Physics Electricity Fill in the Blank Questions There are answers, Physics Electricity Multiple Choi

The voltage indicates the key current of the circuit when the number of forests is large.

i=U (R0+R)=3V (5+10) = Maximum number of voltmeters.

uv=ir=

The voltage represents the R0 voltage when the number is 1V.

u0'=u-uv'=3v-1v=2v

The series voltage is proportional to the resistance of the electrical manuscript, and the resistance value of the resistance wire into the circuit.

r'：r0=uv'：u0'=1v：2v=1：2r'=

At this point the spring is shortened.

10ω=δx：6cm

x = the weight of the object.

g = mass of 15n weight.

m=g/g=

Larger, because the measured number represents the sum of the masses of the plastic disc and the object.

The voltage represents the current of the circuit when the number is maximum.

i=U (R0+R)=3V (5+10) = Maximum number of voltmeters.

uv=ir=

The voltage indicates the number of bright and bright disadvantages, when talking about 1V, R0 voltage.

u0'=u-uv'=3v-1v=2v

The series voltage is proportional to the resistance, and the resistance of the resistance wire to the circuit.

r'：r0=uv'：u0'=1v:2v=1:Bling2r'=

At this point, the spring is shortened.

10ω=δx：6cm

x = the weight of the object.

g = mass of 15n weight.

m=g/g=

What a coincidence, I'm looking for this question too.

First of all, the resistor sheet is connected to the circuit in this way, which is equivalent to connecting the upper and lower parts of the resistor sheet in parallel in the circuit, and the heat generated is directly related to the equivalent resistance after series connection. The supply voltage is constant, as can be seen from q=u 2*t r, the smaller the equivalent resistance, the more heat is released.

In the first diagram, we can see that the resistance of the upper and lower parts is equal, and I assume that the resistance of the upper and lower parts is r, so that the two r's are connected in parallel, and the equivalent resistance should be equal to r 2.

The second picture is also the upper and lower parts in parallel, but this is that the resistance of the upper and lower parts is no longer equal, and the equivalent resistance of the parallel connection at this time is smaller than the equivalent resistance of the parallel connection when it is equal (the specific proof is more troublesome, so I won't talk about it here).

To sum up, the second picture emits a lot of heat, and the answer should be C

Upstairs is the right solution. This question examines the problem of resistor paralleling. In the second diagram, the upper half of the resistance is significantly larger, so the total resistance is on the larger side and the heat generation is larger. There is no need to make detailed calculations, and it is enough to be able to make a rough judgment.

The limit method solves the problem, and the hole is moved to the point away from m, then the line is broken, and A does not heat. So C right.

Answer: Whether the C ra r electric key K is broken does not affect the total current The total current does not change.

When the electric bond k is closed, the current on R0 is 2A, and the current representation is 1 0A.

ra=2r0=12a

C Because the resistance R is very large, the access circuit of Ra and R0 does not affect the current size of the bus, so when K is closed, the current flowing through R0 is 3-1=2 A, so that the resistance of the ammeter is = 2 * 6 1 = 12 ohms.

You see, it's talking about multiples.

RA is many times smaller than R.

I'm using the normal method, and your data is not wrong.

Let the internal resistance Ra of the ammeter be equal to Ra, then:

The power supply voltage is constant, the parallel voltage is the same, and the voltage is divided in series; So:

3(ra+r) = 1ra+1ra r0*r, i.e. 3(ra+r) = ra+ra 6*r

The solution yields r=12ra (ra-18).

Substituting ra=2 3 12 18 into the above equation respectively yields:

Ra=2 3 RRA

So choose C

Select C and look at the equivalent circuit diagram to understand. AC two points are shorted. The following two resistors are connected in parallel, which is 1 ohm, and the total number of resistors plus the surface is 3 ohms. 6 3 = 2a And then half of the current flows through the ammeter is 1 a.

The resistance of the equivalent resistor at each point is 2, and each is 1

If there is 6V between B and C, AC is in parallel, and then it is connected in series with B, the total resistance is the total current I=4A, and the ammeter is the branch current of A, so the indication of the ammeter is 2A selection: B.

When the switch S1 and S2 are both disconnected, R1R2R3 is connected in series, and the power consumed by the resistor R1 is 1WU (R1+R2+R3)] R1=1(1) When only S2 is closed, R1R2 is connected in series, and the power consumed by R2 is 8WU (R1+R2)] R2=8(2).

Only close s1, only r1, resistor r1 consumes power is 36wu r1 = 36(3).

Solution (2), (3) and r1>r2 Get r1=2r2 Substituting r1=2r2 into (1) and (3) Solution r3=9r2=When the manuscript is slowed down, S1 and S2 are both disconnected, the power consumed by R3 p3=[U hail (r1+r2+r3)] r3=(1 8)u r1=

To protect the ammeter, the current cannot be exceeded, according to Ohm's law, i = u r, r total = u i =

r = r total - r =

Because the r is bigger, the i is smaller, so r

To protect the voltmeter, you cannot exceed 3V, according to Ohm's law, i = u r, i = u r = (

r₂=u₂/i₂=u₂/i₁=3/

Because the larger the r, the larger the u, so the range of variation of the r 10 ( ) sliding rheostat is:

When r = 8 ohms, i = u r = u (r + r )=

u₂=i₂r₂=ir₂=

The voltage changes, but the resistance does not change, so the formula p u 2 r is used, originally, p u 2 r

Later, p (u 2) 2 r 1 4 (u 2 r).

p=ui, when the voltage decreases to the original 1 2, the resistance remains unchanged, the current also becomes smaller than the original 1 2, and the electrical power becomes the original 1 4b

b p=u2 r, r does not change, so u2 becomes the original 1 4, so the power becomes the original 1 4

The answer is that the fault in BA is that the voltage cannot say the voltage that passes through, only the voltage at both ends of the conductor.

c has a free charge, and without voltage, it cannot move directionally and cannot form an electric current.

d, there is voltage in the circuit and the circuit is closed before there is current.

Choose, the voltage can not be said to pass through the conductor; c. It should also be able to move in a directional manner; d, it also needs to form a pathway.

The conductor of the selected BA should be a closed loop.

The mistake is that the voltage is not said.

d does not say closed loop.

It is enough to grasp the necessary conditions for the formation of current: voltage, conductor, closed loop, all are indispensable, voltage provides electromotive force, conductor provides free electrons, and closed loop is the path of continuous current.