaving)
(
(
Σ (Entering)
- Σ (Leaving)
=0
(
(
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 (All
currents at the node) = 0
(
Problem 2
I2
I1 = 1 A
I1
I2 = 3 A
I4
I3
I3 = 0.5 A
Find the current I4 in A
0
of
40
120
Timed response
Problem 2
I2
I1 = 4 A
I1
I2 = 3 A
I3 = 0.5 A
I4
I3
Find the current I4 in A
0
of
40
120
Timed response
Parallel Circuits
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
E=
The voltage drops are equal across all the components in the circuit.
Why?
V12 = V23 = V34 =0 (voltage drops across the wires = 0)
φ1 = φ2 = φ3 = φ4 = E;
Similarly,
φ5 = φ6 = φ7 = φ8 = 0 ;
From these: V27 = V36= V45 = 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
Currents in the parallel circuits
What is the total current in the circuit?
IT
I1
I2
I3
E=
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)
Currents in the parallel circuits
IT
I1
I2
E=
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
I3
Equivalent resistance for parallel circuits
IT
I1
I2
I3
REQ
E=
IT = I1 + I2 + I3;
IT = E×(1/R1+ 1/R2+ 1/R3)
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)
If some resistors in the network or a part of it, are
connected in parallel, then the equivalent resistance is:
1/REQP = 1/R1 + 1/R2+1/R3
Equivalent resistance for parallel circuits
IT
I1
I2
I3
E=
1/REQP = 1/R1 + 1/R2+1/R3
Note: G = 1 / R;
GT = G1 + G2 + G3
Another formulation of the parallel connection rule:
the equivalent conductance = sum (all the parallel conductances)
When the circuit contains only two parallel resistors:
The equivalent resistance
1/REQ = 1/R1 + 1/R2
1
REQ
=
REQ =
1 1 R1 + R2
+
=
R1 R2 R1 R2
R1 R2
R1 + R2
Current division in a parallel circuit
E
E
I1 =
R1
I2 =
E
R2
I1 R2
=
I 2 R1
I1 G1
=
I 2 G2