A few difficult physical electrical power problems

The first question is simple! According to the square r of p=u, p is inversely proportional to r under the premise that the voltage is constant, and the more resistors are strung in series circuits, the greater the total resistance value! So the total power is smaller. It involves the above two knowledge points.

Because you can't burn out the bulb, the current in the series circuit is equal everywhere. The resistance of the large is r=u to the 2nd power p So the resistance of 60W is large, first calculate the resistance of 100W60W, then calculate the current, and then calculate the respective voltages, and calculate the total power 2 according to P=UIIt should be that at the same voltage, a 6W bulb can seal a little voltage, and the others have to give 4W 3

The thicker the fuse, the greater the allowable current. The resistance is small, and the Q generated per unit time is small.

The calculation of electric power only needs to remember a few formulas and rules, and the topic says 1600imp kw·h, which means that the electric energy meter consumes 1kw h every 1600 flashes

Then, with each flash, 1,600th of a kilowatt-hour is consumed.

Flashing 32 times, that is, 32 1600th of a consumed = 1/50th of a kilowatt hour is the first empty answer, counting this ignoring time.

wp=1/50th of a kWh 60ths of an hour s=

You don't understand what the power rating means.

The so-called rated power is the value of the power under his normal operation, which is a fixed value, no matter what situation he works, it is that value, so the voltage to the electrical appliance is greater than his rated voltage, and its rated power remains unchanged.

What changes is the actual power, such as the second question: the rated voltage is 10V, the rated current is, working at a voltage of 18V, then his actual power p=U 2 R=18 2 20W=, and no longer 5W.

Pay attention to distinguish between the rated power and the actual power.

Solution: (1) When S and S2 are closed, R 2 and R0 are connected in series U=Ur+UR0=6+6 R2 R0=6+12 R0=6+12 24 12=12V, that is, the power supply voltage is 12V

2) When S and S1 are closed, L and R0 are connected in series.

pl=u²/rl=(6v)²/12=3w

3) When P is on the A-side. There is only r in the circuit

Then pr0=u r0=(12v) 12=12wp at the b side, r0 is connected in series with r.

pr0=i²×r0=(u/r+r0)²×r0=（12/24+12）²×12=

So the range of variation in the power dissipated on the resistor r0 is.

The physics teacher who deliberately copied down the question and asked it should be right.

44W light p=ui i=44 220= r=u i r=220 ohms.

A lamp with 66W i=66 220= r=220 ohms.

When connected in series, the working current is i=220 (1100+, and the power of the first lamp in series is =i*i*r=

When the power of the second lamp is connected in series,,, the total electric energy consumed for half an hour when it is energized, the actual power is 44W 66W when the circuit is connected in parallel at 220V, and the actual power is equal to the rated power when working at the rated voltage.

The resistance of two bulbs can be found according to the rated power, and the current is equal to 220 divided by the sum of the resistances when connected in series. The power is equal to the square of the current multiplied by the resistance. If it is in parallel, the voltage does not change, and the power is naturally the rated power.

When finding power, first see whether the voltage of the resistor is satisfied, and if it is not satisfied, find the resistance, and then

If the bulbs are connected in series, it is necessary to use the calculation of the resistance divider. For bulbs with a known rated voltage and rated power, the resistance can be found based on p = U2 r.

Once the resistance is known, the total current at the specified voltage can be calculated, and the actual power of the bulb can be calculated separately based on the current, p = i 2 * r.

Total power = total voltage * total current, total electrical energy consumed = total power * actual power of each in the 220V circuit in parallel time, which is the rated power of the bulb.

Data calculations, you might as well try it yourself.

Calculation method:

Series connection: first find the resistance of each bulb, according to the formula p=u square r to get r1=220 220 44=1100 r2=220 220 66=

The sum of the resistors after series connection is 1100+ After connecting the voltage of 220V, the current is Because the current intensity in the series circuit is equal everywhere, the voltage obtained by the original 44W flying bulb is U=IR 1100 The voltage obtained by the original 66W flying bulb is.

In this case, the actual power of each bulb can be found according to the voltage and current or the voltage and resistance.

The actual power of the original 44W bulb in series p=ui=132 p=usquared r=132 132 1100=

The actual power of the original 66W bulb in series p=ui=88 p=u2 square r=88 88

The total power of the two bulbs p=u squared r 220 220 (1100+.)

Half an hour of electricity is calculated from one kilowatt hour per kilowatt.

Since so hours = (degrees).

Parallel: The power after parallel connection is 44+66=110w If it is energized for one hour, it is degree or kw·h If it is half an hour, it is degree or kw·h

r1 = 220 * 220 44 = 1100 ohms r2 = 220 * 220 66 = ohms.

When they are connected in series, u1:u2=r1:r2, so u1=132v u2=220-132=88v

Actual power p1 = ui 2 r = 132 * 132 1100 = p2 = 88 * 88

Total power = Q = PT = joules 3600000 = degrees in parallel in a 220V circuit the actual power of each is the identified power 44W and 66W

This is a fill-in-the-blank question, and in the actual exam, you will have to solve it in a flexible and quick way.

The bulb emits light normally, then the current through the bulb is the rated current of the bulb, i=w u=4 2=2a

So, the first blank fills in (2).

Total voltage of the circuit Small bulb resistance r=u 2 p=2 2 4=1 ohm total resistance r1=

So the second one is filled.

Solution: Power electromotive force e=3*

Find the small bulb parameters. Rated voltage U=2V Rated power p=4W then resistance R=U2 P=1 ohm.

When the bulb emits light normally, the bulb voltage is 2V, then the bulb current is i=p u=2A.

So the first empty is 2

At this time, the voltage at both ends of the sliding rheostat u=e-u= current is i(2a), so r=u, i=ohm, so the second empty is.

With the formula p=ui=u 2 r=i 2r The circuit in this question is a series circuit, the current i is the same, and the voltage is.

e=u+u.

1 All 1, rl=6 2=18

2. i=6 18=1 3a ur =1 3 12=4V power supply voltage u=4+8+6=18V

3. i=18 (12+30+18)= pl=4, imax=2 6=1 3a=18 (12+r+18) solution r 24

The resistance of the bulb L is 6 squared divided by 2, that is, the rated current of the 18 ohm bulb is 2 divided by 6, which is 1 3 amps.

The resistance of a sliding resistor is 8 times 1 3, which is 8 3 ohm resistance plus a bulb plus a sliding rheostat. Total resistance. Multiply another 1 3 amps. It's good to do classes behind the power supply voltage

Solution: (1) According to the problem, because the circuit is a series circuit, so i = i lamp = 2w 6v =

According to Ohm's law, r=u, i=6v

2) Because the total voltage in the series circuit is equal to the sum of the voltages of the parts, V = V lamp + V resistance + V slip = 6V + 12 *