kirchhoff's current law (kcl) - ee.sc.eduee.sc.edu/personal/faculty/simin/elct102/09 kcl &...
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![Page 1: Kirchhoff's Current Law (KCL) - ee.sc.eduee.sc.edu/personal/faculty/simin/ELCT102/09 KCL & Parallel circuits... · Kirchhoff's Current Law (KCL) "The algebraic sum of all currents](https://reader033.vdocuments.net/reader033/viewer/2022052313/5b7753c67f8b9ade6f8cbf74/html5/thumbnails/1.jpg)
Kirchhoff's Current Law (KCL)
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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
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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)
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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)
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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
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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.
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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
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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
+
-
+
-
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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!)
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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!)
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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
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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
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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
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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
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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
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The defining characteristic of a parallel circuit is that all components are
connected between the same two wires (ideal conductors).
Parallel Circuits
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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.
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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 =
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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
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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
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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
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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:
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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
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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
+=
+=+=
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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=