A uniform resistance wire with resistance of R, truncated into two sections, and the ratio of its le

The resistance of L1 is 2 3R, before L2 is elongated, it is 1 3R, after stretching, the length is 2 times of the original, because the volume remains unchanged, the cross-sectional area is the original 1 2, so the resistance is (1 3)x2x2=4 and the resistance of 4 3R in parallel after the resistance value is 4 9R

4/9rl1:l2＝2:1，r1：r2=2:1,r1=2/3r,r2=1/3r

A segment of L2 is evenly elongated to 2 times its original length, R2'=4r2=4/3r

rtotal = r1r2'/(r1+r2')=2/3r*4/3r/(2/3r+4/3r)=4/9r

The resistance of R1 is 2 3R, the resistance of R2 is 1 3R, and the elongated length of R2 becomes twice the original cross-sectional area becomes half of the original, so the resistance is 4 times the original, that is, 4 3R, the total resistance is 2 3R, and 4 3R is connected in parallel with 4 9R

The first section is 2 3r, the second section is r 3, if the second section is lengthened by 2 times, it will also be reduced to half of the original, so the resistance of the second section is four times that of the original r 3, that is, 4r 3, and then in parallel with the first section 2r 3, according to the calculation method of parallel resistance, it can be found to be 4r 9

According to your length ratio, i1=2 3r, i2=1 3r, and then the two resistors are associated: i=i1 i2 2 3r*1 3r (2 3r+1 3r)=2 9r, as for you to lengthen the resistance wire, it has no effect on the resistance. Because the overall mass and density is unchanged.

In this case, the length is the original 2 3, the cross-sectional area is the original 3 2, and the resistance is the original 4 9r

If you don't understand, please ask! ^-

The resistance is the original:

l1=2/3r

l2=1/3r*4=4/3r

The resistance after parallel connection is:

1 [(1 (2 3R)+1 (4 3R)]=4 9R in series with a resistance of :

2/3r+4/3r=2r

10000r

The analysis is as follows: because the volume of the resistance wire is unchanged, after pulling it into a uniform filament of d 10, the cross-section becomes smaller than 1/100 of the original (because the area formula is s r, and the square of the god 10 is 100), and the resistance becomes 100 times the original: 100r; At the same time, the length has also become 100 times the original, and the resistance has become 100 times larger

10000r

r= l s ( represents the resistivity of the resistor, which is determined by its own properties, l represents the length of the resistor, and s represents the cross-sectional area of the resistor).

After pulling it into a uniform filament of the ridge d 10, s is equal to the original 1 100 volume, the cross-sectional area is reduced by 100 times, and the length of the leech is increased by 100 times (volume = cross-sectional area length).

r=l s to get r'=10000r

The resistance becomes 10000r, the formula is r=pl s, p is constant.

The resistance r=pl s wire in the elongation or fold of the volume v=sl unchanged, the cross-sectional area is half of the original, so the uniform thickness of the wire, stretch it to 2 times the original, its resistance will become 4 times the original.

Cut it into two equal parts, and twist these two parts, and its resistance will become the original 1 4

The resistance value of the resistive element is generally related to the temperature, material, length, and cross-sectional area, and the physical quantity that measures the influence of the resistance by temperature is the temperature coefficient, which is defined as the percentage of the change in the resistance value for every 1 increase in temperature. The main physical characteristic of the resistor is that it transforms electrical energy into thermal energy, which can also be said to be an energy-dissipating element, and the internal energy is generated when the current passes through it. Resistors usually play the role of voltage division and shunt in circuits.

If a wire of uniform thickness is stretched twice as long, its resistance will become four times greater.

Resistance r=pl s, resistance is related to length and cross-sectional area, the volume of the wire in the elongation or fold of the volume v=sl unchanged, the cross-sectional area is one-half of the original, so a uniform thickness of the wire, elongated to 2 times the original, its resistance will become 4 times the original.

When the electric charge moves in the conductor, it is subject to the collision and friction of other particles such as molecules and atoms, and the result of the collision and friction forms an obstacle to the current of the conductor, and the most obvious feature of this hindrance is that the conductor consumes electrical energy and generates heat (or emits light). This obstructive effect of an object on an electric current is called the resistance of that object.

If the 4r is twice as long, then the length is twice as long (the resistance is twice as long), and the cross-sectional area is half as long (the resistance is twice as long). Therefore, the combined resistance is four times that of the original, that is, 4R

The circuit can be divided into two parts of left and right symmetry, let the midpoint of BC be H, the midpoint of EF is G, and the three points of the symmetry know that the potential of the OGH is equal, and the whole circuit is composed of the same circuit of the left and right parts in series. The left circuit is a parallel circuit with three branches, one is a branch of AB series OB and BH in parallel, the resistance of this branch is 4 3R, one is the AO resistance is R, and the other is the branch of OF parallel FG on the AF series, the resistance of this branch is 4 3R, the total resistance of the three branches in parallel is in Eu, and the total resistance of the left and right sides in series is in Eu.

I can't get on the picture, it's hard to explain clearly. B and F equipotentials, C and E equipotentials, can be regarded as the beginning and tail in parallel, and then connect one of the ob or of of in series to the o point, from the d point to look forward is similar, redundant, also take one, only 8 resistors, two and a series and then ao merge, is the first half. It's easy to understand if you draw a picture.

r 2 + r = 3r 2, parallel r, = 3r 5, after adding 3r 5, the total resistance is 6r 5

Sorry for the mistake, it's r4 5Every resistor is useful.

Why is the circuit divided into three ways? ”

I made a mistake before.,I'm sorry to say more here.,You can see it in the link below.,It's very detailed.,There's a picture.。

First, take AD as the boundary, fold it up and down (because b and f are equal potentials, and c and e are equal potentials), and then connect the resistors of ab and af in parallel to obtain 1r 2. In the same way, the other four parallel resistors are also 1r 2. Then the angular resistance of the three points of bf, ce, and o is converted into a star resistance, because the resistance value is equal, and the converted resistance is 1r 2 of 1 3, that is, 1r 6.

At this time, the potential between the point between BF and CE and the O point is equipotential, so the resistance between them can be regarded as a short circuit, and then the resistance of point A and BF point is connected in series with 1r 6 to obtain: 1r 2 1r 6, which is equal to 2r 3, and then use this resistance to connect in parallel with r between a and o to obtain 2r 5, and the right half is also 2r 5. Finally, in series, we get 4r 5=

It should be evenly lengthened to 4 times the original length. The Cryptont.

When the length of the potato is quadruple, the cross-sectional area becomes one-quarter. The resistance becomes 16 times.

Because the volume is constant.

r= l s, for is different for each substance, l is the length of the resistance wire, s is the cross-sectional area.

Cut the resistance wire into 2 halves, l is the original 1 2, and then the endpoints are tightly wound together and merged, the cross-sectional area becomes 2 times the original, that is, s is 2 times the original, unchanged, so the resistance becomes the original 1 4

I offer two methods:

After being drawn into a uniform filament with a diameter of 10/d.

It becomes a filament of s 100, 100L length.

Method 1: The factors that affect r are length (l) and cross-sectional area (s).

The larger the l, the greater the r (equivalent to series in length, the more you connect in series, the greater the resistance).

The larger the s, the smaller the r (equivalent to parallel [parallel connection, it will be thicker], and the more parallel connections, the smaller the resistance).

Here d becomes 1 10, and the radius becomes 1 10

then s becomes 1 100

It can be seen here that there are 100 lines of original length (1 10*d) connected in parallel.

According to the characteristics of parallel connection, the original resistance is 1 100 of the original length of a line of 1 10*d

So the resistance of the original length of the line of 1 (1 10*d is 100r

Now the length is 100 times the original length, which can be seen as 100 times the resistance of the original length (100 times the original length).

So finally the resistance becomes 10000r

Method 2: According to r=p*l s.

d1=10*d2

Then s1=100*s2 (s=pies squared 4), l1=l2 100

So r1 = r2 10000

Resistance = 10000r

r=p*l/s

r denotes resistance.

P pronounced "rho", which means electrical conductivity, which is related to materials, such as gold, silver and copper, the conductivity is smaller, and our wires are often replaced by aluminum or copper, while iron and other ruler collapsing metals are relatively large.

l indicates the length of the material, indicating that the resistance is proportional to the material.

s denotes the cross-sectional area of the material, indicating that the resistance is inversely proportional to the cross-sectional area of the material.

However, the above formula is established under the premise that the cross-section of the material is uniform, and as for the unequal cross-section, the same is true, the thicker part of the circle resistance is smaller, and the thinner part is larger. You can observe that when the fuse is blown out, the part that burns out is relatively thin.

Analysis: The formula for the work done by pure resistive current is w=uit=i rt, because r1=10 is already known, and the time t=1s is also known, and only the current flowing through r1 can be found.

Solution: r total = r1 + r2 = 10 + 50 = 60 (the total resistance of the series circuit is equal to the sum of the two resistors).

i=u r=6v 60 = the currents are equal everywhere in the series circuit, so the current in r1 is equal to the current in r2.

w1=i rt= is replaced by the formula w=uit, the answer is the same, that's the trouble point).

Because of the series connection, the current is equal, for. From q=i 2rt, work has to be done.

The series current is the same i1:i2=1:1

w=uit=i rt,i,t are the same,w is proportional to r w1:w2=5:6p=ui=i r,p is proportional to r p1:

p2=5:6q=i rt,i,t are the same, q is proportional to r q1:q2=5:6

Ratio of current: 1:1

Ratio of electrical work: 5:6

Ratio of electrical power: 5:6

Calorie ratio: 5:6

Knowing two resistors r1:r2=5:6, put them in series in a circuit, and the ratio of the current through the two resistors l1:l2=1:1

In the same time, the ratio of work done by the current in R1 and R2 w1:w2=5:6 is the ratio of the power consumed by the two resistors p1:p2=5:6

The ratio of heat produced by R1 and R2 in the same time q1:q2=5:6