Basic Electrical Problems, 30, Basic Electrical Problems,

1.[2A+20 B-20 C-30 D-40 The positive and negative gauges here do not indicate the numerical value of the major I choose (d) 3b The electrons in the metal are moving, so they are in the opposite direction to the current.

1. Outside the power supply, generally speaking, it is considered that the current flows from the positive pole of the power supply through the wire and the electrical appliance, and finally flows to the negative pole of the power supply. 2. If the circuit is composed of metal components (i.e., the electrical appliances and wires are all metal materials), then the electrons flow from the negative electrode to the positive electrode. 3. The formation of electric current is formed by the directional movement of a large number of free charges.

And the free charge could be.

The answer on the diagram is correct Judgment method: 1) First of all, it is necessary to make it clear that the voltage and current directions given in the diagram are all positive directions. 2) Judge the actual current and voltage direction according to the positive and negative of the known voltage and current values.

The positive value is the same as the reference direction, and the negative value is the opposite of the reference direction, and the current is known to be -1a is a negative value, so the actual current is the same as the marked current.

2008 high school entrance examination physics "electricity" review question 1, Xiao Ming learned the knowledge of safe electricity use, wanted to make a table lamp, he bought a bulb labeled "36v 36w", when he returned home and started to install, he suddenly remembered that the voltage provided by the socket at home was 220v, so he looked for the manual in the box, but did not find it, but found an accessory (1)...

S2 is closed and R2 is shorted. It's the equivalent of a wire.

Therefore, when S1S2 is closed, only R1 and L are connected in parallel in the original circuit. The parallel buried sail voltage is the power supply electric Kai liquid chain voltage.

R lamp = u lamp forehead i lamp staring grandson forehead = 12 1 = 12 euros. The lamp resistor l emits light normally, so:

U=U Lamp Amount=12V..Supply voltage.

I lamp = i lamp amount = 1a

i1=i-i-lamp=

r1=u i1=12 euros .r1 resistance.

The curve is an equation, the two unknowns are the v and i of the bulb, and then the v and i equations of the bulb can be solved according to the circuit.

The circuit equation 2i=8 (10+r 2), r=v i, ivr are all cautious of the bulb, get 2i=8 [10+v (2i)], solve v=8-20i, this equation and the equation of the graph are combined.

The simple way is to draw a diagram, draw the straight line v=8-20i, and the intersection point of the straight line and the curve is the solution v=2, i=.

The lamp resistor r=v i=20 is wide and prudent3, and the current is 2i=

This topic is not rigorous, the UI curve is not a straight line slippery, so it is a nonlinear resistance, and the resistance value of the non-Senmu linear resistance is equal to the tangent slope of a certain point of the UI curve, and the slope is obviously not calculated in this question. Don't do it, it's a waste of time.

Choose C. Only the switches S and S2 are closed, only L2 will light up and work properly.

A false, if L1 is a short circuit, then the ammeter will be burned out;

B false, if L1 is a short circuit, then the ammeter will be burned out;

D is wrong, L3 has been short-circuited, L2 has been connected in parallel, and it must also be short-circuited.

By the way, this question is really rubbish, saying so many words and testing so few knowledge points)

I think C is right.

Let's look at A: all three switches are closed, and if L1 is a short circuit, the ammeter will be burned out (the resistance of the whole circuit is very small, according to Ohm's law of the closed loop, when R approaches 0, i will approach infinity).

B: It's the same as A (if you understand, it's okay if you don't understand the follow-up question or QQ contact (provided that I do the question correctly and is funny)).

D: Connecting L3 and L2 in parallel The circuit is directly short-circuited, and the consequences are the same as A.

When the sliding blade moves upward, the resistance of the beam lifted by the slag sliding rheostat into the circuit decreases, and the total resistance of the circuit decreases as shown in the figure, so the current representation number becomes larger, so A and B are wrong.

The power supply voltage remains the same, R1 remains the same, the current increases, the voltage at both ends of R1 increases, and the voltmeter voltage decreases.

The sliding vane of the sliding rheostat of a and B slide rheostat moves blindly upward, and its resistance increases, and the series connection increases, and R always increases. U total = power supply does not change, by i = u r, i becomes smaller, the number of the ammeter becomes smaller, voltmeter u + ur1 = u total liter sell, ur1 = ir1 becomes smaller, u is large.

The power consumed by R1 becomes smaller, p=iir

The voltage representation does not change because the voltmeter measures the voltage of the power supply.

The number of current representations increases, and the resistance decreases because the voltage does not change.

L1 brightness does not change because the voltage does not change.

According to p=ui, the voltage does not change, the current increases, so the power becomes larger, c is correct.

1)11000j

2) 66000J 15000 does not give the quality of the goods can not calculate the height.

1) Electro Matching Xiaosun Flow i=Q T=1 Shenghuai 6a Q=UIT=11000J P=W T=UI=110 Pei Chain 3W

2) W total=UIT=66000J Q=I*2R=250J Cargo quality?