A resistor of 1 ohm forms a two dimensional infinite array of infinite distances

The conventional solution is the current analysis method, assuming that the current of size i is connected and sent out at two points of the Japanese diagonal respectively, then the electric potential at infinity can be set to the electric potential of 0, and the electric potential of the access and outlet points is assumed to be +u and -u by symmetry, and the current can be considered to flow from the access point to infinity, and then flow back to the contact point from infinity. In the previous process, the current flows symmetrically, so that the first section of the z-shaped path from the inflow point to the reserve point has the current of i 4, the second section has i 12 (i.e. i 4 is divided into 3 parts), and the third section is still above i 12 (a careful analysis of the diagram shows that this node is two flows in and two out); A similar analysis is performed in the latter process (except that the current is in opposite directions). In this way, the superimposed currents on the three paths are i 3 (i.e. i 4 + i 12), i 6, i 3, and the potential drops to ir 3 + ir 6 + ir 3 = 5ir 6, and this should be equal to the total voltage drop of 2u (i.e. 2u = 5ir 6), so the equivalent resistance r'=2u/i=5r/6

Addendum: The Z-shaped route I'm talking about is the regular route of the Japanese character from one end to the other (through the horizontal in the middle), and you can see for yourself if it's a Z-shape.

Do a star angle transformation and it's OK.

There is a change in the position of the letter symbols, so delete the answer, and I'll think of other ways when I have time.

The second one, 50 after series and 8 after parallel, thank you!

The large resistance becomes the small resistance, and it can only be connected in parallel. Because paralleling increases the cross-sectional area of the conductor, the total resistance becomes smaller.

According to the theorem, in a parallel circuit, the sum of the reciprocals of the total resistance is equal to the sum of the reciprocals of the resistance of each component

Let there be n resistors in the parallel circuit, 1 n r minute = r total and because r total = 5 and r minute = 20, n = 4 is obtained.

tut, is there anything else you want to know can hi me.

Using the formula 1 r branch 1 + 1 r branch 2 = 1 r stem, x 20 = 1 5, the solution is x = 4, and four resistors of 20 ohms are connected in parallel.

1. It is required to convert a larger resistance into a smaller resistance--- which can only be paralleled;

2. The formula for the total resistance of parallel connection is: 1 r=1 r1+1 r2+--1 rn;

3. When r1=r2=---=rn, the total resistance formula is derived as: r=r1 n.

4. The essence of your problem is to find the number of sub-resistors that need to be connected in parallel

n=r1/r=20ω/5ω=4.

This classmate, just for your reference. Good luck with your studies! 】

Method 1: Trigonometric variable star. See image above. Total resistance = (1 + 1 3) (2 + 1 3) + 1 3 = 28 11 + 1 3 = 13 11

The total voltage is V, and the node voltages are V1 and V2 respectivelyIndividual resistor currents can be derived. For v1 node there is v1 1=(v-v1) 1+(v2-v1) 1, and v=3v1-v2

For v2 node, there is (v-v2) 2=(v2-v1) 1+v2 1, and v=5v2-2v1From this we get v1=. Substituting v=5v2-2v1, v2=5 13v, v1=6 13v

So the total current i = v1 1 + v2 1 = 11 13v and the total resistance = v i = 13 11

I'll give you a transformation formula, and you'll understand it at a glance. Sorry for the grass!

Perform a star-triangulation transformation of the resistor, and this problem will be solved.

Two in parallel and one in series to form a total circuit, and then in parallel with the fourth. So that's 3 5

The total circuit formed by two series connections and one parallel connection is then connected in series with the fourth circuit. That's 5 3

Two in series, and then in parallel with one, and then one in series after parallel, 5 3 ohms.

Two in parallel, and one in series, and one in parallel after series, 3 5 ohms.

1. There are three resistors with resistance values of 2 ohms, 3 ohms, and 6 ohms to obtain the ohmic resistance:

2 ohms and 3 ohms should be connected in parallel

2. If you want to get a resistance of 8 ohms: you should connect the resistors of 2 ohms and 6 ohms in series.

Solution: No matter how complex these 8 different resistors are, think of them as one resistor R2

From the parallel calculation formula of resistors, it can be obtained: r1*r2 (r1+r2)=4, and r2=6 can be calculated

then rab=r1'*r2/(r1'+r2)=3

Using the equivalent method, first of all, r1 is in parallel with others, and the equivalent of other resistors is 6 ohms from the situation of the problem.

Then 6 ohms and 6 ohms are connected in parallel, which is 3 ohms, as for how the circuit is connected, it is not the purpose of the examination, but whether you will be able to deal with complex circuits equivalently.