Physical circuit problems in junior 3, physical circuit problems in junior 3

1. The total resistance becomes larger. The total voltage does not change. The current becomes smaller. i=u r2 total resistance becomes larger. The total voltage does not change. But the voltage of each bulb in series becomes smaller (the voltage is at both ends of the bulb. It really can't be said that it passed).

3. In the pure parallel circuit, the branch voltage = the dry circuit voltage. (The branch is connected in series, and the voltage of each bulb in the series becomes smaller).

That's a hot question. Pretty much ... Wow, it changed to ,,, depressed in the end.

1.When the two lamps are connected in series, the original voltage in the circuit remains the same and the resistance becomes larger, so the current becomes smaller.

2.It is not the voltage that passes through the lamp, the value of the voltage on both sides of the lamp must be unequal, which is the direct cause of the current.

3.The voltages of the parallel branches are all equal.

1. The total resistance becomes larger. The total voltage does not change. The current becomes smaller. i=u/r

2. No, the voltage passes through the lamp, and the value of the voltage on both sides of the lamp must be unequal, which is the direct cause of the current, 3. No, the voltage of the parallel branch is equal everywhere.

1. According to Ohm's law, r=u i, so, i=u r, that is, the voltage is certain, the greater the resistance, the smaller the current, and then according to the series circuit, the total resistance is equal to the sum of the resistances of each part, so the sum of the resistance of the two bulbs must be greater than the resistance of one bulb, so the current when connecting two lamps in series is less than the current when only one lamp is connected.

2. There is a doubt about this problem, the voltage cannot be said to "pass", it should be the current, because the voltage must be at both ends at the same time, otherwise the voltage and current cannot be formed.

3. No, because according to the voltage law of the parallel circuit, in the parallel circuit, the voltage at both ends of each branch is equal, which is equal to the total voltage, so it will not.

1. Two lamps are equivalent to two resistors. When connected in series, the total resistance of the circuit is greater than that of only one bulb, and the greater the resistance, the smaller the current when the voltage is constant.

2.voltage, voltage can only be formed at both ends of the resistor. Voltage is not flowing, and current is a different thing.

For example, there is no voltage at both ends of a wire with no resistance. So there is voltage at both ends of the bulb instead of the voltage going through the bulb.

Voltage, which can be understood as the difference between the height of electricity.

3.Id

1. The lamp has resistance, two lamps in series, the resistance becomes larger, the voltage is constant, and the current becomes smaller.

The others were a little bit better than me.

The problem with his solution is that if two small light bulbs are connected in parallel, using a wire to connect one light bulb in parallel will cause a short circuit in the power supply, which is a situation that should be avoided, so there is a problem.

The correct way is to remove one bulb first, close the switch, if the other bulb is glowing, then the two small bulbs are connected in parallel, otherwise it is in series.

Using the equation of equality of the supply voltage, u=u r is obtained from i=u r u=ir(rx+r0)i=i 2(r+r0).

The solution is rx=(r-r0) 2

This kind of circuit problem is used in this way, I hope it will help you.

Pure originality, satisfied?

S is connected to A or B, the power supply voltage remains unchanged, when S is connected to A. The power supply voltage is equal to i times (r0 plus resistance box resistance), when s is connected to b, the power supply voltage is equal to i 2 times (rx plus r0), because the supply voltage before and after is unchanged, so rx = i times (r0 plus resistance box resistance) divided by i 2 and then subtract r0

According to the formula U=IR

When switch S is dialed to A, the voltage at both ends of the power supply U=(Rx+R0)*i, when the switch S is dialed to B, the voltage at both ends of the power supply U=(R+R0)*i 2, because the power supply voltage is unchanged, the above two formulas are connected to obtain (Rx+R0)*I=(R+R0)*i 2, and rx=(R-R0) 2 can be solved

Since the current of the latter is half of the original, the resistance is twice that of the beginning, so (r0+r)=2*(rx+r0).

rx=r/2-r0

Solution: According to the topic.

When the switch is connected to A, Rx is connected in series with RO, and the current is expressed as I1 according to Ohm's law i=u r.

i1=u/rx+ro

When the switch is connected to B, R is connected in series with Ro, and the current is expressed as i2 according to Ohm's law i=u r.

i2=u/r+ro

According to the title, there is. i2=1/2*i1

Substitute the above two formulas to get it.

u/r+ro=1/2*u/rx+ro

Simplify. rx=1/2(r-ro)

Hope it helps.

If the internal resistance of the power supply and ammeter is not counted, the electromotive force e=i 2(r0+r).

e/i=r0+rx

rx can be found

Set voltage U: U r0=1a; u/(r1+r0)=

The solution of the two formulas is r0=12

The current change can only be reduced to.

Simultaneous u r0=1a

u (r1+r0)= solution r0=12

This diagram is very simple, a power supply and a resistor are connected in series.

1) The power supply with a voltage of 220 volts supplies power to a lamp marked pz-220-60 at a construction site in the distance, and the actual power consumed by the bulb is 55 watts due to the resistance of the wire. Then the power consumed by the wire (less than) 5 watts (fill greater than less than equal to cannot be determined).

Because: when the bulb resistance is r and the terminal voltage is 220, the power is 60;

When the line resistance is added, the total resistance is greater than r, the terminal voltage is unchanged at 220, and the total power is less than 60, so it is less than 5W

2) A lamp is connected to a constant-voltage power supply. The power is 100W. If the lamp is connected to a long wire and then connected to the same power supply, and the power loss on the wire is known to be 9W, then the actual power consumption of the lamp is less than 91W.

A 91w c is greater than 91w d and the condition is insufficient to be judged.

Same method as above. 3) Lamps L1 L2 are connected in series. Respectively marked with 6v 6w. 6V 3W, when the switch S is closed, the lamp L1 L2 can emit light, then the indication ratio of the two voltmeters A and B is (A voltmeter measuring lamp L2 B voltmeter measuring total circuit) C 2:

3a 1：1 b1：2 c 2：3 d 3：2

Obviously, simple wordplay yes.

(1) Less than (2) Less than ninety-one watts (3) c

1. R=220V*220V 60W=807 ,P=55W,P=I*I*R=U*U R, for lamp I=,U=211V; For wires, p=u*i=(220v-211v)*<5w.

2. P=U*U R, the original connection total P=lamp P=100W, after connecting the wire, R leads to the total P, now the total P<100W, the lamp P=Total P-line P<100W-9W=91W, so the lamp P<91W.

3、p=u*u/r，∴r1=6ω，r2=12ω；u=i*r,i1=i2=total i, u1 u2=r2 (r1+r2)=12:18=2:3

1, V=IR, set the lamp resistance to R1, if the wire has no resistance, the bulb works at the rated voltage, then R1 I1=V, if the wire resistance is R2, then (R1+R2) I2=V, compare the two formulas, get i2=I1 R1 (R1+R2), power P=i i R, P1=I1 I1 R1, P2=I2 I2 (R1+R2)=I1 I1 R1 (R1+R2), it can be seen that the total power has become smaller, so it is less than 5 watts.

2. The reason is the same as the first question, choose B

3. P=V V R, R1=V V P1=6, R2=V V P2=12, then after series connection, the voltage ratio V1:V2=R2:(R1+R2)=2:3, select C

The bulb L1 happens to be glowing normally, and the pointers of the ammeter A1 and A2 are both pointing at A2 A1

The range of A2 is 0, and the range of A1 is 0 3A

The number of A2 is 1A, and the number of A1 is 1A

u1=p1/

i1 = v1 is 2v

v1 indicator = v2 signal.

V2 is shown as 2V (because it is a parallel circuit).

When only S1 is closed, the R3 slide moves to the far left (R=0), and the R2 slide moves to the bottom (R maximum), the circuit R1 is connected in series with R2. p2=po=8w

When the R3 slide moves to the far right (R is maximum), the R2 slide moves to the top (R=0), R1 is connected in series with R2, the voltmeter V1 measures the voltage of R1 U1, and the voltmeter 2 measures the voltage of R3 U3=:U2=1:4.

According to the characteristics of the series circuit, r1:r3=1:4.

When only S1 and S3 are closed, only R2 (maximum) is present in the circuit. R1 and R3 are short-circuited (purple dots A and B are connected to the positive terminal of the power supply, and there is no voltage at both ends of R1 and R3, so no current passes through R1 and R3). The power consumed by r2 is p2*=p1.

When only S1 and S2 are closed, only R1, R2 and R3 are shorted in the circuit (the purple dots F and E in the diagram are connected to the negative pole of the power supply, and there is no voltage at both ends of R2 and R3, so no current passes through). The power consumed by r1 is p1*=p2.

p1：p2=1：2。

Analysis: From the can be known:

p1=u²/r2 （1）

From , it can be seen that p2=u r1 (2).

p1：p2=1：2

Substituting (1) and (2) into the above equation yields:

u²/r2：u²/r1=1：2

1/r2：1/r1=1：2

2/r2=1/r1

r2=2r1

A false. R1 is connected in series with R2 and the current is equal i1=i2

When R2=2R1 and P2=8W, P1=8 2=4W can be seen according to P=i R, and the total power P=P1+P2=4+8=False.

When only S1 is closed, R3=0, and R2 changes from the maximum to half, the voltmeter V1 measures the voltage U1 at both ends of R1. The currents are equal i=i1=i2

When r2 is maximum, r2=2r1.

u1=i1r1=[u/(r1+r2)]xr1

u/(r1+2r1)]xr1

u/3。When p of r is at the midpoint (half), r2*=r1

u1*=i*r1=[u/(r1+r2*)]xr1

u/(r1+r1)]xr1

u/2。u1：u1*=u/3：u/2=1/3：1/2

u1：u1*=2：3

Answer C is correct.

1. Because it is a parallel circuit.

2. It can be seen from the location.

3 I don't know.

a, the total resistance of the parallel connection becomes larger, the circuit resistance becomes larger, the trunk current becomes smaller, and the variable resistance power becomes smaller.

b. The rated power is fixed and will not change.

c。The trunk current becomes smaller, and L2 is also connected in series in the trunk circuit, so the brightness becomes dim.

A Yes, because the parallel resistance becomes smaller and smaller, when L1 is broken, the left resistance becomes larger, the sliding rheostat voltage becomes smaller, and because p=u r, the power decreases.

L1 burns out, that is, the open circuit, so R always increases, i decreases, p=i 2*r, so the sliding rheostat p decreases.

A is right. r total = r rheostat + (r1 + r2) (r1*r2).

i total 1 = u [r rheostat + (r1 + r2) (r1 * r2)].

When r1=0, i total 2=u (r rheostat + r2); Since R2>(R1+R2) (R1*R2), I always has 2P rheostat 2

That's right. Because the total resistance becomes larger when disconnected, the total current decreases. According to the formula p=i multiplied by the square of r. The resistance does not change, so p decreases.