solution to a complex circuit

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.


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.


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


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



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



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



7/19

When the battery is attached, we get the equivalent circuit.

R3 R4

R9

R2

R8

R1



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


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


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



15 R2

R8

R1


11/19

15 R2

R8

R1



in series with R2 and R6, hence



2 6

20 20

2,..6R = R + 15 + R

= + 15 +

= 55

55

R8

R1


12/19

55

R8

R1

14.67

R1



in parallel with R8 ,hence



6

1 1 1

55

1 1

55 20

4 11

220 220

22014 67

15

55, 8

55, 8

= +R R

= +

+

R .


13/19

14.67

R1


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


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


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=


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


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


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


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

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solution to a complex circuit

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