Big brother and sister, about the power problem, help me solve it

To put it simply: a parallel connection of two 500w electric furnace wires is equivalent to using two 500w electric furnaces at the same time, of course, 1000w.

B Connect two 500W electric furnace wires in series, reverse A, of course it is not right.

c Divide the 2000W circuit wire into two equal sections, take one of them, reduce the resistance by half, and double the power to 4000W

D Connect two 2000W circuit wires in parallel, the same as A, is 4000W

If only the resistance is considered, it should be ACD. These three operations can make the resistor become the original 1 2, according to p u 2 r, u is the supply voltage unchanged, r becomes the original 1 2, and the power is doubled.

This operation results in a 1000W electric furnace wire.

The voltage of the two stages of the electric furnace wire mentioned in this question should be 220V.

500W resistance ohms (P=U2R).

2000W resistance ohms.

A: Two 500W total resistors in parallel: ohms.

p=1000w, a can be.

B: Not one (contrary to A).

C: Take one of the segments, which is actually half of R, i.e.: ohms (contrary to A) D: Contrary to A.

The answer is Note A: Contemptible humble opinion.

The correct answer should be B

The power is 500W. after using method A

The power is 2000W

The power is 2000W

The answer is A, B 250W

c 4000w

d 4000w

According to p=U2r, the resistance changes from the intermediate breaking resistance to the original 1 2

Haha, junior high school physics, oops, I forgot. Tell me how to calculate parallel and tandem.

After that, I think you should understand.

In fact, it is the answers that I get from looking for textbooks that I am most impressed.

<> short circuit of voltage source and open circuit of current source. In A and B, the applied voltage U sets the inflow current to i.

The following relationship exists: u=2i.

According to the series-parallel relationship and KCL, the current of each component is obtained as shown in the figure above.

KCL:I+2U+I+I=0, so: I+2 2I=.

kvl：u=2i+u+2×（i+2u）-2×i=2i+4u=2i+4×2i=10i。

So: req=u i=10( )

Many people often confuse these three relationships, but do you know their real relationship?

The formula p=U2R, P=I2XR, P=IXU It seems that in the series circuit, since the voltage applied to each bulb is reduced by half, the power becomes the rated 1 4, that is, 15W. In fact, the resistance of the bulb is nonlinear, it is greatly affected by temperature, because the 220V bulb does not reach its rated voltage at 110V voltage, can not emit light normally, can not reach the temperature when working normally, its resistance is much less than the resistance at the rated voltage. Since resistance is a variable, it is not possible to calculate the power and current of a series bulb using the above formula.

The electrical power consumed is greater than 15W).

In parallel circuits, the current and power assigned to each bulb are the respective rated current and power ratings. 60w，220v，i=60/220=

The total power and total current of the circuit are the sum of the two.

60W bulb: voltage 220V, current, resistance 807 ohms.

Series: current, power 30W.

Series: current, power 120W.

Two 60W in series is 30W, and two 60W in parallel is 120W

1.Work done per beat: 15000pa*80ml= 75 60= times per second.

p=2.Do work in one day w=

w=mgh h is the lifting height.

h=w/mg=

1).Work done per beating w=p*v=15000pa*80ml= 75 60= times of work done per beat.

p=2).Do work in one day w=

w=mgh h is the lifting height.

h=w/mg=

Blood volume v=sl

Cardiac thrust f=ps (where p is the pressure).

Each time the heart does work, w = fspslpv

15000*80*10^-9 j

Average power p=75*w 60

Please confirm: the rated voltage of a single electric heating pipe is 380V! The rated current of each is 3000 380=, which is the phase current.

Well, the triangle joint. The current measured by the clamp meter is about 13A. Right! Because the current of each wire is multiplied by the phase current by the root number 3, that is.

Contactors from 16A to 20A can be used. 3 pairs of contacts, each contact is connected to a phase power cord, and a vertex of the triangle is connected to the bottom.

From the reasonable design of the wiring process, each tube is directly connected to a lower contact of the contactor with a square copper wire, which acts as a vertex of the triangle and connects another copper wire with a square heating tube at the same time. The other two lower contacts are wired, similar.

Be careful to separate the wires from each other.

The second problem: because the rated current of each tube is, the capacity of the contactor is 10A [each contact is connected to only one tube of live wire]. Use 3 pairs of square copper wires, and the distance between the wires should be separated.

1 The total power of the 3 pipes is 9kw, and the current is, because in a triangle connection, the RMS value of the line current is 3 times the root number of the phase current, which is correct. Choose a contactor of about 30A, which is not easy to burn the contact. 4 square national standard copper core wire.

2 The total power of 3 electric heating pipes, the voltage is 220V, the current is in, and the contactor is at least 30A or more. 6 square national standard line.

1. Calculate the three-phase power according to the current you measured: 13*380*watts, which is similar to the nominal power of the electric heating pipe 9kw.

2. Total power of electric heating pipe: 1500*3=4500 watts, current: 4500 220=amperes.

With 4-6 square millimeters of copper wire, the contactor can be more than 31 amperes.

b, c can be seen from the image before 1 second, f-f=ma=m*10 1, and the image after 1 second can be seen, f=m*10 3

Solve f=m*10+m*10 3=m*40 3 so f:f=4:1, choose b;

According to the definition of work, work = traction force * distance.

The distance is the area of the V-T image.

Traction is only done before 1 second.

wf=f*s1=(m*40 3)*(10*1*1 2) resistance is the work done in the whole process.

WF=F*S2=(M*10 3)*(10*4*1 2) So WF:WF=1:1, choose C