Ask a question about the physical electrical power of the second year of junior high school

p = current squared multiplied by resistance, u is proportional to i.

Therefore, if the voltage is halved, the current will also be halved, and the above equation becomes.

p = 1 4 (current squared multiplied by resistance).

This is equivalent to a 4-fold reduction in power, which is 300W

Resistance R 220 2 1200 When the voltage is 110V, the electrical power W 110 2 R 300W

Solution: Because the rated voltage of the small bulb is 6V, if the voltmeter uses 3V, it will exceed the range and burn out the voltmeter, so he uses the 15V gear; Because the rated power is between 7W-12W, the rated current range is between 7 6A-2A, so the ammeter uses 3A gear.

According to the diagram, it can be seen that in this circuit, the voltage at both ends of the small bulb is 5V, and the current passing through is 10 3 The resistance of the small bulb is 10 3. a。

Small bulb rated power = u r = 6 (10 3) = a error. b。

Power at 5V voltage of small bulb = UI = 5

b correct. c。

Because the voltmeter uses a range of 15V instead of 3V, the measured voltage is not a 1VC error. d。

Because option b calculates the power of the small bulb at 5V, D is also wrong.

To sum up, the answer is.

(1) Efficiency = q w 90 = 118800 w w = 132000j

2) w=pt p=w t=132000 (5*60)=440w

3）∵p=u^2/r ∴r=u^2/p=220^2/440=110ω

4）∵i=u/r ∴i=198/110=

5) w=uit*efficiency w=198*

In fact, there is nothing to analyze, as long as you can master the formula and be familiar with it, you see my thinking above is very clear, the front is the formula, and the back is to bring the known into it, OK, you still give me the one who gave me ls, after all, he answered first, o( o

The electrical energy consumed by the electric kettle w thermal efficiency n = the heat absorbed by the water q , so the electric energy consumed by the electric kettle w = 118800 90% = 132000j

W=pt, so the rated power p=w t=132000 300=440wp=ui, so the current i=440 220=2a, the resistance r=u i=220 2=110

The actual current i=u r=198 110=

q=uitn=198×

It's simple.

90% of the electricity consumed by the Water Margin is absorbed by water, that is, the electrical energy consumed is 118800 90% j power = electric energy 5*60s

Resistance = U2p

At 198V, i=u r

q=i^2rt

The ratio can also be...

It's not easy to play so many complex symbols, so I'm here to help you, a good student who loves to learn.

Solution: The electric energy consumed by the electric kettle is w=q suction 90 =118800j 90 =132000j;

Its rated power is P=W T=132000J (60S 5)=440W;

Obtained by p=u r: resistance r=u p= (220v) 440w=110;

If the actual voltage is 198V, because the resistance does not change with temperature, the current through the electric kettle i'=u' r=198v 110=;

The electric energy consumed by the electric kettle in 1 minute w'=u'i't'=198v, and the heat generated by the electric kettle in 1 minute q'=90 w'=90 21384j=.

Finally, I wish you the best of luck!

c p = u u r, with a piece of the same material but thinner electric furnace wire, the broken one is taken and repaired in the original length, the thinner the resistance of the resistance wire, the greater the resistance, and the voltage is 220V unchanged, and the electric power decreases with the increase of resistance.

p1=u 2 (r1+r2) in series

When connected in parallel, p2=u 2 (r1*r2 (r1+r2)) according to the question r1=r2

p1/p2=...=1/4

I lamp = p u = 12 w 6v = 2a, r lamp = u p = (6 v) 12 = 3

The resistance box started with the R-box, and later the R-box-6; Later the normal luminous current of the bulb was 2A, and at first it was.

The power supply voltage is U

u (r box +3) =

u (r box - 6 + 3) = 2

The solution yields u=36v, r-box=21

The total power consumed by the circuit. p total = ui = 36v * 2a = 72w

Let the power supply voltage be U, and the resistance of the resistance box at this time is R, and we can know that R lamp = (U lamp 2) P lamp = 3

I light = P light U light = 2a

The total voltage of the original circuit u=(r+6+r lamp) (i lamp.

At this time, the total voltage of the circuit u=(r+r lamp) i lamp

Synoptic r=15

At this time, the total power consumed by the circuit p=(ilamp 2) (r+rlamp)=72w

Let the resistance of the resistor box be r1: the resistance of the small bulb is r=6 2 12=3 ohms i=12 6=2a, 2x(3+r1-6)= r1=21 e=36v p=ui=36x2=72w

When S is closed, S1 is in a state of insulation when it is disconnected, the power is 40W, and only R1 plays a role at this time, so when calculating the resistance value of R1, the power should be 40W when the power is insulated, such as the heating power of 440W, and the equivalent resistance of R1 and R2 is calculated in parallel.

If there is any uncertainty, please specify it, I will answer it and hope that it will be adopted.

The work done by an electric current per unit of time is called electrical power. It is a physical quantity used to express the speed or slowness of electrical energy consumption.

The main thing is to use proportions. The actual voltage is U, the actual power is P, and the resistance is R, and the watt-hour meter is 3000 revolutions per kWh. That is to say, 3000 revolutions per hour, and the work done is 1 kilowatt hour.

Therefore, the actual power p=[(121 3000)*1] (20 60)=can be seen from p=u*u r, u*u u(amount)*u(amount)=p p(amount) can be substituted into the data u=242v

Actual power: 121W, voltage: 242V, remember to use p=i r when the current is unchanged, and p=u r when the voltage is unchanged

A: 121w; 242v

First, find out how much electricity the small bulb consumes in 20 minutes, which is 121r 3000r per kWh, according to p=w t. That's 121r 3000r per kWh one-third =

Then find the resistance from p=ui, i=u r to get p=u2 r r=u2 p to get 220v*220v 100w r=484 ohms.

By r=u*u p u*u=pr u=root pr u=242v

Divide 121 by 3000 to find the electricity, the teacher taught, and then divide the number at the beginning of autumn by the time, and that's it.

Because the rated current of the small bulb is (3 6)=, as long as the current flowing through the bulb is, the resistance of the bulb can be found from 6V, 3W is (6*6) 3=12 Euro 9 (12+6)=

Therefore, it is enough to connect a resistor with a resistance value of 6 ohms to make the bulb glow normally.

The formula p=ui is involved

p=u squared r

u=ir

The bulb resistance U2 P is found to be 12, according to the series voltage divider characteristics, the resistance value of a string is 6 resistance, and the voltage at both ends of the small bulb is 6V, which can emit light normally.

When the bulb emits light normally, the current is ampere, and it should be connected in series in a 9-volt circuit, and it should be connected in series with a 6-ohm resistor.

r=u square p

The resistance of 40W is greater than that of 60W.

Tandem i.

i squared * r = p

p1>p2

It's been 5 years since I graduated, and I don't know if that's how it counts.

It can be calculated by using the formula of power. Just calm down and take a good look at the formula.

> P2 series circuit, the ratio of electrical power is equal to the ratio of resistance.

The resistance r of a l1 is 1210 and the resistance r of l2 is . Yes, because in series, the current is the same, and p=i 2*r. So P1 >> P2

When the switch is disconnected, R1 is connected in series with R2, the power of R1 is P1 = IU1, and the voltage of R1 is U1 = P1 I = 1 I

The voltage of R2 is U2=IR2, and the resistance of R2 is R2=U2 I=2 i, and U1 and R2 are substituted into , respectively.

When the power supply voltage U=U1+U2=P1 I+2V=1 I+2 is closed, R1 is short-circuited, and only R2 works. The indication of the ammeter is 1a

Supply voltage u=i r2=1 (2 i) ....

1/i+2=1×2/i

2/i-1/i=2

1/i=2i=

Although this requirement is a bit harsh, the process is as follows.

Suppose there are two unknowns: the supply voltage U, the resistance R2 of resistor 2 has U=R2 for the closed switch, the voltage is conserved U=U1+2 when the switch is disconnected, and the current series circuits are equal.

i=2/r2=1/u1

Three positions number three equations, and you can solve them.