Junior high school physical current comparison questions, junior high school physics questions. Curr

1) Since i=u r, the current ratio is the voltage ratio, which is equal to i:i'=2:32) Let r2=x, when disconnected, u total = i (10 + 20 + x) closed r2 short circuit, u total = i'(10+20)

From the above two equations, x=15

1) The current measured by the ammeter is the current flowing through R3, and the current ratio is the voltage ratio, because i=u r, so the current ratio is equal to i:i'=2:3

2) Let R2=X, when S is disconnected, U total = (10+20+X)*i closed R2 short circuit, U total = (10+20)*i'

Comparing the above two equations, we get x=15,

s Break i=u (r1+r2+r3) s close i=u (r1+r3) Ratio of the two to 2 to 3 Find the relationship between r2 and r1 r3 List the current voltage equation for r1 Simplify it.

It's been a long time since I've done an electricity problem

Analysis: S is disconnected, three resistors are connected in series; S is closed, R2 is short-circuited, and R1 is connected in series with R3.

1 solution: from i=u r, the current ratio at both ends of r3 is the voltage ratio, because the series current is equal, so the current at both ends of r1, the voltage is i:i'=2:3

2 solution: can be listed by the condition: s is disconnected, u·r3 (r1+r2+r3)=2 ur3=2r1+2r2+2r3

s closed, u·r3 (r1+r3)=3 ur3=3r1+3r32r1+2r2+2r3=3r1+3r3 r2=(r1+r3) 2=(20 +10) 2=15

Solution: When the series ruler is taken, the current in the circuit i=u (6 +rx)=2v rx

When connected in parallel, the root is buried according to Ohm's law u rx=2a

Solving the equation system consisting of two formulas of liquid wax yields rx=3 u=6v

(1) The ammeter burns out, and after reconnection, the current flows from L1 to the negative pole of the ammeter, and then flows into the positive pole of the ammeter, resulting in current.

The meter is reversed, thus burning out the ammeter.

2) When measured for the first time, the positive pole of the ammeter is connected to 3A, and the positive pole of the ammeter is connected to the second time, resulting in different deflection angles of the ammeter, but the current value measured twice is the same.

The factory motor works at 380 volts.

Whether it is a current or voltage measurement, it is necessary to zero before the measurement, so that the finger is aligned with the zero scale mark, and the value of the range and the index value (each small square) should be clarified.

When Xiao Yang used the ammeter to measure the current, he found that the pointer was deflected in the opposite direction of the scale, which was because the positive and negative binding posts of the ammeter were reversed; Another student noticed that the pointer was deflected a lot, exceeding the graduated part, because the range used was too small.

380v;

gears, scales;

The pen of the ammeter is reversed, and the gear used by the ammeter is small.

Charged bodies can attract small and light objects that are not charged.

1 When an object has more positive charge than negative charge, then the object is positively charged and vice versa3 The current is opposite to the flow direction of the electrons.

4 Yes, quarks, but quarks cannot exist on their own.

5. The movement of the positive charge produces a non-inverting current, and the movement of the negative charge produces a reverse current. So it's a mess if you don't do that (the forward movement of a positive charge is equivalent to the reverse movement of a negative charge, and you should know that in t3).

1 charge gathers together to produce positive electricity, 2 charges gather together to produce negative electricity, and 3 charges do irregular motion in the current.

4 is not 5, because the negative charge moves in the opposite direction, so the positive charge moves in the direction of the current.

The first question ABC can achieve the goal, because all she asks is that the circuit is switched on and on, and the purpose can be achieved.

The second question a requires that the flow from 5 to 6 is the flow direction of the current, and the current starts from the positive pole of the power supply, so it must first pass through 5 from the connected circuit + pole

Hope it helps you, dear, praise.

The first question is "You can't achieve your goal", right? The title is wrong! D should be chosen.

The second question is A, the current flows from the "+" binding post, and finally flows back from the "-"! So BC is not right. d is short-circuited.

If you don't understand, ask me again.