There are two junior high school circuits, and one junior high school circuit problem

1.The resistor R1 is connected in parallel with R2 in the circuit, R1=6, R2=4The total resistance of their parallel connection is that when energized, the ratio of the voltage between the two ends of R1R2 u1:u2=1:1, and the current of R1R2 is only greater than i1:i2=2:3

2.The resistance R1 R2 is connected in series in the circuit, the voltage at both ends of R1 is 3V, the voltage at both ends of R2 is 9V, R1+R2=600, find:

1. The total voltage of the circuit is U=3V+9V=12V

2 The current in the circuit i=U (R1+R2)=12V 600 =3The resistance of the resistors R1 and R2 R1=U1 I=3V, R2=U2 I=6V,

2.Total circuit voltage = U1 + U2 = 12V

Current in the circuit = u r = 12v 600 = 20mar1:r2 = u1:u2 = 1:3

r1=1/4x600=150ω

r2=600-150=450ω

2.The total voltage in the series circuit is equal to the voltage at both ends of each appliance and 3+9=12, which is directly derived from Ohm's law 12 600=

From u=IR, it can be seen that the voltage ratio is equal to the resistance ratio when the current is equal.

So R1 150

r2 300

Rice cooker, m. Expression:Procedure: p w t (m n).

Meanwhile: p ui, i p u p 220v (m n).

The ideal voltmeter has an infinite resistance, so ignore the voltmeter above.

p shifts to the right, the resistance of the circuit increases, i=u r, the current decreases, and the sum of the resistance of the ammeter and r1 does not change, set to r',u'=i*r', hence u'Minish. And the voltage of R2 U2=U-U', u does not change, so u2 increases.

Bigger, P divides the sliding rheostat into two parts, after the power is turned on, the current is like this through the ammeter and then to the resistance R1, because you are a junior high school, the voltmeter is regarded as an open circuit, the wire resistance is 0, the current from P to the left part of P is back to the negative pole, in fact, there is no current on the right part of R2, and the left part of the sliding left part becomes longer and the resistance becomes larger, and the voltage is also larger.

You can think of the wire to point P and the right part of the resistance of R2 in parallel, the wire has no resistance, and the current is all shared, and then flows back to the negative terminal from the left half.

Such a question, first of all, we must analyze the circuit, to see whether it is in series or in parallel, and then look at the voltage and current measured by the voltmeter and the ammeter, the voltmeter is regarded as an open circuit, and the ammeter is regarded as a wire, so that the analysis is easy to analyze, the sliding rheostat R2 and R1 are connected in series, the voltmeter is connected in parallel at both ends of R1, the voltage at both ends of R1 is measured, and the ammeter measures the current of the entire circuit, and then we analyze the sliding rheostat to move to the right, and the resistance is increased, so the resistance of the whole circuit increases. According to Ohm's law, i = u than r, the power supply voltage does not change, the resistance increases and the current decreases, so the current representation decreases, according to Ohm's law, U1 = I1R1, so the voltage at both ends of U1 also decreases, and the total voltage of the series circuit is equal to the sum of the voltages at both ends of each appliance, U=U1 + U2, the total voltage remains the same, U1 decreases, so U2 increases.

I don't know if you understand this, there is a fast way, the series circuit is divided, the resistance of the sliding rheostat increases, and the voltage it divides increases, which is also a law, relatively fast!

Smaller. Because the slide is moved to the right, the resistance of the entire circuit becomes smaller, and the current does not change, so the voltage becomes smaller.

There are two methods, depending on which one is convenient for you.

1) When both the ammeter and the voltmeter are ideal, P slides to the right, causing the resistance of R2 to become larger, and the expression of the voltage on R2 is U=U total * R2 (R1 + R2) = U total * 1 [(R1 R2)+1], when R2 becomes larger, the voltage at both ends of R2 also increases.

2) When both the ammeter and the voltmeter are ideal, the sliding rheostat divides the voltage, so when R2 becomes larger, the voltage obtained by R2 also increases, that is, the voltage at both ends of R2 becomes larger.

When it gets bigger, the sliding rheostat slides to the right, and the resistance value becomes larger, so the circuit current becomes smaller, the resistance of R1 does not change, so the voltage division of R1 becomes smaller, and the power supply voltage does not change, so the sliding rheostat voltage becomes larger.

Larger, in the series circuit, the voltage is proportional to the resistance, that is, r1:r2=u1:u2 ignores the internal resistance of the power supply, so move the dicing to the right, the resistance of the sliding rheostat becomes larger, and the voltage at both ends becomes larger.

Big. R2 moves to the right, the total resistance in the circuit becomes larger, the total current becomes smaller, the voltage applied to both ends of R1 becomes smaller, the total voltage remains the same, and the voltage at both ends of R2 becomes smaller.

First of all, you need to know what a short circuit is!

Short circuit: The power supply flows directly from the positive electrode to the negative electrode without passing through the electrical appliance!

Therefore, there will be no short circuit in the circuit in the diagram.

After the switch is closed, the power supply goes from the positive pole to the switch, then to L2, and then to L1 back to the negative pole, then a closed loop is formed, and both L2 and L1 emit light. L2 and L1 are connected in series in the loop, and if one of them is extinguished, the other is also extinguished. Because when one of the bulbs goes out, the circuit will be disconnected and the circuit will not be formed.

A voltmeter connection in a circuit can be "considered" as an open circuit (but it is not actually an open circuit, but only because of its large resistance and very, very small current).

When the switch is closed, most of the current flows through the switch to the negative end of the power supply of the bulb L2 and the bulb L1.

When the switch is disconnected, the data measured by the voltmeter is equivalent to the supply voltage.

First of all, it is important to be clear: which element the voltmeter measures the voltage.

The diagram shows that the voltage measures the voltage of the bulb L2, as the title suggests, since the voltage representation number is 6V, it means that the voltage at both ends of L2 is 6V, and L2 emits light. If it is a L2 short circuit, then the voltage representation should be 0V.

If L1 is not short-circuited and L2 is short-circuited, the voltage representation is not 6, but smaller than 6.

Because the voltmeter is equivalent to a resistor, the current goes to L2. I don't know if that's the case, but I've been away from learning circuits for more than a decade.