Problems of physics and electricity knowledge in junior high school, and problems of physics and ele

When A and B are ammeters, A measures R1 current, B measures R3 current, and parallel: A: B = 3:

2 Illustrates that the ratio of resistance to the two is 2:3. But no matter what the resistance ratio is, when A and B are voltmeters, B is connected in series with R3, A is connected in series with R1, because the internal resistance of the voltmeter is large, and the partial voltage is large, and its indication is close to the power supply voltage, so A:

B = 1:1 (or understood as equal parallel voltages). In Figure B, the partial voltage of the lamp is too small and will not be lit, and the voltmeter has an indication close to the power supply voltage.

Classmate, the question is not very good.

A and B are ammeters, A measures R1 below, and B measures R3 aboveIf it is a voltmeter, it is the R2 voltage that is measured.

A and B are ammeters, and the ratio of A and B is 3:2, (indicating that R1:R3=2:3,), which can be replaced by a voltmeter, then the two representations are 1:1, because they are connected in parallel, we don't care about anything else, just look at the parallel.

Figure 2: The light will not turn on because the internal resistance of the voltmeter is too large and the current is too small.

A - r1 current, B - r3 current.

The bulb is not lit, the voltmeter has an indication, and it is close to the mains voltage.

Hope it helps. If you don't understand, please ask.

1. When A and B are ammeters, they measure the current flowing through R1 and R3 respectively; When they are voltmeters, the voltage of R2, which is the supply voltage, is measured with a ratio of 1:1;

2. When the switch is closed, the bulb will not light up, and the voltmeter has an indication, that is, the power supply voltage.

You can enter a limited number of words (100), and you can only answer the short answer as follows:

From the inscription, it can be seen that the maximum resistance of the rheostat and the resistance ratio of the bulb = 2:1, then when the slider p is moved to the far right, the voltage at both ends of the bulb is U=18V*1 (1+2)=6V, so the resistance of L R=(U2) P=36 6=6 ohms, so the rated power of the bulb = 9 2 R=

You need to look at the diagram to know the change in resistance when the slide moves to the right side; to complete the calculation.

Whether the rightmost end is the largest or the smallest, how can it be analyzed without a legend!

How many ohms are a sliding rheostat?

First of all, you need to understand what a short circuit is in principle. According to Ohm's law, i=u r, therefore, there is a formula to know that the smaller the resistance, the greater the current of the loop, and the imaginary current that can be anthropomorphized is always willing to take the path with the least resistance, so when the resistance of a section of the circuit is very small (close to zero, such as a wire), the current of this circuit becomes very large, resulting in a small or almost no current in the loop with large resistance, which causes a short circuit.

Therefore, to see whether there is a short circuit, only to see whether there is a wire connected in parallel at both ends of the meter, that is, to see if there is a circuit with zero resistance (i.e., wire) that is more "labor-saving" than the circuit where the meter is located.

See if there is a wire at both ends of this appliance, connected together, if there is, it will be short-circuited.

A short circuit in an electrical appliance means that the current does not pass through the electrical appliance but goes back to the negative electrode through another path The short circuit of the ammeter has no indication The short circuit of the voltmeter has an indication but the deflection is small and the value is low In fact, I think that electricity is not very much used by rote memorization Understand the conceptual stuff and then imagine it more It's almost the same When I was in junior high school in electricity, I relied entirely on my interest And then there is imagination Come on.

It is recommended that you draw and analyze, it is very intuitive, thank you.

Electrical appliances, wires, power switches, can form a circuit, if there is no electrical appliances in it, it is directly connected, it is a short circuit. If the ammeter and voltmeter are short-circuited, there is a wire connecting them.

When S is disconnected, the L1 current is 3W 6V=

When the L1 voltage is 6V and the L2 resistance is R, then when the total voltage of the circuit is U=6+S, L1 is short-circuited, and L2 emits light normally, so the rated voltage of L2 is U12W power.

12=u²/r

The solution is r=12 ohms, u=6+

The diameter is one-tenth of the original one, and your length must become 10 times the original one, so that the resistance becomes 10 times the original because of the length, because the cross-sectional area becomes 10 times the original one, so it becomes 100 times the original 4300 ohms.

The resistance of a conductor is inversely proportional to his cross-sectional area and directly proportional to his length. Refer to the textbook First of all, the volume of the copper wire does not change before and after, it starts with v, v=s*l, where s is a very interface agent, if the initial cross-sectional area is s1, and later s2, then by the square of s=, then, s2:s1=1:

100, so later the ratio of the wire length to the initial length is.

l2:l1=100:1, the original resistance is 43 ohms, and the diameter is one-tenth of the original, then the ratio of the resistance to the initial resistance is .

r2:r1=10000:1, then the elongated resistance is 430000 ohms.