Electricity problems in the third year of junior high school, help solve !!

The first question should be that there is voltage at both ends of the voltmeter, and the internal resistance of the voltmeter is infinite, according to the law of voltage division of the series circuit, the voltage at both ends should be the power supply voltage when L is broken, and the resistance is infinite, so the ammeter does not find the number, that is, the current is 0; Let's take a look at other answers, L short circuit, then the voltmeter measures the voltage at both ends of a wire, and the wire resistance is 0, the voltage should be 0, and the current due to the current limit of the resistance R will not be very large But there will be a reading, which does not agree with the topic, if R is short-circuited, the current will also have an indicator, and at this time L should also have voltage at both ends, not in line with the topic, L is broken, then the voltmeter and the ammeter do not have an indicator. Fixed selection b

The second question should be L1 short circuit, only in this case the voltage measured by the two voltmeters will be the same when only one light is on, L1, L2 any lamp is broken, no light is on, L2 short circuit, then the voltmeter V2 is 0, V1 has a reading, so choose a L1 short circuit.

l Circuit opening is correct, it can be understood that the filament of the small bulb is broken, it will not light up, and the ammeter has not passed through the current, so the indication is 0 After the switch is closed, isn't the voltmeter equivalent to being connected to the positive and negative poles of the power supply? It's just an extra rheostat.

The three points of DCE are located at half of the voltage drop between AB, that is, the three points of DCE are equivalent, and there is no voltage between DC and CE, that is, there is no current, so these two lines can be removed when calculating the resistance, and what remains is a simple series and parallel connection. The final result is 1 (1 2r + 1 2r + 1 2r) = (2 3)r

Let the power supply voltage be U, the bulb resistance r1, and the line electrical slow balancing cover which beam r2 because the ratio of the bulb power before and after = 25:16

So the ratio of the current passing through the bulb = 5:4

Because the power supply voltage does not change, R1:(R1+R2)=4:5, so R1:R2=4:1

The ratio of the power of the series circuit = the ratio of the resistance to the material, so the line resistance is 4W

It must be d, 25-16 is how much!

4 * 100-4 * 20 = 320w = , saving electricity every day, electricity cost yuan, saving 48 degrees of electricity in 30 days, electricity cost 24 yuan.

Power saving: 4x (100-20).

Save electricity bills: yuan.

The watchtower lord adopts, thank you.

Save 48 degrees of electricity and save 24 yuan in electricity bills.

Solution: Connect the resistor R1 to a circuit with constant voltage, and the current through R1 is 2A and the total voltage of the power supply is U

then u=ir1=2r1

After connecting the resistors R1 and R2 in parallel, the current in the dry circuit is I1I1=U R=U(R1+R2) (R1*R2)=2(R1+R2) R2

If R1 and R2 are connected to the circuit in series, the voltage at both ends of the resistor R2 is 8V, and the current through the circuit is I2

i2=8 r2, i2=u (r1+r2)=2r1 (r1+r2) The two equations are equal to 4(r1+r2)=r1*r2 (1)i1:i2=9:2

i.e. [2(r1+r2) r2] [8 r2]=9 2 gives r1+r2=18 (2).

Solve the equation (1) and (2), and r1 gives r1=6 ohms and r2=12 ohms.

The basic idea: set the power supply voltage to U R1 R2

Establish the equation: u=2r1 r1r2 r1+r2 :r1+r2=9:2 r1:r2=(u-8): 8

From the above three equations and (r1, let i1:i2=9k:2k

According to r1:r2=r1:r2.

9k-2):2=(u-8):8 The solution is u=36k, so u 2k=18 u 9k=4

i.e. r1r2 r1+r2=4 r1+r2=18 gives r1=6 r2=12

Let the power supply voltage u, then u=2*r1 .1）

u=r1r2/(r1+r2)*i1 ..2) and U (r1+r2)*r2=i2r2,..3)u/(r1+r2)*r2=8...4)

i1:i2=9:2

Thus the solution r2=1, r1=2 or r2=2, r1=1, and the problem requires r1< r2, so r1=1 ohm (the calculation process is to divide equation (3) by equation (4), and then find the ratio of ri and r2 according to the known problem, and solve the answer in various formulas in the back generation) < p >

1 Solution When the normal luminescence, the voltage at both ends is u=220v, the current is i=p u=40 220=, and the resistance is r=u, squared, p=(220*220), 40=1210 ohms.

Abnormal light emitting under 110V voltage, and the current flowing I=U2 R=110 1210=

Actual power p=u2 square r=10w

2 Solution Use r=u squared p to find the resistance of two bulbs r1=3 ohms and r2=ohms.

So the total current is I=U (R1+R2)=6 (3+) when connected in series on a 6V power supply

The actual power of bulb 1 p1 = i square * r1 = 16 3w the actual power of bulb 2 p2 = i square * r2 = 8 3

If it is connected in parallel, the rated voltage is reached on the 3V power supply, so it emits light normally.

3 Solution The rated current of the bulb is i=p u=3 6=

To work properly, the resistors in series should share the voltage of 8-6=2V.

So the resistor to be connected in series r=u1 i=2 ohms.

4 Solution: Use i=p u to calculate the rated current of the two bulbs are i1=2a, i2=3a because it is connected in series, so in order to make the lamp not damage, the maximum allowable current in the circuit is 2a, and the allowable voltage u=i1*(r1+r2)=2*(64 16 +144 36)=16v

5 Solution Because the power is the same, it is the same bright when working normally, and R1 = 12 ohms and R2 = 3 ohms are calculated by using r=u square p, and the current is equal when connected in the same circuit, and the power p with large resistance is obtained according to p = i square * r, so r1 is bright.

The electrical energy consumed by the bulb is converted into electrical energy and light energy, and the electric energy consumed by the electric soldering iron is all converted into heat energy, because they consume the same electric power, so the electric soldering iron produces more heat at the same time.

When the filament is attached, the length becomes shorter, the resistance becomes smaller, and the power p becomes larger according to p=u squared r, so it is brighter.

1。When"220V 40W" bulb glows normally, then the voltage at both ends of the bulb is 220V, the current is 40 220, and the electric power is 40W

If it is connected to 110V, does the bulb glow normally and how much current flows? What is the actual power?

It can't emit light normally, and the bulb resistance = 220 * 220 40 = 1210 ohms.

So at 110V, the current = 110 1210 = 1 11A

Power = 110 * 1 11 = 10W

2。"3V 3W" and 3V 6W"Two bulbs are connected in series on a 6V power supply, find the total current, the actual power and voltage of the two bulbs? If connected in parallel on a 3V power supply, will the two bulbs glow normally?

Resistance of 3W = 3*3 3 = 3 ohms.

6W Resistance = 3 * 3 6 = Ohms.

So the total current = 6 (3+.)

3W voltage = 3*4 3=4V, another = 6-4=2V

Normal glow in parallel.

3。A "6V 3W" bulb is connected to an 8V power supply, in order to ensure normal operation, how much resistance should be connected in series or parallel?

Bulb current = 3 6 =

Resistance divider = 8-6 = 2V

Resistance = 2 ohms.

4。The existing lamp L1 "8V 16W" and lamp L2 "12V 36W" are connected in series in the circuit, in order to keep the lamp from damage, what is the maximum allowable current in the circuit? What is the maximum allowable voltage?

L1 rated current = 16 8 = 2A

L2 rated current = 36 12 = 3A

So the maximum current is 2A

At this time, the normal working voltage of L1 is 8V, and the L2 voltage = 12*12 36*2=8V, so the maximum voltage is 16V

5。"6v 3w"and "3v 3w.""Which one is bright to work properly? Which one is lit in series on the same circuit?

220V 660W "Electric Heating Iron and"220V 660W" electric lamp, which produces more heat?

The electric light at home is broken, the filament has to be connected, and the electric light is brighter than before, why?

The resistance of 6V is larger, so the partial voltage is larger, so the power is greater, so it is brighter.

After connecting the filament, the filament becomes shorter, and the resistance will become smaller, so the power will become larger, so it will be brighter.

1) The resistance of the sensor is: r transmission = u2 p = (3v) 2 0 9w = 10 The rated current of the sensor is: i transmission = p transmission u transmission = 0 9w 3v = 0 3a2) requires the safety of each part of the circuit, then the maximum current of the circuit is required:

i=iTransmission=0 3AAt this time, the maximum value of power supply voltage: um=u+u0

U is the rated voltage of the sensor, and U0 is R0 to the maximum value.

, the voltage at both ends, i.e., the maximum supply voltage:

3) When the actual detection is set, the voltage added between A and B is U, and the actual resistance of the sensor is According to the first experimental record data: U

According to the second experiment, the data were: u

It is solved by , and the two formulas:

r=u= so you can use anything.