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INC 111 Basic Circuit Analysis
Week 2
Kirchoff’s Law
Kirchoff’s Current Law (KCL)
“The sum of all currents entering a point is equal zero.”
I1
I2
I3
I4
I1 - I2 - I3 + I4 = 0
-I1 + I2 + I3 - I4 = 0
or
KCL Metaphor
From the pipe that is full of water,
the amount of flow-in water must
be equal to the amount of flowout water.
This is because water cannot
disappear.
How to use KCL
Place a circle anywhere any size that you want
5Ω
3V
2Ω
1Ω
2A
Kirchoff’s Voltage Law
“The sum of all voltages in a closed loop is equal zero.”
+
V2
-
+
+
V1
V3
-
-
V1 – V2 – V3 = 0
or
-V1 + V2 + V3 = 0
KVL Metaphor
C
2m
hA B  hB C  hC  A  0
5m
B
3m
A
(3)  (2)  (5)  0
How to use KVL
Form a loop in the circuit
5Ω
3V
2Ω
1Ω
2A
Example: Voltage Divider
X mA
6K
10V
KCL: Current is equal on all
points in the circuit because there
is no separate path
KVL:
4K
 10  6000 x  4000 x  0
6000 x  4000 x  10
10
x
 1mA
6000  4000
Example: Voltage Divider
1mA
10V
10V
6K
4V
10V
4K
0V
0V
R in series Equivalent
X mA
X mA
6K
10V
10V
10K
4K
10
x
 1mA
6000  4000
10
x
 1mA
10000
Resistor Reduction
Series
(อนุกรม)
R1
Parallel
(ขนาน)
R2
R1+R2
R1
(R1*R2)/R1+R2
R2
or
1
Rtotal
1
1
 
R1 R2
Example
10V
10V
7.5mA
10V
10V
2K
4K
5mA
0V
0V
0V
2.5mA
R in parallel Equivalent
10V
2K
4K
10 10
I

2K 4K
1
1
 10  (

)
2K 4K
 7.5mA
10V
1.333K
10
1
I
 10 
Req
1.333K
 7.5mA
Voltage & Current Sources
Combination
3V
6A
2V
4V
1V
5A
4A
7A
Example: Current Divider
1.333V
1mA
1mA
1.333V
1.333V
2K
4K
0.667mA
0V
0V
0V
0.333mA
Voltage Divider
R1
+
VR1
-
R2
+
VR2
-
V
R1
VR1 
V
R1  R 2
R2
VR 2 
V
R1  R 2
Current Divider
IR1
I
R1
IR2
R2
R2
I R1 
I
R1  R 2
R1
I R2 
I
R1  R 2
Example
Find V1
10V
2K
2K
+
V1
-
4K
2K
V1 
10V  3.333V
2K  4K
Voltmeter and Ammeter
Voltmeter measures the difference of
voltage between 2 points.
V
A
Put 2 connectors to 2 points that we
want to know the voltage difference.
Ammeter measures the current pass
through wire.
Cut the wire, and put the two openends to the meter.
Voltmeter
6K
10V
4K
V
Ammeter
A
6K
10V
4K
Voltmeter and Ammeter
Resistance Characteristic
V
Same as open circuit ( not connected)
A
Same as conductor (wire)
Dependent Source
The amount of voltage (current) supplied depends on
other voltage (current).
• Dependent Voltage Source
• Dependent Current Source
+
-
Example
Find I
+
I
5V
Vx
4Ω
1V
2Ω
+
-
3Vx
 5  4 I  1  2 I  3Vx  0, (Vx  4 I )
 5  4 I  1  2 I  12 I  0
18I  4
I  0.222 A
Example
Find I1
4Ω
A
2Ω
B
I1
5V
2V
1A
1Ω
C
4Ω
A
I1
5V
2Ω
B
I1+1
2V
1A
1Ω
C
KVL on the outer loop
 5  4 I1  2( I1  1)  2  1( I1  1)  0
7 I1  4
I1 
4
A
7