09 KCL & Parallel Circuits09 KCL & Parallel circuits
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Transcript of 09 KCL & Parallel Circuits09 KCL & Parallel circuits
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=