Physics and electricity questions in junior high school as soon as possible Thank you 3

18V 18W" can find the bulb resistance as 18 ohms.

When the sliding blade P of the sliding rheostat moves to the A terminal, the lamp L emits light normally, indicating that the power supply voltage is 18V

When the slide p is moved to the midpoint, the voltage representation is 15V, and the maximum resistance of the sliding rheostat can be calculated to be 180 ohms according to the voltage divider principle.

The power of the bulb p=i r i=18 18+x Let x be the resistance value in the rheostat access circuit, r=18 ohms.

So p=[18 (18+x)] 2*18, the larger the x, the smaller p is.

To ensure the safety of each element of the circuit, the maximum current of the ammeter: when the access resistance of the sliding rheostat is 0, it is 1A, which meets the requirements;

Voltmeter: The maximum voltage cannot exceed 15V, that is, when the sliding blade p of the sliding rheostat moves to the midpoint, so the maximum resistance of the sliding rheostat cannot exceed 90 ohms, that is, the X range is 0-90

When x = 90 p is minimum, the minimum power is.

rl=U2 p=18 ohms.

When the sliding vane p of the sliding rheostat is moved to the A terminal, r slip = 0, so when the sliding vane p of the sliding rheostat with a supply voltage = 18V is moved to the midpoint, r slip rl = u slip ul = 5 1 r slip = 90 ohms r slip total = 180 ohms.

When the voltmeter is fully biased, rslip = 90 ohms, so rslip is less than or equal to 90 ohms.

When the ammeter is fully offset, i = u r = 18v 18 ohms + r slip = 3a r slip = -12 ohms so r slip is greater than or equal to 0

When r slips to the maximum, i.e., 90 ohms, pl takes the minimum.

plmin=u2r=15v2 90ohm=

When set at point A, the sliding rheostat voltage U(A), R1 voltage U(R1A), R2 voltage U(R2A).

When set at point B, the sliding rheostat voltage U(B), R1 voltage U(R1B), R2 voltage U(R2B).

The series circuit has r1 r2=u(r1a) u(r2a)=u(r1b) u(r2b)=k (introduced unknowns).

List 4 equations u(a)+u(r1a)=11 . Equation a

u（r1a）+u（r2a）=6。。。Equation B

u(b)+u（r1b)=10。。。Equation c

u（r1b)+u（r2b）=9。。。Equation D

Subtract the voltage of R2 with k and push the equation b to U(R1A)*(1+K)=6

Equation d pushes out u(r1b)*(1+k)=9

u(r1a) u(r1b=2 3....Equation e

The equation a+b pushes out u(a)+(2+k)u(r1a)=17... Equation f

The equation c +d pushes out u(b)+(2+k)u(r1b)=19 . Equation g

The power supply voltage remains unchanged, u(a)+u(r1a)*(1+k)=e=u(b)+u(r1b)*(1+k).

Equation g-f=u(r1b)-u(r1a)=2 .Equation h

Equation h and equation e solve the system of equations, u(r1b) = 6 u(r1a) = 4

Substituting Equation b u(r2a)=2 Substituting Equation a u(a)=7

So the supply voltage = 4 + 2 + 7 = 13V

Isn't it a list of equations and then mathematically eliminated elements?..

AB point is the sliding rheostat at both ends?

1) When S is closed, S1 and S2 are disconnected, the circuit is a series circuit.

When the slide is at the midpoint: UL=6V, then the bulb emits light normally, i=p UL=3 6=, RL=UL I=6 When the slide is at the B end, i'=

From these two cases we can get the system of equations:

e=i(rl+1/2rp)

e=i'(rl+rp)

Bring in the data to solve: e=8v, rp=8

Close S1, S2, the circuit is a parallel circuit, the current must increase, so i''=i+2=

ip = e 1 2rp = 8 4 = 2a

then i0=i''-ip=

r0=e/i0=8/

So the supply voltage is 8V, and the resistance value of the fixed value resistor R0 is 16

(When S is closed, S1 and S2 are disconnected) is equivalent to L and the sliding rheostat R is connected in series, because the voltmeter measures the voltage of the lamp L at this time, and it is 6V, so the lamp L is working normally, and the circuit current is the rated current through (the bulb L is marked with the words "6V 3W") can be calculated. i= p u=3w 6v= As the sliding rheostat slide moves to the left, the resistance is getting smaller and the current is getting bigger. (The indication of the ammeter has changed, so the original current should be .

No matter how the current changes, the total voltage is constant. The expression for the total voltage after moving before and after moving is U = lamp) = lamp) R lamp is calculable according to (bulb L is marked with the words "6V 3W") R lamp = U square p = 12 ohms so lamp) = lamp) in the slip r=8 ohms Derive back to the original supply voltage = 8V

Keeping the position of the slider P unchanged, closing S1 and S2, the indication of the ammeter changes by 2A. ) becomes r0 in parallel with r, so the judgment current becomes larger again. By a+2a= r, r0 both ends of the voltage are 8v slip into 4 ohms, so the current through r i=u r=8v 4 ohms = 2a through r0 so r0=u i =8v = 16 ohms.

It's too inconvenient to write, I don't know if I can understand it?

S is closed, when P is at the midpoint, the lamp and the rheostat are connected in series, the voltmeter is 6V, so the lamp is normal, the current, when S1S2 is closed, half of the rheostat is connected in parallel with the fixed value resistance, and the total current can be obtained by combining the power supply voltage and the rheostat resistance.

Solution: (1) The indication of the voltmeter is 6V, and the voltmeter is connected to both ends of the power supply, indicating that the circuit fault may be the lamp open circuit or the resistance value of the sliding rheostat connected to the circuit is 0, and the bulb is not lit and the ammeter has no indicator, and the circuit fault should be the bulb open circuit

Because the indication of the voltmeter is, so the bulb emits light normally, and the current indicates the number i =, so the rated power of the bulb p = u i = (2) from the voltage at both ends of the bulb u = i1r, and when the switch s is pulled from position 1 to position 2, the connection mode of the three resistors in the circuit has not changed, so the voltage at both ends of the bulb is still i1r at this time, and l is connected in parallel with r, the bulb current i = i2-i1, and the bulb resistance rl = i1r i2-i1 is obtained by Ohm's law

Due to the thermal effect of the current, when the bulb emits light, the resistance of the filament will increase with the increase of temperature, so when the voltage at both ends of the bulb is not equal to its rated voltage, the resistance obtained is not equal to the resistance when it emits normal light, so that the power obtained is not the rated power;

Therefore, if you want to find the rated power of the bulb, you must make the voltage at both ends of the bulb equal to the rated voltage

Therefore, the answer is: (1) the filament is broken;

i1r i2-i1, the bulb does not emit light at the rated voltage, the temperature change of the filament causes the resistance of the filament to change, the switch s is pulled to position 1, and the slider p is moved so that the current representation number is i1=u0r

The answer to the last empty is: u0 i i r i or i i i r (not the rated power, but the actual power, to get the rated power let i is enough). The penultimate empty answer should be:

i₂－i₁﹚×r ÷i₁

1) Because the small bulb emits light normally, the voltage is 6V, and R2 and the bulb are connected in parallel, so the voltage of R2 is also 6V

According to the conditions, it can be seen that the small bulb emits light normally p=ui, and the current is because the total current is, so the current r2 is .

According to Ohm's law, r2 = 6v ohms.

2) This is very simple, according to the first question you can know the power supply voltage when 6V, the second question is R1, R2 in series, assuming that the resistance value of R1 into the circuit is X, then, P1 = Ir1 i=u (x+r2) so so x=20 or 5 are in the resistance range.

So current = 6 (20+10) = or 6 (5+10) = this is definitely true, and the third question is 1 3 watts.

Wrong! 1) I lamp=p u=3w 6v=

IR2=I-I Lamp=

R2=U IR2=6V ohms.

2) U1 = P1 R1 4V under the root

i=u1 r1=4v 24 ohms

3) R light = U p = 6V 3w = 12 ohms.

i=u (r1+r lamp).

P light = i r light =

1) When S1, S2 and S3 are all closed, R1 is short-circuited, R2 is connected in parallel with the bulb, and the bulb happens to emit light, indicating that the power supply voltage is 6V.

il=p, u=3, 6a=

i2=i-il=

r2=u i2=6 euros=10 euros.

The second question is not very easy to do, as if there is a shortcoming.

rl = u amount 2 p amount = 36 euros = 72 euros.

The power of the small bulb is the smallest, and R1 is the largest.

i=u r=6 (72+24)a=1 16ap min=i2r(1 16)2*72w=9 32w

Solution: The passing current of the bulb is obtained i=, the current passing through R2 is I=A, and R=u i gives R2=10 ohms.

Total = i total * u total = power of r2 + power of sliding rheostat = 6i = squared, solve the 2nd equation, and get the current approximately equal to.

Lamp = U square p = 12 ohms, to be smaller, the smaller the power of the bulb, only the greater the resistance of the sliding rheostat, that is, r slip = 24 ohms. Find the current in this state i=6 (24+12)=1 6 ampere. P lamp min = i square * r light = 1 3 watts.

Solution: (1) The lamp and R2 are connected in parallel"6v 3w"The lamp emits light normally, then the power supply voltage is 6V, RL=UE PE=(6V) 3W=12, I lamp=U R=6V12=,I2=I- IL=,R2=U I2=6V;

2) The sliding rheostat (with RP below) is connected in series with R2, and the current in the circuit is I

By the title: p=pp+p2,p=ui,pp=,p2=i r2,r2=10 ,u=6v;

i.e.: 6v i=

Solution: i=or.

3) When the sliding rheostat is connected to the maximum resistance of the circuit, the power consumed by the small bulb is the smallest, and R2 and L strings are at this time.

Join; r=r2+rl=24ω+12ω=36ω

Then the current in the circuit i = u r = 6v 36 = 1 6a

The power consumed by the small bulb pl=i rl=(1 6a) 12 =1 3w=

1) Small bulb current i1=p u=3 6=

At this point, the current of r2 i2=

r2=u/i2=6/

2) u=u1+u2

u1=p/i u2=i*r2

Substituting u=6 p= r2=10 to get the solution.

i= or i=

3) When r1=24, the small bulb power is the smallest.

Small bulb resistance r=6

At this time, the current i=6 (12+24)=1 6

Small bulb power p=i2r=1 3w

1), fully closed, then R1 is short-circuited, R2 and the bulb in parallel, the bulb happens to emit light, indicating 6V 3W, according to P=UI, find i=, the power supply voltage is 6V, because it is in parallel, left and right I2=, R2=U2 I2=6V ohms. In fact, you can also find the total resistance directly from 6V and subtract the bulb resistance, and you can get the R2 resistance.

2）……Classmate, you have two S1, I don't know where S2 ......In fact, don't feel confused when you encounter this kind of problem, just draw an equivalent simplified diagram according to different situations

Above ·

The normal glow must be 10W.

So the circuit current is 10 10 = 1A

The resistance of a lamp with 10v 20w is 10 10 20=5, so the actual power of a dim lamp is 1 1 5=5w

The lamp resistance of 10V 20W is 10 10 20=10 The total resistance of the circuit is 10+5=15

The current is 1A, so the voltage is 15V

From r=u 2 p, we know that r1=10 2 10=10 ohms, r2=10 2 20=5 ohms.

If the current of the series circuit is the same, the voltage with large resistance will be large, so the lamp L1 will emit light normally, and the voltage will be 10V; L2 emits a darker glow.

When lamp L1 emits light normally, the current in the circuit is i= p u=10w 10v=1a.

So the actual power of the dimmer lamp L2 is R2*I 2=5*1 2= 5 (W).

The supply voltage of the circuit is 10V + R2*i = 10 + 5 * 1 = 15 (V).