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Question 1: Select B

The reason is that the resistance of lamp A is 10 ohms, and the resistance of lamp b is ohms.

Question 2 The first diagram shows that the total resistance Rab decreases When the slide is at the A terminal, the resistance is 5 ohms.

When the resistance of the slide vane at the B terminal is 20 9 ohms.

The second figure shows that the total resistance Rab increases first and then decreases, when the resistance of the slide is 3 ohms at the A end, when the resistance of the sliding blade above R3 is 4 ohms, the total resistance Rab reaches a maximum of 4 ohms, and when the resistance of the sliding blade at the B end is 55 16 ohms.

Question 3 The process is that if the resistance of the voltmeter is r, then the current through the voltmeter v1 is 8 r

If the current through the voltmeter v2 is 5 r, then the current through the voltmeter v3 is 3 r, then the reading of the voltmeter v3 is 3 r *r = 3 volts.

The fourth question can be obtained from the diagram The resistance of the two lamps changes with the voltage But the diagram does not indicate which is a and which is b The question is not too strict If you don't do it according to the above diagram The resistance of lamp A is ohm, and lamp b is 484 ohm When the two lamps are strung together, the total resistance of the lamp is European, and the power of the electrical appliance in the series circuit is proportional to the resistance, the total power is ohm, the power of lamp A is watts, and the lamp b watts.

Question 5 A Question 6 A The reason is that R1 and lamp L1 are connected in parallel, R2 and R3 are connected in parallel, lamps L3 and R4 are connected in parallel, and then all in series, when R1 is disconnected, the resistance of the whole circuit becomes larger, and the current becomes smaller.

Question 7 Voltmeter is an ideal voltmeter, so the internal resistance of the voltmeter is infinity The voltmeter measures the voltage of EF.

When the current in the circuit is 4 ref at the illustrated position, then e r total = 4 ref

When the current in the circuit is 3 ref at the left of the position shown, then e r total +r = 3 ref

Total solution r = 3r

When on the right side of the diagram position, the current is e r, total -r = e 2r and the voltmeter reads e 2r *ref

e/3r *ref =4 e/4r *ref =3

e 2r *ref = 6 volts.

Question 8 The process is as follows: Let the resistance of the voltmeter be r, the internal resistance of the power supply is r, the resistance of parallel connection is r1, and the original reading is v

12 (r+r)=i The total voltage of the voltmeter and shunt resistor after parallel connection R2=Rr1 (R+R1).

12/(r+r2）=3i

V=IR The current of the shunt resistor is 3i-i 3=8i 3

then r=8r1

r2=8r1/9

r=24r1/9

v=9 volts.

There are so many questions, it took a long time, ask me again if you have questions.

Question 7 The title is shifted to the right l The reading of the voltmeter u2 is to find the reading of the voltmeter when l is left.

This way you can use the original diagram.

Or don't change the title of the voltmeter and the power supply, swap the positions of A and B, and swap the positions of C and D.

The answer is 14 3

Is the leftmost resistor 3 or 7? I can't see it clearly.

uab = -i3*r3 = u2 - i2*r2= u1 - i1*r1；

i1+i2+i3=0；

Okay, let's do it yourself.

Using the superposition principle, Thevenin's theorem reduces the circuit to an equivalent voltage source circuit, and the result is.

The equivalent voltage U0=(2-4J) (1+3J), the equivalent impedance Z0=(3+4J) (1+3J), I=U0 (Z+Z0), U=U0*Z (Z+Z0), let Z=X+YJ, to obtain the maximum effective power of the load, the current and voltage are required to have equal phase angles, and the calculated X=1, Y=0, that is, the load should be 1 ohm pure resistance, and the maximum effective power P=20 55J

There are two cases, one is modulo matching and the other is conjugate matching. w=1；

Disconnect the two ends of zl, and the endpoints are a, b;

The equivalent circuit can be used to simplify the circuit, and it is not easy to talk about the above diagram.

Using Thevenin's theorem, it is obtained: uoc=-j 2 2,v,zo=;

When conjugate matches: z=zo*=

pm=uoc 4ro=(1 2) when 4 modes are matched: r=|zo|=√（

pm=uoc²|zo|／[ro+|zo|）²zo²]=1／2×

AC is too troublesome, you divide it into two questions maybe someone does. Scores are not attractive.

Question 2: The equivalent circuit of Thevenin is:

Open circuit voltage: U1 20X(2, solution: U1 40V, therefore: Uoc, equivalent impedance:

ZEQ 20 J10 Euros.

1. There are 2 voltages in the three-phase circuit, which are 6 voltages, of which AB, BC, AC is the line voltage, 380V AN, bn CN, and the phase voltage is 220V

2 What power does a transformer have? I don't understand...

The unidirectional conduction of 3 PN junction is the same principle used in DC circuits, that is, 2 diodes, and should mainly be electrical appliances such as proximity switches.

4 Encoder?? I haven't seen anything that was eliminated many years ago since I started automating, so it's a simple compilation for PCB, PLC servers, etc.

Programmers, which have no practical value, but now they all use laptops. (Hey, that's what you study in college now??) It's too up-to-date with the times, and you've learned this stuff well.

No, hey, China Education!!

5 counters? There are too many types, depending on how to divide them, there are mechanical ones, and there are digital ones. There are also only counting, and there are action contacts, as well as power-off memory, and multiple reset, etc.。。。

I don't understand college, but I know a little bit about electricity, for example, the three-box electricity belongs to 380 alternating current, and the output current of this transformer becomes larger after changing the pressure electricity, which can also be said to be stable current.

A power supply, an external resistor, why the voltage of the external circuit is equal to the electromotive force.

This statement is incorrect.

The voltage of the external circuit U = electromotive force e - the voltage on the internal resistance u is inside, that is, you said U outside + U inside = E.

If the internal resistance is calculated as 0, so u inside = 0, so the external circuit voltage u outside = electromotive force e.

Under normal circumstances, the internal resistance is not equal to 0, so the electric potential between the two poles can be expressed by ur=i*r.

Since the internal resistance is assumed to be 0, the electric potential between the two poles cannot be expressed by ur=i*r, and this condition is no longer true.

Therefore, the internal resistance equal to 0 does not mean that the poles are on the equipotential plane.

The internal resistance is equal to 0, which can only be said to be less resistance to the current between the two poles.

No, the two stages of the internal circuit will not have equal potential, as you said, if the internal resistance is zero, there will be a current when the external resistor is connected. EMF e=

U outside + u inside, u inside = 0, so e = u outside! Your key misconception is that there is no current! As long as you connect to the external circuit to form a loop, there will be a current ......

I don't know if I can help you, I don't know if I can ask, hehe......

The electromotive force of a power supply is the ability of a power supply to convert other forms of energy into electrical energy, which is numerically equal to the work done by a non-electrostatic force to transfer a unit of positive charge from the negative pole of the power supply to the positive electrode through the inside of the power supply. It is a function that can overcome the resistance of the conductor resistance to the current and make the charge flow in the closed conductor loop! If the external circuit is cut off and there is no current, the charge cannot enter the negative plate of the battery, and the internal circuit will have no current!