Junior high school circuit questions, a junior high school circuit multiple choice question

First, the voltmeter has an indication display. Because the resistance of the voltmeter is very large, it is equivalent to an open circuit, and it is necessary to connect the positive and negative poles of the electrical appliances in parallel. L2 emits light because the circuit is normal.

The second: the ammeter measures the current of the whole circuit, in this figure, the current of L1 and L2 is the same, the ammeter used to measure the current must be connected to the positive pole and the negative pole to the negative pole, otherwise, the pointer reversal will damage the ammeter.

On the left, L1 is short-circuited, and the voltmeter does not indicate the number, because it measures the voltage of the wire. L1 does not emit light when it is short-circuited, and L2 emits light normally, because L1 is short-circuited and equivalent to a piece of wire.

On the right, the series circuit is equal current divider, the current reading is the same, the current of Li and L2 and L1+L2 is the same, as long as you don't connect the meter to the right or minus.

Because the current of the bulb when it is working normally is i=p u=3w 3u=1a, and the bulb resistance is r=u i=3v 1a=3

Under the premise of ensuring the safe operation of the circuit, that is, the bulb can not be burned out, so the maximum current in the circuit can only reach 1A, according to P=I 2*r, when the current reaches 1A, the electrical power is the maximum.

The supply voltage is 6V, the current is 1A, and the resistance in the circuit is R=U I=6V 1A=6

The maximum power consumed by the circuit is p=i 2*r=(1a) 2*6 =6w, according to the relationship between the voltage distribution of the series circuit and the resistance.

When the whole circuit of the sliding rheostat is connected, the voltage at both ends reaches the maximum total resistance of 12, and the current is i=u r=6V 12 = The maximum voltage at both ends of the sliding rheostat is u=ir=

Solution: p amount = 3w; u = 3v

r=(u amount2) p=3 2 3=3 i amount=p amount=3 3=1a

i i amount.

i=u (r+r string)=6 (3+rslip) 1a rslip 3

P total = U2 (R+r slip) 6 2 (3+3) = 6W, i.e. P total max = 6W

U slip U slip (r+r slip) = 6*3 (3+3) = 3V, that is, U slip min=3V

U slip max=your slip total (r+r slip total)=6*9 (3+9)=, i.e. u slip max=

Calculate the resistance of the bulb first:

According to the formula: p= u 2 r can be obtained: 3 = 3*3 r solution: r = 3

Then the maximum power of the circuit is analyzed by the circuit power formula:

p total = u total 2 r total from here it can be known that the voltage is constant, the resistance is the smallest, and the total power of the circuit is the maximum.

But also to consider safety, the maximum withstand voltage of the small bulb is 3V, so according to the principle of voltage division of the series circuit, when the resistance value of the sliding rheostat is 3, the small bulb just reaches the maximum power, that is, the safety limit. At this time, the total resistance of the circuit is 6, so the maximum power is: 6*6 6=6W

It can be solved by the principle of voltage division of the series circuit, when the sliding rheostat reaches the maximum resistance, the voltage at both ends is the largest, and the value of the total resistance on the voltage magnitude and its own resistance ratio is multiplied by the total voltage: that is: 9 (9+3)*6 ==

i=p u-lamp=1a.The sliding rheostat is jacked up, that is, the maximum point current of the circuit is 1, pmax=1x6=6wIt's not complicated.

The correct statement is (d).

The analysis is described below:

1.Circuit analysis: when the switch is closed, the whole circuit is a parallel circuit, the bulb L is connected in parallel with the resistance R, and the ammeter A1 is located on the trunk road, that is, the total current of the measured circuit; The current A2 is located on one of the branches of the parallel circuit and is connected in series with the bulb L (i.e., the current of the measurement L); The voltmeter V is equivalent to being connected in parallel at both ends of the power supply, so the voltmeter measures the total voltage of the power supply.

2.As can be seen from the diagram, when the switch is closed and the slider p moves to the left, the resistance of the sliding rheostat to the circuit becomes smaller (i.e., rx becomes smaller). According to the formula for calculating the total resistance of parallel circuits: 1 r total = 1 rl + 1 rx analysis, the total r decreases.

1) Under the condition that the power supply voltage U is unchanged, the total current of the circuit i = u r total, i becomes larger, that is, the ammeter A1 indicator becomes larger. So, option c is wrong, exclude.

2) From the above analysis, it can be seen that the voltmeter V measures the total voltage of the power supply, so the voltage representation remains unchanged before and after the slide P slides. So, option b is wrong and excluded.

3) According to the above analysis, it can be seen that the bulb is alone on a branch of the parallel circuit, the power supply voltage is unchanged, that is, the voltage UL at both ends of the bulb does not change, and the resistance of the bulb does not change, according to the bulb power: PL=UL R, therefore, before and after the slider P slides, the actual power of the bulb does not change, therefore, the brightness of the bulb also does not change, therefore, option A is wrong, exclude.

4) From the above analysis, it can be seen that the total resistance r of the circuit always becomes smaller, while the power supply voltage does not change, according to: p total = u r total, it can be seen that p always becomes larger. So, option d is correct.

I hope it helps you, and if you have any questions, you can ask them

I wish you progress in your studies and go to the next level! (*

After the switch is full, it is a parallel circuit, L and R are connected in parallel, A1 measures the total current, A2 measures the current of L, and V measures the total voltage.

When the slider P moves to the left, a bulb L becomes brighter wrong, the voltage of L is equal to the power supply voltage, and the resistance does not change, so the power does not change, and the brightness does not change.

b Voltage indicates that the number becomes large and wrong. v measures the total voltage, so it does not change.

c The A1 indication of the ammeter becomes smaller, and the error is incorrect. The current of L does not change, but the current of R increases, so the total current becomes larger, and the A1 indicator becomes larger.

d The total power consumed by the circuit becomes larger Yes, the left slide, the resistance of r becomes smaller, and the total resistance becomes smaller, p u r, because u does not change, r becomes smaller, so p becomes larger.

I don't know what to ask about.

From the circuit structure, it is known that the sliding rheostat R and the lamp L are in parallel relationship, A1 measures the total current of the trunk circuit, A2 measures the current of the lamp L, and V measures the voltage of the lamp L, which is actually the total voltage.

When the slider p moves to the left, r becomes smaller, then.

A. The bulb L turns bright. False, the voltage of L is equal to the power supply voltage, its resistance does not change, the power does not change, and the brightness does not change.

b. The number of voltage representations becomes larger. Wrong. v measure the total voltage, is constant.

c. The A1 indication of the ammeter becomes smaller. Wrong. The current of L does not change, but the resistance of R becomes smaller, the current becomes larger, the total current becomes larger, and the A1 indicator becomes larger.

d. The total power consumed by the circuit becomes larger. Right. Just now, the c term judged that the total current becomes larger, and p becomes larger because p ui, u does not change, and i increases.

In the circuit, the rheostat and the bulb are parallel circuits, A1 measures the total current, A2 measures the current of L, when the slider P moves to the left, the rheostat resistance becomes smaller, the equivalent resistance of the circuit becomes smaller, then the current of the rheostat circuit increases, A1 becomes larger, the current of the bulb path remains unchanged, the bulb does not change, and the voltmeter measures the power supply voltage, unchanged.

Because p=u*u r, the resistance of the rheostat becomes smaller, the equivalent resistance also becomes smaller, and p increases, so D is selected

Pick D! The current passing current of lamp L becomes smaller and will not become brighter, so A is wrong voltmeter to measure the power supply voltage, which is unchanged, so B is wrong.

The total resistance becomes smaller, so A1 becomes larger, and C is wrong.

u does not change, the resistance of the whole circuit becomes smaller, the current becomes larger, so the small power consumption becomes larger, d is correct with the same student, please advise if there is a mistake.

The lamp (including the ammeter 2) is connected in parallel with the sliding rheostat, and the voltage at both ends is unchanged, so the current through the flow and so on does not change--- the brightness of the bulb l does not change; The voltage at both ends of the lamp is constant, and the voltmeter is the same. As the current passing through the sliding rheostat increases, the A1 meter actually becomes larger. The total power consumed by the circuit is equal to the product of the voltage u and the total current i at both ends of the parallel, u does not change, i increases, so the total power becomes larger.

Choose D. for this question

First, analyze the circuit composition: the parallel current between the lamp l and the sliding rheostat starts from the positive electrode and passes through the A1 meter to the bulb and the sliding rheostat are divided into a branch.

The voltmeter is connected in parallel in the two sections of the power supply, and now P is moved to the left, the resistance is smaller, and the current through it increases, so the number of A1 indicates that the total current of the circuit increases, and the power obtained by P = UI increases, and the bulb and the sliding rheostat are connected in parallel, and the voltage at both ends of the bulb remains unchanged, so the passing current is unchanged, so D is selected

Pick D! A1, A2 ammeters are considered short-circuited, V is considered open-circuit, and when P moves to the left, R becomes smaller.

1. The voltage indication number is the bulb voltage, and the resistor voltage is U'=u-ul=resistor resistor r=u'i=oh.

2. The actual total power of the whole circuit is p=ui=10v * total power is equal to the total voltage multiplied by the current.

When applying Ohm's law, it is required that the voltage U and current i should be for the same conductor R, and in the answer to question (1), Wang Gang mistakenly took the voltage at both ends of the bulb as the voltage of the resistor, (1 point), so the resistance value of the bulb was calculated instead of the resistance value of the resistor. (1 point).

2) In questioning, Wang Gang confused the rated power and the actual power, when the voltage at both ends of the bulb is, the power consumed by the bulb is less than the rated power. (1 point).

The correct answer should be:

1) The voltage at both ends of the resistor r U(R) = U - U (L) = min) According to Ohm's law, the resistance value of the resistor r r = U (R) i = min) 2) The actual power consumed by the whole circuit p = ui = 10 v * min).

Solution: When both S1 and S2 are disconnected, the circuit is shown in the following figure.

When both S1 and S2 are closed, the circuit is shown in the following figure.

When S1 is closed and S2 is disconnected, the circuit is shown in the following figure.

This can be seen from these three figures.

u2 u3 supply voltage, and u1 u2 1 2, so ur1 ur2 u2 u1 u1 u1 1 1 1 1 so r1 r2i1 u2 r1 r2 u2 2r2

The output power of the power supply in the figure is equal to u2 i1 u2 2r2 1 because i1 i2 1 6, so i2 6i1 so the output power of the power supply in the figure is equal to u2 i2 u2 u2 6i1 3u2 r2 two.

Since i3 u2 r2 in the figure so.

The output power of the power supply in the figure is equal to U3, I3, U2, R2, 3W, and the three are substituted into one and two respectively.

The output power of the power supply in the figure is equal to U2 I1 U2 2R2 The output power of the power supply in the figure is equal to U2 I2 U2 6I1 3U2 R2 9W

The output power of the power supply in the figure is equal to U3 I3 U2 R2 3W, so it can be seen that only when S1 is closed and S2 is closed, the power of the circuit is maximum, 9W.

That's what you want.

1、i1=u/(r1+r2);

2、u1=r2*u/(r1+r2);

3、u2=u;

4、i2=u/r2+u/r3;

Bring in i1:i2=1:6, u1:u2=1:2 to get the solution: r1=r2; r3=;

3w=u 2 r2;

So when S1, S2 are both closed, there is maximum power.

p=u^2/r2+u^2/r3=9w;

According to u1:u2=1:2, r1=r2 is stated

According to i1:i2=1:6, r2=2 r3 is obtained

1) When S1 and S2 are closed, R1 is short-circuited, R2 and R3 are connected in parallel, the voltage at both ends of the resistor is closed with the switch S1, and when S2 is disconnected, the power of R2 is 3W, and the power of R3 can be found to be 6W, a total of 6+3=9W

2) When S1 and S2 are disconnected, R1 and R2 are connected in series, and the voltage at both ends of R2 becomes half of the original, so the power of R1 and R2 is.

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9w, look upstairs, God-man too.

The ampere meter should be divided into 30 compartments.

Because now a resistor is connected in parallel between the two ends of this ammeter, and its resistance is equal to 1 4 of the resistance inside this ammeter

So the current passing through this whole is 5 times that of the original.

Choose b and you help your teacher say that the internal resistance of the ammeter is 1 4, not 4 times. . . Let him read the next question again.

If the resistance value of a resistor in parallel is equal to 1 4 of the internal resistance of the ammeter, the total resistance becomes the original 1 5

Therefore, B should be chosen

r=(r 1 4r) (r+1 4r)=(1 5)r The voltage is unchanged, and the current is inversely proportional to the resistance, so B is selected

r r meter = 1 4, so the current through the original ammeter is 1 5 of the total current, when the inner coil of the ammeter passes through the ampere current, the pointer is deflected one grid, then the current through the new meter circuit after parallel connection is, that is, the current in the circuit under test is, and not.