Physics calculations. About electricity, a problem of physical calculations in electricity

。Volt.

Slip on the far left, ammeter is the largest; Slip the rightmost voltmeter is the largest. You can find the maximum by a column of equations.

If there is no solution, it can't, and if there is a solution, it is.

1 Ohm's law gets: r2=u i=

2 It is easy to know from the question: r1=10 r2=30 u2=e r2 r1+r2 u2= substituting the solution to obtain e=6v

3e=6v When r2=r1=10, the indication of v is e, r2, r1+r2=3v is the maximum value of the 3v range.

When r2=0, i=e r1=, which is the maximum value of the 0 range.

Hit with all hands, hope, if you don't understand, you can ask!

1) Rheostat resistance r2==30 ohms.

2) Power supply voltage = + 10 = 6 volts.

3), the voltmeter has two ranges of 0 3 and 0 15, to meet the maximum range, that is, the voltage of the sliding rheostat is 3 volts, the voltage of R1 is 3 volts, the current of the circuit is, and the general current (0; 0 3a), that is, if the ammeter is satisfied, the voltmeter cannot be satisfied. No!

1) As can be seen from the graph: u2 = concatenation isoflow; i2= r2=u/i=

2) r total = r1 + r2 = 10 + 30 = 40 u = r · i = 40

3) I don't know.

UL=2V is just 6V in series, and i= at this time

rl=ul/i=2/

pl=ur^2/il=7*

It should be just looking at the picture, or you can't do it).

Let the current be i, i(2r+r)=220

ir=215

w=iv=215i=

Solution: The current through the transmission line i=60a

r=ohm. The total resistance of two transmission lines 2r = ohms.

The total power lost by the two transmission lines due to heat generation w0 = 2r * i * i = coul.

i=p uactual=60a

U line = U-U actual = 5V

R-line=U-line, i=

P heat = i square r line = 288W

1、p＝ui，i＝u/r

De: r u 2 p 2 2 2 2

2. The power ratio of the lamp is 4:1, that is:

ia^2r：ib^2r＝4:1

So: ia:ib 2:1

3. At A, the lamp glows normally: IA P U 2 2 1A At B, the current is times the current at A: IB

At a, the voltage of the resistor is: U4 and the current: 1A (U4) R0

At B, the voltage is expressed as 4:8 U':11, get: u', Current:

To solve the above two equations, we get: U 7V, R0 3

Solution: (1) Bulb filament resistance rl=u p=(2v) 2w=2 Answer: The filament resistance is 2

2) When the slider p is placed at point a, the bulb emits light normally, i=p u=2w 2v=1a

When the slide p is placed at point b, p=1 4 2w=

p=i²r, i'= square root =

Answer: When the sliding vane p is placed at point A and B respectively, the ratio of the current in the circuit is 2:1 (3) When the sliding vane p is placed at point A, the power supply voltage U=4V+1A*r When the sliding vane P is placed at point B, the power supply voltage U=

4+1*r₀=，rb=r₀+6 ①

When the slider P is placed at point B, the ratio of the voltmeter V1 to the voltmeter V2 is 8:11::11

2+rb)：(rb+r₀)＝8：11

8r 3rb+22 substituting , the solution is r = 8

Answer: The resistance value of the resistor r0 is 8

The 20,000 mAh battery should refer to 20,000 mA ·h, and the power, the general battery is 5 volts (here refers to the battery of the mobile phone power bank, if the No. 5 battery is volt).

Current i=p u=

Time t=q i=20a·h

That is, it can be used for 200 hours.

60 hours in the case of a size 5 battery.

Note: The above calculation is an ideal situation, excluding energy losses such as heat loss, and the actual value should be smaller than this value.

If the potential of point A is less than zero, then the current of R1 is i1=100 (10+5)=20 3A The potential of point A is 50-i1*R1=-50 3V, and if the potential of point A is greater than zero, then the current of R1 is I1'=100 (10+5)+50 (10+20)=25 3A The potential of point A is 50-I1'*r1=-100 3V, not true. So the potential at point A is -50 3V

1: In the circuit with the power of the electrical appliance and the power supply voltage of 220V, can the fuse with the fuse current of 6A be selected? (process) known:

P=,U=220Vthen:I=P U=2400W220V=The rated current of the general fuse is half of the fusing current, and the rated current is now 3A, so the fuse with the fusing current of 6A cannot be selected. 2：

When the rated voltage is 220V, the rated power is 40W, what is the resistance of the bulb? Since r=U2 p, then r1=(220) 2 40=1210

r2=(220)^2/40=

r3=(220)^2/100=484ω

3: In the resistance box commonly used in the laboratory, the rated power of each resistor is specified as, try to find the rated current of 100 ohms and 10 ohm resistors? Because p=i2ri=

p r) then i1=

: An electric water heater with a rated voltage of 220V and a resistance of ohms, how much electric power does it have when it is working normally? How much heat is produced during a working hour?

Its electrical power has. p=u 2 r=(220) 2 When the electric water heater works normally, the heat release q=pt=2000w*1800s=: the resistance of the transmission line is 1 ohm, the electric power transmitted is 100kw, and the power lost on the transmission line due to heat generation is transmitted with a low voltage of 400v?

If I switch to 10,000V high-voltage power transmission, what is the power loss? P=100KW=100000W When 400V is used for low-voltage power transmission, i1=p U1=100000W 400V=250A power lost on the transmission line P1=I 2R=(250A) 2*1 =62500W When 10000V is used for low-voltage power transmission, i1=p U1=100000W 10000V=10A power lost on the transmission line P1=I 2R=(10A) 2*1 =100W

Let the power supply voltage be V, and the maximum resistance of the sliding rheostat is R

Column equations. Get vr (r+20)=6

vr/(r+40)=4

So r=20 v=12

Units bring their own.

You forgot to give the circuit diagram, didn't you?

The power supply voltage can be set to U, then when the switch is disconnected, the current i1=u r2;

When the switch is closed, the current i2 = (u r2) + (u r1) according to the title, i1 i2 = 3 4

Substituting r2=10 to find r1=30

The resistance of the filament remains unchanged, the power of the filament is the original quilt, the current is the original times, the voltage of the filament is the original times, that is, 220*, then the voltage of the wire connected to the lamp is 220V-198V=22V, so the power consumed by the lamp is 22V*(P U)=22*100 220W=10W

Of course, there are other algorithms for this problem, such as finding the resistance of the wire connected to the lamp.

The current is divided into 198V on the lamp and 22V on the wire

So 9w