Physical and electrical power in the second year of junior high school, and physical electrical powe

Start by calculating the resistance of the bulb.

w=u^2/r

r = u 2 w = 220 2 100 = 484 ohms.

A resistor r is connected in series in a circuit'

r+r')/r = 240/220

1+r'/r=1+20/220

r'=(20/220)r

44 ohms.

That is, a 44 ohm resistor is connected in series in the circuit, which causes the bulb to emit light normally.

p=ui to find the rated current of the bulb is i=100 220 a

In the calculation of that resistor, it must be connected in series, and the current is the same, and the voltage it shares is 20 volts

u=ir r=u i=20*220 100=44 ohms.

A 44 ohm resistor is connected in series.

ur = u total - ul = 240 - 220 = 20 v

i=p/u=100/220=5/11a

r=u i=20*11 5=44 ohms.

r1 = 220 2 100 = 484 ohms.

According to the title, u1+u2=240v, u2=20v, so u1:u2=r1:r2=220:20=11:1, so r2=484 11=44 ohms.

A 44 ohm resistor is connected in series.

Connect a 44 ohm resistor in series.

To divide the voltage in series to make it glow normally, to reach the rated voltage, it is necessary to connect a resistor with a voltage of 20 in series. i=p u to find i, divide the voltage of the resistor to be connected in series by i (in series circuits, the currents are equal everywhere) to find the resistance.

Add a 20 ohm resistor.

Like, the process was forgotten.

Whether the resistance of the sliding rheostat at the far left end is 0 ohms, there is no figure I don't know, if so, at this time, the small bulb emits normally and reaches the rated voltage, so the voltage at this time is the power supply voltage volts.

At this point, there is. The bulb resistance is.

OhmsWhen the slide is moved to the far right end of the rheostat, the resistance reaches a maximum of 20 ohms, and the current in the series circuit is.

i=u r=ampere total circuit power is.

p=ui wattsIf in the other case, the resistance of the sliding rheostat is 20 ohms at the far left, and the resistance becomes 0 ohms when it is moved to the far right.

The supply voltage is.

U=IR=Volts, if the resistance reaches a minimum of 0 ohms when the slide is moved to the far right end of the rheostat, then the small bulb is subjected to volts and will undoubtedly burn out, so the second case will not exist.

Since there is no circuit diagram, I assume that there are two scenarios for your circuit diagram.

The first case:

1) When p is moved to the far left, the resistance of the sliding rheostat is 20 ohms, then the total voltage is .

2) The filament resistance is r = ohms.

3) When moving to the far right, the resistance of the sliding rheostat is 0. Then, the power at this point is p=u r=

The second scenario:

1) When p is moved to the far left, the resistance of the sliding rheostat is 0 ohms, then the total voltage is .

2) The filament resistance is r = ohms.

3) When moving to the far right, the resistance of the sliding rheostat is 20 ohms. The power at this point is p=u r=

1.At the leftmost end of the slide: there is only a bulb in the circuit, and the supply voltage is equal to the bulb, and at this time the bulb is working normally, so u=.

2.From Ohm's law: At this time the bulb works normally i=, u=, so r=ohm.

3.When the slide is on the far right, the sliding rheostat is connected in series with the bulb, and the total resistance is 25 ohms in the sum of the two, at this time in the circuit i = ohms = , so p = (ohms =

The power supply voltage is U, the current is I, and the filament resistance of the small bulb is R lamp 1From the meaning of the title: u resistance = r resistance i = 20*;; Because the bulb glows normally; So u light =; So U = U Resistance + U Light = 10V; +

2.Because the bulb emits light normally, the indication of the ammeter is, and because only the words are visible on the bulb, R lamp = U lamp.

i = ohms. 3.Because when the slider is moved to the far right end of the rheostat, the r resistance of the r lamp = 0 ohms, which is only the bulb connected to the circuit, from (2) the r lamp = 5 ohms, so p total = u square r lamp = u total square r lamp =

Without a circuit diagram, slip cannot be judged.

Ask: When the sliding rheostat moves the slider p to the far left, is the resistance of the input circuit 0 or 20?

Is the resistance of the sliding rheostat P the largest or the smallest at the leftmost end?

Robot, your circuit shouldn't be short-circuited, right?