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page 15

+

 

 

+

 

 

 

 

 

 

 

R1

 

 

 

 

 

 

I

 

V1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

-

 

 

 

 

 

 

 

R2

 

+

+

 

 

 

 

 

 

 

 

 

 

V2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

V

 

-

 

 

V

 

Rs

 

 

 

 

 

 

 

 

 

I

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Rn

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Vn

 

 

 

 

 

-

 

 

-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

First, sum the voltages about the loop,

 

 

 

 

 

 

V = – V + V1 + V2 + …

+ Vn = 0

 

 

Next, use ohms law to replace voltages with the current, and then relate the values to the equivalent resistor Rs.

V + IR1 + IR2 + …

+ IRn = 0

 

V

 

 

 

 

 

R1

+ R2 + …

+ Rn = Rs

 

-- =

 

 

I

 

 

 

 

ASIDE: We can use a “single subscript notation” to indicate a voltage or current. To do this we need to add a direction arrow for current, or use ‘+’ and ‘-’ for voltages.

3.1.2 Node Voltage Methods

page 16

If we consider that each conductor in a circuit has a voltage level, and that the components act as bridges between these, then we can try some calculations.

This method basically involves setting variables, and then doing a lot of algebra.

This is a very direct implementation of Kirchoff’s current law.

First, let’s consider an application of The Node Voltage method for the circuit given below,

page 17

Find Vo

 

 

10

 

20

+

 

 

 

 

 

 

 

 

 

 

 

30V

40

-

 

 

 

 

 

 

 

 

5

 

30

 

 

 

 

 

 

 

 

 

 

First, we will add labels for nodes, currents and voltages,

 

 

10

I1

node

 

 

20

 

 

n1

 

I2

 

 

 

 

I3

 

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

 

 

 

 

 

30V

40

 

 

 

V12

 

 

 

 

 

 

 

-

 

 

 

 

 

 

 

 

 

 

 

 

5

 

 

 

 

 

-

 

30

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

node n2

Next, sum the currents at node 1,

In1 = I1 I2 I3 = 0

+

40Vo

-

+

40Vo

-

Then find equations for the three currents based on the the difference between voltage at nodes 1 and 2. For the center path,

V12 = I340

 

I3 =

V12

----------

 

 

 

40

page 18

For the left branch,

 

 

 

 

 

 

 

 

 

 

 

V

12

= – 5I

1

+ 30 – 10I

1

 

 

V12 – 30

30 – V12

 

 

 

 

 

 

 

I1

=

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

–15

15

For the right branch,

 

 

 

 

 

 

 

 

 

 

 

V

12

= 20I

2

+ 40I

2

+ 30I

2

 

 

V12

 

 

 

 

 

 

 

I2

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

90

 

Now, combine the equations,

 

 

 

 

 

 

 

0 = I1 I2

I3 =

30 – V12

V12

V12

 

-------------------15

– -------

– ----------

 

 

 

 

 

 

 

 

 

 

90

40

 

V12 = 19.5V

Finally, find the current in the right circuit branch, and then the voltage across R2,

I2 =

V12

=

19.5

= 0.217A

-------

---------

 

90

 

90

 

VR2

= I240 = ( 0.217) ( 40) = 8.7V

ASIDE: When using a voltage between nodes we are using the double subscript notation. When doing this we label nodes and then the voltage is listed as ‘Vab’ where ‘a’ is positive relative to ‘b’.

3.1.3 Current Mesh Methods

If we consider Kirchoff’s Voltage law, we could look at any circuit as a collection of current loops. In some cases these current loops pass through the same components.

We can define a loop (mesh) current for each clear loop in a circuit diagram. Each of these can be given a variable name, and equations can be written for each loop current.