Circuits (MTE120) (Spring 2010)
Akrem El-ghazal
http://pami.uwaterloo.ca/~akrem/
University of Waterloo,
Electrical and Computer Engineering Dep.
Example (An extra session example)
Find all currents and voltages in the following circuit.
Solution:
The Original circuit
Step #1:
Find the total resistor seen by the voltage source and use it to find I .
To find the total resistor start at the opposite end of the circuit from the source.
The equivalent circuit #1
By combining resistors the original circuit is reduced to the following circuit
The equivalent circuit #1
Circuits (MTE120) (Spring 2010)
Akrem El-ghazal
The equivalent circuit #2
By combining resistors the above circuit (The equivalent circuit #1) is reduced to the following
circuit.
The equivalent circuit #2
Step #2
From the last equivalent circuit (The equivalent circuit #2):
I Step #3
48
12 A
22
Retrace the simplification of the circuits and apply Ohm’s law or KCL or KVL or current divider or
voltage divider rules to find all currents and voltages.
From the last equivalent circuit (equivalent circuit #2):
Since we find I , we can easily find V , using Ohm’s law:
V 2 x I 2x12 24 v
Another way to find V , is by using the voltage divider rule:
From the equivalent circuit #1:
The voltage across 4Ω is V 12 v. Therefore, we can find I by using the Ohm’s law:
V 24
I 6A
4
4
The 4Ω and 6Ω are in parallel, so they have the same voltage. Therefore, the voltage across the
6Ω is 12 v. Then, the current I can be calculated by using the Ohm’s law :
V 24
I 4A
6
6
I can be found by applying the KCL at node (A):
I I I I 0
12 6 4 I 0
I 2 A
Since we know I4 , we can easily find V by using Ohm’s law:
V 2 x I 2x2 4V
Circuits (MTE120) (Spring 2010)
Akrem El-ghazal
Another way to find V , is by using the voltage divider rule:
V V x
2
2
24x
4v
2 10
2 10
From the original circuit:
We can find I by using the Ohm’s law:
V 4
A
3
3
And, we can find I by using the KCL at node (B):
I I I 0
4
2 I 0
3
2
I A
3
Another way to find V is by using the Ohm’s law:
V
4
2
I A
24 24 3
I Now, we can easily find V by using the Ohm’s law:
V 4 x I 4x
2 8
V
3 3
Another way to find V is by using the voltage divider rule:
4
4
8
V V x
4x
V
24
24 3
Example (An extra session example)
Find the value of the voltage source V0 , if I4=0.5 mA.
Circuits (MTE120) (Spring 2010)
Solution
Since I is given, we can calculate the current I , using the current divider rule:
3
I I x
63
3
0.5 mA I x
63
I 1.5 mA
Now, we can easily calculate the voltages V and V using Ohm’s law:
V 2x10 xI 2x10 ! 1.5x10" 3 v
V 6x10 xI 6x10 ! 0.25x10" 3 v
To find I , apply the KVL around the loop shown in the figure below
V V 3K xI 1k I 0
3 3 3x10 x I 1x 10 x I 0
I 1.5 mA
To find I , apply the KCL at node a:
I I I 0
I 1.5 mA 1.5 mA 0
I 3 mA
Finally, to find V& , apply the KVL around the loop show in the figure below
Akrem El-ghazal
Circuits (MTE120) (Spring 2010)
Akrem El-ghazal
V& I x6k V V I x4k 0
V& 3x10 x6x10 3 3 3x10 x 6x10 0
V& 36 v
Example (An extra session example)
For the following circuit, determine the value of R , given that Req =9 Ω.
4Ω
A
R
24 Ω
12 Ω
B
5Ω
8Ω
30 Ω
Req=9 Ω
Solution
The first step is to redraw the circuit so that you can easily see the parallel and the serial connections.
Circuits (MTE120) (Spring 2010)
Akrem El-ghazal
4Ω
A
R
24 Ω
12 Ω
c
B
8Ω
d
5Ω
30 Ω
Req=9 Ω
Now, it is clear that:
(1) The 24 Ω and 8 Ω resistors are connected in parallel.
(2) The short circuit (Zero Ω) between c and d is connected in parallel with the 30 Ω.
Remark:
The equivalent of a short circuit connected in parallel with any resistor is a short circuit.
Short
circuit
Zero Ω
30 Ω
Rt
Rt
R( Now, we can redraw the circuit:
0 x 30
0Ω
0 30
Circuits (MTE120) (Spring 2010)
Akrem El-ghazal
4Ω
A
R
24//8= 6 Ω
12 Ω
c
B
d
5Ω
Req=9 Ω
The 4 Ω and 6 Ω are connected in series, so the equivalent is 10 Ω
Redraw the circuit:
Redraw the circuit:
Circuits (MTE120) (Spring 2010)
Akrem El-ghazal
Now, it is clear that 12 Ω, R, and 10 Ω are connected in parallel and the equivalent resistance
Req between A and B is:
R *+ 12 //R//10 5
1 1
1 "
R *+ - . 5
12 R 10
1 1
1 "
9- . 5
12 R 10
1 1
1 "
4- .
12 R 10
1 1
1
1
- .
12 R 10
4
R 15 Ω