How to do this electrical physics problem? A physical electricity problem

Hello, there are two aspects of knowledge to be tested in this physics question:

First, the judgment of the circuit connection method. That is, to determine how the resistors are connected to each other, to know whether they are connected in parallel or in series.

Second, the solution of the total resistance. The total resistance of the series circuit is equal to the sum of the resistance values of each resistor, and the reciprocal of the total resistance of the parallel resistor is equal to the sum of the reciprocal resistance of each resistor.

Once these are figured out, the solution is not very complicated. Take a look at the process of solving this question:

There are four resistors, and there are four ways for the current to pass, so these four resistors are connected in parallel.

2.Calculate rab

Hope it helps!

This question tests the principle of series and parallel, just let the two switches be connected in series.

The circuit diagram that meets the above design requirements is as follows:

The power supply, lamp, S1, and S2 are all connected in series to form a loop.

There are too many electrical appliances connected in parallel, and the total resistance is too small, resulting in excessive trunk current (excessive load leads to excessive trunk current); A short circuit occurs in Figure B, resulting in excessive current.

When the voltage on r is 200V, R0:R=20:200, R0=20, then: R=20 ohms;

Resistance r of each water heater'=200x200 1000=40, then the number of water heaters that need to be connected in parallel in the circuit: n=40 20=2, at this time all water heaters can work normally;

When the number of water heaters connected in parallel is m, the parallel resistance: r'=40/m ω

Total power of water heater: p=i r'=(220/(2+r'))²r'When r'=2, p is maximum, pmax=6050w, m=20

At this time, the circuit current: i=220 (2+2)=55A is less than the rated current, and the circuit is safe.

(1) A: The total power is too large.

B: Short circuit.

2) P total = 220 * 10 = 2200W

1>2 units: P solid = 2000W

2>p total instantaneous = 220 * 60 = 13200W

n = p total instantaneous p = 13 units.

Very unsafe.

2000 is the second question.

The first question is 42000+2000

i 2rt is the heat energy generated by the current, write it directly, don't write the total work of uit can not be calculated directly, because you don't know the power of the motor, that is, you don't know what the mechanical energy output is.

Don't talk about heat, talk about heat.

You calculate 2000j as the heat generated, because the formula you use is i 2rt, which is calculated as heat production. The total work done in 10s is equal to mechanical energy + heat generated = 2000J + 42000J = 44000J

You are thinking wrong, it should be like this, mechanical energy = work done by electric current + heat produced.

Therefore, the work done by the current in 10s w=i 2*r*t=2000jThe heat produced by the motor q=42000j-w=40000j, the mechanical energy is 4200j, and the efficiency of the motor conversion is generally below 70%.

, the heat emitted by the resistance, which is thermal energy, is the answer to the second question!

The first question should be the result of the second question plus 42000j, 2000 + 42000 = 44000j

When the switch is disconnected, the capacitor is not connected to the circuit, so the potential is 0

After closing, it is connected in parallel with R3.

In this case, the voltage is the voltage of R3.

q=cu3 ①

Regardless of the resistance of the capacitor in series with it, the distribution of the voltage satisfies.

u1=u4+u3 ②

r3+r4)r1/(r1+r3+r4)+r]i=e ③u4/u3=r4/r3 ④

The charge flowing through is.

q=q is solved by the five formulas on the synthesis.

q=36 29*10 -4c (I can't see which r4 is.) The answer is that r4 is the topper-most resistor).

q=2*10 -4c (this is the answer when r4 is connected in parallel with the capacitor).

Set the voltage to up=u r

u²／r1=6①

u r1+u r2=9 so u r2=3 gets; r1:r2=1:2 2r1=r2

After series p=u (r1+r2)=u 3r1=6 3=2, so the total power consumed by the two resistors after series connection is 2 watts.

The total power is 2W

Analysis: Oops, it's too hard to write a formula on this, so I won't write it. I hope you forgive me It's a column equation, and when you connect to r1, an equation contains unknown quantities r1 and u squared.

Another equation when parallel r2 contains the unknown quantities r1, r2, u squared.

Calculate R1 and R2 with an unknown amount of U squared

In the end, the total power is an equation that can be summed up to cancel out the u-squared. All right!

Set the voltage to u

When the switch is closed, the motor does not rotate, and the resistance of the motor r=u i=u 2 When the motor rotates normally, the output work of the motor is w=ui and the heat generated by the motor is q=i r

The heat generated by the motor is the loss of electrical energy, and the loss rate is a=q w*100%=i r ui *100% Bring r=u 2 in, and sort out a=i 2*100% =

Efficiency of conversion of electrical energy to mechanical energy =1-30%=70%.

The efficiency problem of the motor is because when the motor converts electrical energy into mechanical energy, due to the internal resistance of the motor, Joule heating will inevitably be generated, and the conversion of the latter part of the energy has no use value for people. Therefore =w e=(e-q) e=1-(i'^2*r)/(u*i')=1-i'*r/u=1-i'*r/(i*r)=1-i'/i=70%

where i and i'The first and second measured currents are represented respectively, R is the internal resistance of the motor, and U is the electromotive force of the power supply.

Let the current in the loop be i when s is disconnected; The current in the loop is i' when the s is closed (ignoring the internal resistance of the power supply, if it is high school, it should be considered).

S is disconnected: R1, R2, R3 are connected in series, V measures the voltage at both ends of R1 and R2, i.e. U=I(R1+R2).

S closed: R2 is short-circuited, R1, R3 are connected in series, V measures the voltage at both ends of R1, i.e. U'=I'R1

u/u'=i（r1+r2）/i‘r1 (1)

p=ui=i 2r (this circuit is a pure resistive circuit, this formula can be used).

p3 quietly collapses p3'=i^2r3/i'2r3 gets i i'=2 Kai 3 can be solved by substituting this formula into equation (1).

r2/r1=11/4 （2）

By the power supply voltage does not change, there is: i(r1+r2+r3)=i'(r1+r3) solution (2) gives r1 r3=2 9

p1’/p3'=r1 r3 so p1'=2w