Thevenin s theorem problem, Thevenin s theorem problem, solve

Upstairs is the nodal method, not Thevenin's theorem.

1.Disconnect the 12 resistors, disconnect the 15A current source, and short-circuit the 6V voltage source, and the equivalent resistance seen from the 12 ports is 6.

The current source is disconnected, and the voltage at both ends is calculated to be 6V when the 6V voltage source works alone, with the left negative and the right positive.

The voltage source is short-circuited, and the voltage at both ends of the 12 is 90V when the 15A current source works alone, with positive left and negative right.

4.The two voltages are superimposed, and the equivalent voltage source voltage of Thevening is 90-6=84V, and the resistance is 6. a

Let the voltage of the common connection point of the three resistors be U1; The voltage at the 6V voltage source is U2; The voltage at the 6V voltage source-terminal is U3; The voltage of the two 4 resistors common connection point (bottom) is 0 (reference point); The current flowing through the right4 resistor is i3 (from top to bottom). u3/4=i3①，u2/4=i1②，i1＋i3=15③，u2-u3=6④；Syndicate: U2=33, U3=27, I1=33 4, I3=27 4;In addition(u1-u2) 12=i2 ,(u1-u3) 6 i2=15 ; Lianli:

i2=14/3、u1=89。So i2 = 14 3.

First, disconnect r to find the ruler rotten open circuit voltage uoc:

u2=10×5/(5+5)=5（v）

u3 = 1 1 = 1 (v).

uoc=u2+e2-e3-u3=5+5-2-1=7(V) two, the power supply is set to zero, and the equivalent resistance req:

req=5||5+1=

3. i=uoc (req+r)=7 7=1(a)<>

Fourth, back to the original picture.

The current flowing through E3 is obtained from KCL = I+IS, P=E3 (I+IS)=2 2=4(W).

The sock is trapped in the direction of the current facing downwards, which is the associated reference direction, so E3 absorbs power and is the load.

Understanding of Thevenin's theorem:

1. Remove the load of the original circuit, and equip the remaining part with an active two-terminal network. This two-terminal network can be a voltage source and resistor in series, or a current source and resistor in parallel.

2. Finally, add the load.

Thevenin's theorem: A linear resistor single-port network n with an independent power supply can be equivalent to a single-port network with a voltage source and a resistor in series in terms of terminal characteristics. The voltage of the electric grip voltage source is equal to the voltage UOC of the single-port network when the load is open, and the resistance RO is the equivalent resistance of the single-port network NO obtained when all independent power supplies in the single-port network are zero.

Solution: There is no current in the resistor at the bottom2 resistor, so the voltage of this resistor is zero.

So: UAB = 3 + (-3) + 0 = 0 (v).

b) Solution: Similarly, there is no voltage drop in the lower 2 resistor, UAB = 2 + 2 + (-4) = 0 (v).

1. Find the open-circuit voltage, which is equal to the resistance value of the current source 1a current multiplied by 1 and 2 resistors in parallel and then connected in series with two 1's resistors in parallel

uoc=1x(1//2+1//1)=1x(2/3+1/2)=1x(4+3)/6=7/6 v

2. Remove the source resistance, open the current source, and the resistance in the AB end is the resistance value after the 1 and 2 resistors are connected in parallel and then connected in series with the resistors of two 1s.

rab = 1 2 + 1 1 = 2 3 + 1 2 = 4 6 + 3 6 = 7 6 3, the voltage source of 7 6V in series 7 6 resistors constitute the Thevenin equivalent circuit.

<> for complex circuits where there is a controlled source, the following methods can also be employed:

First of all, calculate the output voltage UO when the port is open and bright, and the delay is the potential in the Thevenin equivalent circuit;

Then calculate the current IS that flows through the port during a short circuit, and UO IS is the resistance in the Thevenin equivalent circuit [IS is the current source in the Norton equivalent circuit].

Obviously, the current in the loop is the current source current is=2a, and the direction is counterclockwise.

So: uoc=uab=is r+us=2 5+5=15(v).

Then short-circuit the voltage source and open the current source to obtain:

req=5ω。

The Thevenin equivalent circuit is a UOC = 15V voltage source with 5 resistors in series.

Why is the current in the loop IS? Isn't there also a voltage source?

Do you use Thevenin's theorem to find the current i of the right2 resistance?

Solution: Disconnect the 2 resistors from the circuit, and set the upper end as a and the lower end as b.

Then connect the 1A current source at the left end in parallel with 2 resistors, which is equivalent to 2 2=2V voltage sources, and connect 2 resistors in series, and the direction of the voltage source is positive and negative.

The current of this circuit is: i1 = (10-2) (4 + 2 + 2) = 1 (a) in the direction of counterclockwise.

Therefore: uoc=uab=-4i1+10=-4 1+10=6(v).

Then the voltage source is short-circuited and the current source is opened, and the req=rab=4 (2+2)=2( ) is obtained

Davenin: i=uoc (req+r)=6 (2+2)=.

Are you trying to unravel it with numbers?