The second circuit problem of the junior high school, the third circuit problem of the junior high s

Solution: According to the problem, there is a permissible power supply with a total output power of 6 watts. And R1 = 2 ohms, the maximum sliding rheostat is 50 ohms, the maximum resistance of the main circuit is 50 10 60 ohms, the minimum current is 6 60 A, and the power consumption on R1 is watts.

That is, the main loop current range is, so the power consumption range on R1 is.

Analysis: U source 6 volts, i1 amps, u1 2 volts.

From "when the switch is closed and the sliding rheostat slide is moved to a certain position, the current is expressed as and the voltage is expressed as 2V", and the resistance value of resistor R1 is R1 U1 I1 2 ohms.

According to the sliding rheostat marked "50, 1A", the current in the circuit can not exceed I2 1A.

It can be seen that when the current in the circuit is i2 1 amps, R1 consumes the most power, and the maximum power of R1 at this time is.

P1 large i2 2 * r1 1 2 * 10 10 watts.

When the sliding piece of the rheostat moves to the far right end of the diagram, the resistance value of the rheostat into the circuit is the maximum, which is r2 50 ohms, then the current in the circuit is the smallest, and the minimum current is.

i3 u source (r1 r2) 6 (10 50) amps.

Then the minimum power consumed by R1 is P1 Small i3 2 * R1 *10 Watts.

That is, in order to ensure the safety of the circuit, the allowable power consumption range on the resistor R1 is in the closed range of watts to 10 watts.

Ohm's law gets: r1=u1i

2) The voltage on the sliding rheostat u2=u-u1=6v-2v=4v;

The power dissipated on the sliding rheostat is p2 = u2i = 4 v 3) When the voltage is 3 V, it is obtained by Ohm's law, so the current flowing through the resistor R1 is the largest i=U1. r13v

Therefore, the power consumption on R1 is the greatest at this time.

p1 max = u

1 maximum. r13v)2

When the sliding rheostat resistance is maximum, the power consumed on R1 is minimal.

Minimum current in the circuit i minimum = total your.

6v；Minimum power consumed on r1 p1 min = i1 min 2r1 = (;

A: The allowable power consumption range on R1 is:

When the R2 resistance value is maximum: (the current of the series circuit is equal everywhere) i=U R=12V (12 ohms + 18 ohms) = the minimum current amount in the circuit) i=U R=12V 12 ohms = 1a (the maximum current in the circuit) can be obtained u1=ir1=ohm=

U1 = IR1 = 1a 12 ohm = 12v can be obtained

So the voltmeter ranges (to 12V).

1。When r2 = volts.

2。When r2=18. r=12+18=。v1 = 12 * 12 30 = volts.

3。Range of variation: volts.

1。In a series circuit, if a bulb is short-circuited, how will the number of spot current representations change when measuring the voltmeter of this bulb?

In this case, the number of voltage representations is 0 and the number of current representations becomes larger.

2.In a series circuit, L1 and L2 are connected in series, if the L1 bulb is short-circuited, how will the number of point current representations change when measuring the voltmeter of L1? The voltage is expressed as 0, and the indication of the ammeter becomes larger, which is the current of L2.

How will the number of point stream representations change in the voltmeter measuring L2? The current indicates that the number increases, and the voltage indicates that the number increases.

How will the total voltage representation for L1 and L2 change? No change.

3.In a series circuit, if a bulb is disconnected, how will the number of spot current representations change when measuring the voltmeter of this bulb? The number of voltage indications becomes larger, close to the power supply voltage, and the number of current indications is 0.

4.In a series circuit, L1 and L2 are connected in series, if the L1 bulb is broken, how will the number of point current representations change when measuring the voltmeter of L1? How will the number of point stream representations change in the voltmeter measuring L2? Total voltmeter for measuring L1 and L2.

How will the indication change?

5. In the parallel circuit, L1 and L2 are connected in parallel, if the L1 bulb is broken, how will the number of point current representations change when measuring the voltmeter of L1? How will the number of point stream representations change in the voltmeter measuring L2? How will the total current representations of L1 and L2 change?

6.In the parallel circuit, L1 and L2 are connected in parallel, if the L1 bulb is short-circuited, how will the number of point current representations change when measuring the voltmeter of L1? How will the number of point stream representations change in the voltmeter measuring L2? How will the total current representations of L1 and L2 change?

In addition, can you give me an example of the circumstances under which the ammeter and voltage representation will become larger or smaller in a series-parallel circuit? (The series circuit includes the total voltage and the current and voltage of each consumer, and the parallel circuit includes the total current and the current and voltage of each consumer) and how will the ammeter and voltage representation change with the sliding resistance in the series-parallel circuit?

Say as much as you can, and give extra points for a good answer.

1. The voltmeter becomes zero and the ammeter becomes larger.

2. The L1 voltmeter becomes zero and the ammeter becomes larger. L2 voltage and current are increased. The total voltage is theoretically constant.

3. The voltage becomes larger and the current becomes zero.

4. The L1 voltage becomes larger and the current becomes zero. L2 becomes zero. The total voltage is theoretically constant.

5. The voltage of L1 is unchanged, and the current of zero L2 is unchanged. The total current becomes smaller.

6. The voltage of L1 is constant, the current becomes larger, and the total current of L2 is constant.

1. The voltage divider in series does not shunt, so when one of the bulbs is short-circuited, the voltage of this bulb is 0, and the current will not change.

2L1 voltage is 0, current does not change, L2 voltage increases, current does not change, L1L2 total voltage does not change.

3. The voltage becomes a large current of 0

4L1 voltage becomes maximum current 0 L2 voltage 0 current 0 total voltage remains unchanged.

5l1 constant current 0l2 voltage constant high current.

The total current is constant.

6L1 current changes to a voltage of 0 (due to the parallel short-circuit power supply is directly filtered by + through the short-circuit wire to the pole L2), the voltage does not change.

7. According to the principle of series divider, shunt distribution, and shunt distribution, you can get the answer you want.

1.The number of voltage representations becomes 0, and the number of current representations becomes larger.

1 with 1ll 2 The voltage becomes larger, and the current also increases. The total voltage does not change.

3.The voltage becomes larger than the supply voltage, .

The current is 04.The ammeter is all 0The voltmeter is all about the supply voltage.

5.The voltmeter does not change, the current l 1 becomes 0l 2 unchanged, and the total current becomes 6It's dangerous, l 1 ammeter burns out, and the other meters are all 0 before the circuit burns out, good luck.

The equivalent circuit is shown in the figure, L2 is in the dry circuit, L1 and L2 are in series, and you close the switch just to short circuit L1. You can remove the wires first, look at the circuit diagram, L1 and L2 are connected in series, and then connect the wires to both ends of L1.

The series and parallel connection of the components of this type of circuit is not from the diagram you draw, but from both ends of the power supply;

You first draw the power supply on the left, the components on the right, and then look from the left to the right, do you change the series and parallel connection of the circuit?

L2 is in front of the node and is the electrical appliance of the trunk circuit, and the switch and L1 are on the two branches, so L1 will be short-circuited, but L2 has no effect. As for whether the current at both ends of the bulb is equal, which one are you talking about?

When S is disconnected, L1 and L2 are connected in series.

This is not parallel, L1 and L2 are not simply parallel, I want L1 to be connected in parallel with the switch and then connected in series with L2. Do you want to know which bulb is the same and under what circumstances the current is different?

Short circuit is divided into partial short circuit and individual electrical short circuit, you know the closing switch short circuit but it only shorts L1, who shorts and who does not work then only refers to L2 on.

This is a simple relay and thermistor application circuit. The whole circuit is divided into two parts: the magnetic circuit part (DC 12V power supply part) and the contact part (AC 220V power supply part).

1: When the bath bomb circuit is closed, the current flowing through each bulb is 250 220 A, so the total current is equal to the sum of the currents flowing through the four bulbs, i.e. 250 x 4 220 = A.

2: When the current in the relay coil is 50mA, the resistance R1=(12V 50mA)-R2=220 ohms; The temperature at this time was about 30 degrees Celsius;

When the current in the relay coil is 40mA, the resistance R1 = (12V 40mA)-R2 = 280 ohms; The temperature at this time was about 20 degrees Celsius;

In summary, the room temperature can be controlled between 20 and 30 degrees Celsius.

Hope mine is useful to you.

1): When the bath bomb circuit is closed, the total power of the bath bomb p=4*250=1000 WP=u*i = 220*i i= a

2): When the current in the relay i1=50mA, i.e. i1=u1 r1+r2) 20+r1) r1=220

When the current in the relay i2=40mA, i.e. i2=u1 r1+r2) 20+r1) r1=280

Check the above table: R1 = 220, T = 30 degrees, R1 = 280, T = 23 degrees, so the bathroom temperature can be controlled in the range of 23-30 degrees.

Connect the physical circuit according to the circuit diagram: first remove the voltmeter, start from the positive pole of the power supply, connect it in turn, when there is a branch, the binding post connects two wires, and finally returns to the negative pole. Then connect the voltmeter to both ends of the DUT.

Pay attention to the range of the ammeter and voltmeter and the positive and negative binding posts, because we are connected according to the current flow direction, so the positive binding posts flow in and the negative binding posts flow out.

Draw the circuit diagram according to the physical drawing: it also starts from the positive pole of the power supply, when there is a binding post connected to two wires, it means that there is a parallel connection, that is, a branch, and it is the key to find the branch point and the convergence point.

Series circuit: Features: sequential connection, no branching. It can follow the direction of the current, and the switch that passes through is the switch that needs to be closed.

Parallel circuits: Feature: Bifurcation of current.

This kind of thing cannot be said in general, it is easy to analyze with examples, and it is faster to understand through examples.

It's too general, the closing switch depends on how many circuits there are after closing, one loop is in series, and two or more are in parallel. The best way is to start from the positive pole of the power supply, let the current flow, draw an arrow to indicate the direction of the current, and there will be a current bifurcation and convergence port in parallel, and the parallel connection is originally to play the role of shunt. If you really can't understand the circuit and can't figure out the relationship, it is recommended to draw an equivalent circuit diagram.

The resistance of the conductor is related to the temperature, the higher the temperature, the greater the resistance, so, after heating in Figure A, the resistance of the wire increases, the total current decreases, and the resistance of the bulb does not change, so the electrical power of the bulb decreases, so it becomes darker.

Figure B heats the bulb, and the resistance of the bulb increases, so the power distributed to the bulb increases when the output power of the power supply remains the same, so it becomes brighter. Light and dark are related to electrical power.

The power distributed on the bulb p=r (r+r)*p0, r represents the resistance of the bulb, r represents the resistance on the wire, p0 represents the output power of the power supply, when r increases, p decreases, there is the first problem, when r increases, p increases, there is the second problem.

Heating the metal causes an increase in conductor resistance.

Heating the air ionizes and conducts electricity!

The junior high school questions are all full, what's the use of counting this thing? It's the same as what Zhao Benshan said about the problem of releasing water from a pool and draining a pool.

I don't think there's a solution to this problem at all. Assuming that the sliding rheostat is 0 ohms at the minimum, then the voltmeter should be 0 volts, that is, 0° deflection. Do you think about how much resistance the string needs to make a voltage of 18V out of 0A current? It has to be big enough, or infinity.

When the sliding rheostat is at its maximum, the voltmeter is about 15 volts, which is a deflection of 100°. Do you think about how much resistance the string can make a voltage of 3V out of 3A current? It takes 1 euro. Fulfill both of these conditions?

In other words, if the sliding rheostat moves towards a small time, the total resistance will decrease, so the total current will inevitably increase, but what will happen to the voltage drop across the sliding rheostat? will get smaller ... Isn't that a paradox? Is the deflection angle the same?

Set the meter pointer to turn x, the fixed resistance value is r, and the sliding rheostat value is r

U=18V,I=U (R+R) V=I*R,V=15x I=3X 15x=3X*R,,, then R=5

The voltmeter is connected in parallel at both ends of the fixed value resistor

The setpoint resistor r0, the sliding rheostat r

At this time, the current size of the loop is: i=18 (r+r0) and the voltage at both ends of the sliding rheostat: u=18*r (r+r0) because the deflection angle of the voltmeter and the ammeter is the same, then.

15：u=3：i

r=5

Is there a bit of a problem with the title, should the voltmeter be connected to both ends of the fixed value resistor?