solution to a complex circuit
TRANSCRIPT
-
8/13/2019 Solution to a Complex Circuit
1/19
Solution to
Review Series (3/3)
REDG
If there are errors, I would be glad if you would let me know.
-
8/13/2019 Solution to a Complex Circuit
2/19
THE PROBLEM
Determine the current and voltage across each resistor in the ci
Each resistor has a resistance of 20 ohms. The circuit is connecte
volt source.
-
8/13/2019 Solution to a Complex Circuit
3/19
First thing--- redraw the circuit so that it appears less complex.
Notice that R3, R4 and R5 are connected in series but just bent mi
between each resistor. That series connection makes parallel wit
equivalent circuit of that section is reflected on the right.
R3 R4
R9
SIMPLIFYING THE CIRCUIT
-
8/13/2019 Solution to a Complex Circuit
4/19
Consider R2 and R6, they can be straighten and align with R9 , the
circuit reveals that they are in series with that circuit on the red d
The equivalent circuit for that six resistors is the one on the right
confusing if we use the first illustration.
R3 R4
R9
R2
R3 R4
R9R2
or
SIMPLIFYING THE CIRCUIT
-
8/13/2019 Solution to a Complex Circuit
5/19
This time, look at R8, it is clear that it is connected in parallel with
previous circuit. Therefore, redrawing the whole things gives you
equivalent circuit on the right?
R3 R4
R9
R2
R8
SIMPLIFYING THE CIRCUIT
-
8/13/2019 Solution to a Complex Circuit
6/19
What about R1 and R7? Isnt it that they are connected in series w
preceding circuit? (Just like how R2 and R6 are connected with their prece
The one on the right is the equivalent circuit of the whole circuit.
R3 R4
R9
R2
R8
R1
SIMPLIFYING THE CIRCUIT
-
8/13/2019 Solution to a Complex Circuit
7/19
When the battery is attached, we get the equivalent circuit.
R3 R4
R9
R2
R8
R1
SIMPLIFYING THE CIRCUIT
-
8/13/2019 Solution to a Complex Circuit
8/19
Information that can be derived from the circuit:
R3 R4 R5
R9
R2 R6
R8
R1 R7
Fac
In s
on
In p
acrequ
Information Reason/explanation
I3=I4=I5 They are arranged in series
I2=I6They are arranged in series
I1=I7=ITThey are arranged in series
Information Reas
I1=I8+ any of I2 or I6 The cis thebran
I2=I9+ any of I3 or I4or I5
The cis the
bran
EXTRACT INFORMATION
-
8/13/2019 Solution to a Complex Circuit
9/19
The equivalent resistance of the
section that is in red.
3,4,5R = 20 + 20 + 20
= 60
EQUIVALENT RESISTANCE R3 R4 R
R9
R2
R8
R1
Now we can simplify the
illustration like the one on the side
60
R9
R2
R8
R1
-
8/13/2019 Solution to a Complex Circuit
10/19
60
R9
R2
R8
R1
60 9
60 9
4 1
60 15
, 9
,
1 1 1= +R 60 R
1 1= +
60 20
R =15
EQUIVALENT RESISTANCE
The 60- equivalent resistance is
in parallel with R9, hence
Now we can simplify the
illustration like the one on the side
15 R2
R8
R1
-
8/13/2019 Solution to a Complex Circuit
11/19
15 R2
R8
R1
EQUIVALENT RESISTANCE
The 15- equivalent resistance is
in series with R2 and R6, hence
Now we can simplify the
illustration like the one on the side
2 6
20 20
2,..6R = R + 15 + R
= + 15 +
= 55
55
R8
R1
-
8/13/2019 Solution to a Complex Circuit
12/19
55
R8
R1
14.67
R1
EQUIVALENT RESISTANCE
The 55- equivalent resistance is
in parallel with R8 ,hence
Now we can simplify the
illustration like the one on the side
6
1 1 1
55
1 1
55 20
4 11
220 220
22014 67
15
55, 8
55, 8
= +R R
= +
+
R .
-
8/13/2019 Solution to a Complex Circuit
13/19
14.67
R1
EQUIVALENT RESISTANCE
The 14.67- equivalent resistance
is in series with R1 and R7 ,hence
Finally, we can represent the
previously complex circuit to its
simplest form
1 7
20 20
2,..6R = R + 14.67 + R
= + 14.67 +
= 54.67
54.67
-
8/13/2019 Solution to a Complex Circuit
14/19
14.67
R1
CURRENT AND VOLTAGE
Consider a 54.67- resistor
impressed with a 90-V source. The
current can be so easily calculated.
Suppose the resistance is split into 3 notnecessarily equal portions like this. Take
note, because they are in series, they have
equal current but not voltage.
1 65
VI =
R
90=
54.67
. A
54.67
I= 1.65AV=90V
I= 1.65A
V=24V
I= 1.65A
V=33V
V=IR
-
8/13/2019 Solution to a Complex Circuit
15/19
14.67
R1
CURRENT AND VOLTAGE
If the 14.67- resistor breaks into
two branches, then the voltage
across each branch is 24V. Current
can easily be computed.
VI =
R
I= 1.65AV=24V
I= 1.65AV=33V
55
R8
R1I= 1V=
I= V=
I=
-
8/13/2019 Solution to a Complex Circuit
16/19
CURRENT AND VOLTAGE
If the 55- resistor is split into 3
not necessarily equal resistors,
then the current of each resistor is
equal to the current when they are
combined. But their individual
voltage can be easily calculated.
55
R8
R1I= 1V=
I= V=
15 R2
R8
R1
I= 0.44A
V=6.6V
I= 0.44AV=8.8V
I 0 44A
-
8/13/2019 Solution to a Complex Circuit
17/19
CURRENT AND VOLTAGE
If the 15- resistor is split into 2
branches, then the voltage of each
branch must be 6.6V and the
current of each can easily be
derived.
15 R2
R8
R1
I= 0.44AV=6.6V
I= 0.44AV=8.8V
60
R9
R2
R8
R1 I= 0V=6
I= 0
V=6
I 0
-
8/13/2019 Solution to a Complex Circuit
18/19
CURRENT AND VOLTAGE
If the 60- resistor splits into 3
equal resistors, then each resistor
will have the same current as
when they are not split (0.11A)
and divide the voltage equally
(2.2V).
60
R9
R2
R8
R1 I= 0V=6
I= 0V=6
R3 R4 R
R9
R2
R8
R1
I= 0.11AV=2.2V
I=V
I= 0.11AV=2.2V
-
8/13/2019 Solution to a Complex Circuit
19/19
CURRENT AND VOLTAGE
To summarize, we get this
R3 R4 R5
R9
R2
R8
R1
R V I
R1 20 33 1.65
R2 20 8.8 0.44
R3 20 2.2 0.11
R4 20 2.2 0.11
R5 20 2.2 0.11
R6 20 8.8 0.44R7 20 33 1.65
R8 20 24 1.20
R9 20 6.6 0.33
Its time to go back to the information w
if we have correct values.I3=I4=I5
I2=I6
I1=I7=IT
I1=I8+ any of I2 or I6
I2=I9+ any of I3 or I4 or I5
One way to check if you are