Electricity in junior high school science, and electricity in junior high school. Ask for an explana

First of all, you need to clarify the conditions under which Ohm's law applies: it must be a pure resistance to apply.

If it is not a pure resistive circuit u" IR

Then according to the topic: when the voltage applied to the resistor is volt, the current is ampere, and the motor does not rotate.

If the motor does not rotate, it means that all the electrical work at this time is used to generate heat.

Then the conditions for the application of Ohm's law are true.

So according to the formula p=ui=u2 r, then r=u2r=oh.

Then the title says: When the voltage applied to the resistor is 2 volts, the current is ampere.

At this time, the motor rotates normally, then the motor is not a pure resistance circuit at this time.

We can only calculate the sum of electrical power based on p=ui.

So p=ui=2*

Then the heating power can only be calculated by p (heat) = i r.

So p (heat) = i 2r=

So the output power at this time is p (out) = p total - p heat = so the efficiency of the vacuum cleaner n = p out p total =

So the efficiency of the vacuum cleaner is 60%.

r = Euro p (total) ui 2* p (heat) = i 2r = p (vacuum cleaner) = p (total) - p (heat) =

1。The voltage of the parallel circuit is equal to that of the branches, and the rated voltage of the LED is to make it work normally, it must be connected in parallel, if it is connected in series, 10 lamps divide the voltage, and the voltage ratio at both ends of each lamp is too small.

2。The total voltage of the two dry batteries is 3V, if they are connected in parallel, the voltage at both ends of each lamp is greater than the rated voltage, according to P=U2 R, the resistance is unchanged, the voltage becomes larger, the actual power becomes larger, and the total power becomes larger.

The voltage of the two sections of each electrical appliance in parallel is equal.

If the new battery is 3V, if the voltage is large, the power consumption will be large.

Why ...... in parallelAccording to the diagram, it should mean that the current flows through a bulb to the negative pole, and then each bulb flows through.

If one is broken in series, it will be all broken, but not in parallel; High power should be consumed more by wires and other components.

Because the dry battery is each cell, two cells, so the LEDs are connected in parallel.

This is because when there is still current in the circuit passing through the wires and rheostats, it is converted into heat loss.

Answer: 2160j

Analysis: According to the circuit diagram, when the switch is AB, turn off the electric blanket; When connected to BC, only R2 is connected to the circuit, which is a low temperature gear; Connect CD, two resistors in parallel, high temperature.

Therefore, at low temperatures, only R2 is connected to the circuit, and the power:

P low u r2 12 12 4 36w

According to: p=w t

Heat: Q W PT 36W 60S 2160J Pay attention to the working process of this circuit, the switch can only be connected between AB, BC, CD, these three connection methods, none of which is R1 connected to the circuit alone, so it is not that when the resistance is the largest, it is the lowest power, and the situation with the largest resistance cannot be realized.

Give a hint and please figure it out yourself.

—80W is just a reference number.

2. According to P=V square R, when the switch S is only connected to R1, it is low temperature, and the minimum power consumption is 24W, 3. Heat = P*T (t-seconds) ......

3. When the electric table mat is at a low temperature, the heat generated in one minute q=i*irt=(12*12 4)*6*60=***j

The reason is that R2 is only 4 ohms, and if R2 is ohms, the minimum power of 30W is not met

Because of the low temperature gear, it is connected to R1, so w=i rt=(2a) 6 60s=1440j

You're not thinking the right way.

Think about it as I'm taught.

In series, the voltage is constant, your power supply has not changed, of course the total voltage has not changed, understand?

In the second step, in the circuit, the current = the total voltage divided by the total resistance, right? So, if the sliding r becomes larger, and the original r1 remains the same, does the total resistance become larger? So, the current of the entire circuit = total voltage divided by total resistance, it gets smaller, right?

In the third step, the voltage of R1 = current multiplied by its resistance, the resistance does not change, the total current becomes smaller, does that become smaller?

Fourth, the voltage of R1 becomes smaller, but the total voltage is certain, because the power supply does not change, then, the voltage of the sliding R = the total voltage minus the voltage of R1, you see if it has become larger!

2.Because the resistance of the sliding rheostat increases, and the resistance is proportional to the voltage in the series circuit, the voltage of the voltmeter 1 also increases, and because the total voltage of the series circuit is equal to the sum of the voltages at both ends of the resistor, the voltage of the voltmeter 2 decreases.

The sliding rheostat is connected in parallel with the voltmeter V2.

Because the resistance of the sliding rheostat increases, and the resistance is proportional to the voltage in the series circuit, the voltage of the voltmeter 2 increases (the voltmeter v2 measures the voltage of the sliding varistor), and because the total voltage of the series circuit is equal to the sum of the voltages at both ends of the resistor, the other voltmeter v1 (measuring R1) becomes smaller.

In series circuits.

u1+u2=u (supply voltage).

u1：u2=r1：r2

R1 remains the same, R2 increases - U2 increases - U1 decreases.

R2 increases, the total resistance increases, and the total current decreases (R1 and R2 are connected in series), V1=R1*I1; When r is constant, the current becomes smaller, and the voltage at both ends of r1 becomes smaller; v2 becomes larger; v total = v1 + v2;

1) L1 emits light normally, the voltage of L1 is 8V, the current of L1 is 2W 8V = R is connected in parallel with L1, the voltage of R is equal to L1 is 8V, the current of R = the current of the dry circuit (the number of current representations) - the current of L1 =

r=u/i=8v/

2) L emits light normally, the voltage of L is 5V, the current of L is 2W 5V=S1 is closed, S2 is disconnected, L is connected in series with R1, and the current of R1 is equal to the current of L.

Voltage of r1 = current of r1 r1=

Power supply voltage = voltage of l + voltage of r1 = 4v + 5v = power of 9vl p = u r, the power becomes, is the original 1 4, the resistance of the lamp remains unchanged, the voltage becomes the original 1 2, for; Current of lamp l = p u =

S1 is disconnected, S2 is closed, L is connected in series with R2, and the current of R2 is equal to the current of L.

Voltage of R2 = Power supply voltage - Voltage of L =

Resistance of r2 = voltage of r2 current of r2 =

3) R is connected in parallel with L, voltage of R = power supply voltage = 6V, current of R = 6V 30 = current of L = current of the dry circuit (number of current representations) - current of R = voltage of L = power supply voltage = 6V

l Rated power = current of l Voltage of l =

Voltage of w=uit=r Current of r Time=6V J

First of all, all electrical appliances are connected in parallel, p=ui, i=p u40 220*6*12+150 220*6+100 220*6=2)p lamp=12 6 40w+6 100w=3480ww lamp=p lamp t=3480 30=

3) The circuit limits the current, which can be derived from the meter:

imax=20a

Idle current quota: i=

imax-i=

Idle power quota: p=u i=220

The number of 25W lamps that can be increased: n=

p p25 = only).

If you don't understand, you can ask me.

Fill in sequentially: 2 15V

The series currents are equal.

The R2 current is i2=i1=1a

R2 voltage U2 = I2R2 = 1 * 6 = 6V

R1 voltage U1 = I1 * R1 = 1 * 4 = 4V

The power supply voltage U = U1 + U2 = 4 + 6 = 10V

If the fixed value resistor R1 is known, it is marked with "10 1A" and the other fixed value resistor R2 is marked with "20 When they are used in series during the test, then their current ratio is 1:1 and the voltage ratio is 1:2

When the two fixed-value resistors in Example 2 are used in series, the maximum current allowed to pass through the circuit and the maximum voltage allowed to be applied to both ends of the circuit are respectively.

Knowing that r1 = 4 ohms and r2 = 6 ohms, they are connected in series in a circuit, and the measured current through r1 is 1 ampere, then.

1) What is the voltage at both ends of R2? 6v

2) What is the power supply voltage? 10v

The ratio of the two resistors is r1:r2=2:5, and when they are used in series in a circuit, the ratio of the current passing through them is 1:1

The ratio of voltages across them is 2:5

The current ratio is 1:1 and the voltage ratio is 1:2

Maximum current, maximum voltage 15V

1*6=6v

6+1*4=10v

This is a sufficient and necessary condition. It's about both. The diode through the current is, the rheostat is the total current can be.

Hypothesis: When the rheostat is, is the diode current likely to be 0A? Is it possible to have a rheostat when the diode passes through the current?

It is only possible that the question requirements can be guaranteed.

The diode passes through the current, and the rheostat is.

The total current can be:

Hypothesis: When the rheostat is, is the diode current likely to be 0A? Is it possible to have a rheostat when the diode passes through the current?

It is only possible that the question requirements can be guaranteed.

Upstairs is the right solution, just like the barrel principle.