How to analyze complex circuits and solve complex circuits

There are three key points in circuit simplification: analyzing the series-parallel relationship, finding the path and direction of the current, and seeing the measurement object of the meter.

Complex circuits are generally disgusting to look at. But don't forget, no matter how disgusting it is, it is also a mixed connection added up by series and parallel, and finding the relationship between the various appliances is the first step to simplify. From this, the arterial road section is drawn, and the most disgusting parallel section is vacated.

Empty is absolutely indistinguishable (nonsense - so analyze it by current. After the diversion on the main road, see which electrical appliances it passes through, when the second diversion is made, and draw them one by one according to the circuit path. If it's not simple enough, keep simplifying.

One point to say here is that the parallel connection after parallel connection is completely parallel, that is, if the shunt is directly shunted again, it means that the circuit and the front after the shunt of these circuits are completely parallel.

The measurement object of the electric meter generally refers to the voltmeter (because the ammeter is directly in the circuit, and the voltmeter must be unique).The measurement object of the voltmeter is mainly to see the position of its two wires in the circuit, and the middle part of the contact in the road is its measurement object. There are vicious and inhumane questioners (personal attacks.

The exam test corns) is to give you a small trick in the big part, such as the circuit diagram of the voltammetry resistor is connected to the two ends of the battery switch rheostat ammeter, in fact, it is connected to both ends of the resistor, there is no difference, I don't believe that I can draw a physical diagram to try.

When the <> circuit is in steady state, the inductance is equivalent to a short circuit, and the capacitance is equivalent to an open circuit, as shown above. Therefore: i7=0.

The node voltage method is adopted, and the lowest end is a common node, and the three node potentials are u1, u2 and u3 respectively.

List the junction voltage equation according to KCL:

Node 1: (u1-us8) r8+(u1-us1) r1+(u1+us4-u2) r4+(u1+us3-u3) r3=0;

Node 2: (u2-u1-us4) r4+(u2-u3+us5) r5+u2 r6=0;

Node 3: (u3-us5-u2) r5+(u3-u1-us3) r3+(u3-us2) r2=0.

Substitution parameter: <>

The simplified equation is: 11u1-2u2-4u3=76;

u1+10u2-3u3=-8；

2u1-3u2=8u3=60。

Solution: u1=12, u2=4, u3=12.

The current of each branch is calculated as follows:

i1=（u1-us1）/r1=（12-21）/3=-3（a）。

i2=（u3-us2）/r2=（12-14）/2=-1（a）。

i3=（u1-u3+us3）/r3=（12-12+6）/3=2（a）。

i4=（u1-u2+us4）/r4=（12-4-2）/6=1（a）。

i5=（u2-u3+us5）/r5=（4-12+2）/2=-3（a）。

i6=u2/r6=4/1=4（a）。

i7=。For node 1, according to kcl: i11=i1+i4=-3+1=-2(a). Or: i11=-(Na i3+i8)=-2+0)=-2(a).

For node 3, according to kcl: i12=i2-i5=-1-(-3)=2(a) or i12=i3-i7=2-0=2(a).

In the steady state, the inductor is set to a short circuit, and the capacitor is set to an open circuit, and then the basic circuit principle can be applied to calculate.

The most basic method for analyzing and calculating complex circuits: the branch current method.

The branch current method is one of the most basic methods for calculating complex circuits. It does this by applying Kirchhoff's law of current.

and the voltage law list the required equations for the junction and the loop, respectively, and then solve the unknown branch current; It is the most straightforward and intuitive method for calculating complex circuits, and the premise is to select the reference direction of the current.

When the physical quantity of multiple branches (especially all branches) is required, the A branch current method, the B loop current method, and the D node voltage method can be selected.

However, when only the physical quantity of a certain branch is required, it is more convenient to choose the superposition principle, Thevenin's theorem, and Norton's theorem.

1. DC circuit.

1. Definition of simple circuit: No matter how many components there are in the circuit, as long as the circuit can be analyzed and calculated by Ohm's law, the circuit is a simple circuit. The simplified method is to simplify the circuit by using the calculation method of series and parallel.

2. Definition of complex circuits: Any circuit that cannot be analyzed and calculated by Ohm's law is a complex circuit.

2. AC circuit.

The analysis method of simple circuits and complex circuits of AC circuits is the same as that of DC circuits, but the calculation needs to be calculated by trigonometric functions or complex numbers.

3. Electronic circuits.

Transistors are included in electronic circuits, and the transistors need to be simplified to the equivalent circuit of the H-parameter before analysis.

Recognize the components in the split unit first.

Both resistors are not shorted, but are connected in parallel because the voltage at both ends of both resistors is the same, and both have electromotive force, so both have current. It won't be a short circuit.

In this circuit, 4 resistors and 12 resistors are connected in parallel and then connected in series with 6 resistors.

The total resistance of this branch is 1 (1 4+1 12)+6=9 so that the current of 6 resistance i=3 9=

There is no short circuit, the so-called short circuit is that the two ends are directly connected by the wire.

The 2 ohms above the voltage source don't need to be looked at, and the voltage source is directly connected in parallel, which can be ignored, and the 4 ohms and 12 ohms are connected in parallel, and then connected in series with 6 ohms, so the current is one-third of a ampere.

Classmates: Circuit analysis doesn't care about how the current flows, don't think about it in high school thinking mode, get a circuit diagram, and think about how the circuit flows, hehe, it's not necessary and impossible, the circuit is very complicated, how the circuit flows, and you can't see it at once, circuit analysis only cares about the reference direction, and doesn't care about the so-called actual flow direction at all.

Redraw the diagram and you'll understand. Current flowing through a resistor of 6 ohms = 3 9 (a).

The current flowing through 6 is 1 3A; The 4 resistors are connected in parallel with the 12 resistors and then connected in series with the 6 resistors.

None of the resistors were shorted.

These two problems are essentially finding the Thevenin (Norton) equivalent circuit.

a) Solution: Because ports A and B are disconnected, so i=0, then the controlled current source 5i=0, which is equivalent to an open circuit.

Let the right end 30 resistor current be i, and the direction is downward; The current of the left end 30 resistor series 40 obtained by KCL is: I-5, and the direction is to the right (top). kvl：（30+40）×（i-5）+30×i=。

So: uoc=uab=30i=30.

Short-circuit the voltage source and open the current source. The voltage U0 is applied from A and B, and the current flowing from A is .

The current of the 30 resistor at the right end is: U0 30. direction downward; The current of the 30 resistor at the left end is: i0-u0 30, and the direction is downward.

The resistance current of 40 obtained by KCL is: 5I+I0-U0 30=-5I0+I0-U0 30=-4I0-U0 30, and the direction is to the left.

kvl：40×（-4i0-u0/30）+30×（i0-u0/30）=u0。

Simplify: u0=-39i0, so req=rab=u0 i0=-39( ).

So the simplest equivalent circuit is: uoc = voltage source series req = -39 resistor.

b) Solution: Using the node voltage method, set the two nodes and the common node o in the diagram. Column Write Node Voltage Equation:

Node 1: u1 10+(u1-12u) 30+2i+(u1-u2) 40=0;

Node 2: 2i+(u1-u2) 40+5=u2 (40+30).

Supplement two controlled source equations: i=u1 10, u=-(u2 (40+30)) 40.

Solve the system of equations: u1=-17v, u2=, i=, u=.

uoc=uab=u2×30/（30+40）=。