Rush to the beginning of the second physics Ohm s law and the push down of electrical power, all, it

Basic: i=u r p=ui p=w t

Derivation: from the formula: u=ir r=u i

From the two formula: u=p i i=p u will be one generation into two: p=ui p=i 2 r p=u 2 r

It is obtained by the three formulas: w=pt t=w p and the two are substituted into three pieces: w=pt w=i 2 rt w=u 2 rt

Ohm's law is experimentally derived, not derivative.

Ohm's law i=u r u=ir r=u i

The formula is p=ui i=p u u=p i

Define the formula w=pt p=w t

The derivation formula u=ir p=ui gives p=i*i*r i=u r p=ui gets p=u u r

w=uit r=u*u p u=root rp

p=i*i*r is suitable for series circuits because the currents are equal.

p=u u r is suitable for parallel circuits because the voltages are equal.

Note: Multiplication and division are in order.

Let the bulb resistance value be rl, yes.

The bulb is connected in series with R1 rl*(V (R1+rl)) 2 =3 (1).

The bulb is connected in series with R2 rl*(v (r2+rl)) 2 =3 4 (2).

From 3w 6v 6v rl, find rl 12 known condition r2=n r1

1), (2) divide the two equations, and (n-2)r1-12=0 is solved to obtain n with two values that meet the constraints, i.e., 4 and 3

n 4, r1=6, r2=24, the voltage source should be 9v;

n 3, r1=12, r2=36, the voltage source should be 12v

The conditions are not complete, right?? For example, is the power supply voltage constant?

Amount 2 p=12,rl*u (r1+rl)=6v,rl*u (r2+rl)=3v,(r2+rl)=2, (r2+rl)=2, since r1:r2=1:n, let r1=x, i.e., nx+12=2*(x+12),(n-2)*x=12,n is a positive integer less than 5 12=1*12=2*6=3*4, so n=3,4

x=12,r1=12,r2=36

n=4,x=6,r1=6,r2=24

1. Correctly analyze the series and parallel connection of the circuit, and draw the equivalent circuit diagram.

Second, carefully examine the problem to clarify which quantities are known, what quantities are unknown, and which quantities correspond to the same circuit or the same conductor, and third, and then find the relationship between known quantities and unknown quantities according to the characteristics of series and parallel circuits, Ohm's law, and electric power formulas.

Fourth, pay attention to the bulb, electric heater, fixed resistance, when the circuit changes, current, voltage changes, their resistance value does not change.

Fifth, the calculation method, (1) generally use the formula method, through the direct substitution of the formula or the formula after the deformation of the calculation, can be solved, 2) some to use the ratio method, such as the series partial pressure, parallel shunt method.

3) If the first two methods cannot be solved, the solution should be considered with an electrical equation or a system of equations.

This is the problem solving technique of electrical calculation and analysis, but you must combine the problem and experience it carefully to understand the mystery, I hope these will help you.

Memorize the formula, just look at the circuit diagram.

1. Correct analysis.

The series and parallel connection of the circuit are drawn.

Equivalent circuit diagram.

Second, carefully examine the question to clarify which quantities are known, what quantities are unknown, and which quantities correspond to the same circuit or the same conductor.

Ohm's Law. Electrical power.

formula, find the sum of known quantities.

Relationship of unknown quantities.

Fourth, pay attention to the light bulbs

Electric heaters, fixed-value resistors.

In the circuit changes, the current,

voltage changes when they.

The resistance value does not change.

Fifth, the calculation method, (1) generally use the formula method, which can be solved by directly substituting the formula or calculating the formula after deformation, and 2) some need to be used.

Ratio method, such as series divider, parallel shunt method.

3) If the first two methods cannot be found, consider using them.

Electrical equations. or a system of equations.

This is the skill of solving the problem of electrical calculation and analysis, but you must combine the problem and experience it carefully to understand the mystery, I hope these will help you.

1. The current and voltage of the series circuit have the following rules: (such as: R1, R2 in series).

Current: i=i1=i2 (the current is equal everywhere in the series circuit).

Voltage: U=U1+U2 (the total voltage is equal to the sum of the voltages everywhere).

Resistance: r=r1+r2 (total resistance is equal to the sum of each resistance) If n resistors with the same resistance value are connected in series, then there is r total = nr

2. The current and voltage of parallel circuit have the following rules: (such as: R1, R2 in parallel).

Current: i=i1+i2 (the trunk current is equal to the sum of the currents of each branch).

Voltage: u=u1=u2 (the main circuit voltage is equal to the voltage of each branch).

Resistance: (The reciprocal of the total resistance is equal to the reciprocal sum of the various parallel resistors) or .

If n resistors with the same resistance value are connected in parallel, then there is r total = r

Note: The total resistance of a parallel circuit is smaller than that of any one of the branches.

The formula for calculating electrical work is: w=uit (unit w joule (j); u volts (v); i Ann (a); t seconds).

5. When using w=uit to calculate electrical work, pay attention to the following: w, u, i and t in the formula are in the same circuit section; The unit of calculation should be unified; Any three quantities known can be found to be a fourth quantity.

6. The following formula can also be used to calculate electrical work: w=i2rt; w=pt；w=uq (q is the amount of electricity);

Well, okay, I'll help you figure it out myself. Are you in a hurry? If you're not in a hurry, I'll help you push it down on the weekend. After all, I'm also a student... Russia also has to write homework ==,

Answer: Ohm's law formula: i=u r

Export formula: u=ir r=u i

Derivation of the derivation of the derivation of the formula.

Because: i=u r

Multiply both sides of the equation by r at the same time, ir=u

So: u=ir

Divide both sides of the u=ir equation by i: u i=r, i.e., r=u i