A small bulb marked with "2.5V & nbsp; & nbsp; 0.5W" must be connected in series with a large resistor in order to connect it to a 6V power supply for normal lighting?

∵ when UL = 2.5V, the lamp will light normally, ∵ ur = 6v-2.5v = 3.5V, and ∵ lamp L and R are connected in series ∵ I = & nbsp; plul = URR, that is: 0.5w2.5v = 3.5vr ∵ r = 17.5 Ω; a: to connect it to 6V power supply, a 17.5 Ω resistor must be connected in series

There is a small light bulb marked with "2.5V & nbsp; & nbsp; 0.5W". In order to connect it to a 4.5V power supply, it should be in the circuit &Join one The resistance in Ω

The power supply voltage is 4.5V, the rated voltage of the bulb is 2.5V, so a resistor should be connected in series, and the current of the series circuit is equal. The current through the circuit is: I = plul = 0.5w2.5v = 0.2A. The size of the series resistor is: r = u − Uli = 4.5V − 2.5v0.2a = 10 Ω; so the answer is: string; 10

Connect a light bulb marked "6V, 3W" to a 10V power supply. In order to make the light bulb shine normally, you need to connect one in series What is the power consumed by the resistor Tile

The voltage of the bulb is 6V and the power is 3W. The current in the series circuit is equal. According to P = UI, the current in the circuit is: I = IL = plul = 3w6v = 0.5A. The total voltage of the series circuit is equal to the sum of the partial voltages. The voltage at both ends of the resistor is ur = u-ul = 10v-6v = 4V. According to Ohm's law, the current in the circuit can be calculated

As shown in the figure, pull out the square closed coil with side length of L and total resistance of R from the uniform magnetic field with magnetic induction strength of B at a uniform speed of V (magnetic field direction, perpendicular to the coil plane) (1) with the pulling force F? (2) The work done by pulling force F? (3) Power PF of tension f? (4) Heat released by coil q? (5) Power of coil heating? (6) The quantity of electricity passing through the conductor section q?

(1) Because the coil is pulled out at a constant speed, the magnitude of F = f a induced electromotive force is: e = BLV. According to the closed Ohm's law, I = BLVR. Then the pulling force is: F = f a = b2l2vr (2) the work done by pulling force F is: W = FL = b2l3vr (3) the power of pulling force F is: pf = FV = b2l2v2r (4) the heat released by the coil is equal to the work done by pulling force F, then q = w = b2l3vr (5) the heating power of the coil is: P heat = FV = b2l2v If 2R (6) the quantity of electricity passing through the conductor section is q = it = BLVR and L = VT, then q = bl2r answer: (1) the pulling force F is b2l2vr. (2) the work done by pulling force F is b2l3vr. (3) the power PF of pulling force F is b2l2v2r. (4) the heat released by coil q is b2l3vr. (5) the power P heat generated by coil is b2l2v2r. (6) the quantity of electricity passing through the conductor section q is bl2r

In the horizontal uniform magnetic field with magnetic induction intensity B = 0.2T, the length of one side is L = 10cm, the number of turns is n = 100, and the resistance is high
The square coil with r = 1 Ω rotates at a constant speed around the OO 'axis perpendicular to the magnetic induction line. The rotation speed is 100 / PA, and there is a resistance R = 9 Ω. The brush contacts with two slip rings, and an ideal voltmeter is connected at both ends of R to calculate the work done by driving the coil to drill holes within 1 minute. The picture shows that the coil is parallel to the magnetic induction line, and the voltmeter measures the voltage of resistance R

The peak value of electromotive force is calculated first
E=nbsw
Find the effective value of current again
Finally, the work done by current is the work done by external force

The distance between two horizontally placed metal plates is D, which is connected with an n-turn coil by a wire. The coil resistance is r, and there is a vertical magnetic field in the coil. The resistance R is connected with the metal plate, and other resistances are ignored. As shown in the figure, there is a positively charged oil drop with mass m and electric quantity g between the two plates, which is just in a static state, then the change of magnetic induction intensity and magnetic flux in the coil The rate of change of quantity is ()
A. The magnetic induction is upward and increasing, △ Φ Δ t = mgdnqb. The magnetic induction is downward and increasing, and △ Φ Δ t = mgdnqc. The magnetic induction is upward and decreasing, △ Φ Δ t = MGD (R + R) nqrd. The magnetic induction is downward and decreasing, △ Φ Δ t = MGD (R + R) NQR

According to the balance condition, if the electric field force of the oil drop is vertical upward, the upper plate of the metal plate will be negatively charged and the lower plate will be positively charged

What resistance does the motor have besides coil resistance?
Maybe you don't quite understand what I said. For example:
A toy motor is marked with "rated voltage 6V, coil resistance 0.5eu" in the manual. When it runs under the push of 4 dry batteries, the measured current through the coil is 0.3A
I find it very strange that 6V divided by 0.3A equals 20 ohm, but the title says that the resistance is 0.5 ohm. I know that coil resistance is different from 20 ohm resistance, but where does this 20 ohm resistance come from?

Using DC, there will be inductive reactance!
I think on the one hand, the battery has internal resistance, and on the other hand, because the DC motor is used, the contact resistance of the brush and commutator can not be ignored. From the outside of the motor, there will be sparks on the contact surface of the brush, which will also produce a voltage drop

About the resistance of motor coil
There is a smooth slope with a length of 2.4 meters. One end is placed on the ground and the other end is supported on a platform with a height of 1.2 meters. A weight of 200 kg is placed on it. The voltage applied to the motor is 220 v. when the motor lifts the weight along the slope at a speed of 1 m / s, the current passing through the motor is 5 A, and the resistance of the motor coil is calculated regardless of the friction of the motor shaft

Input power of motor P = UI = 220 * 5 = 1100W output power of motor (take g = 10) 200kg = 2000np = f * V = 2000 * 1 * (1.2 / 2.4) = 1000 (V is the upward speed) power consumed by motor resistance 1100-1000 = 100wp = UI u = RI P = RI * I resistance of motor coil r = P / (I * I) =

How to detect the resistance of motor coil?
For a 1000W brushless DC motor, the stator resistance given by the manufacturer is 10-15 ohm. I directly use the bridge test to measure the three leads of the motor in pairs, which is only tens of milliohm. Is the method wrong?
I mean to test the stator resistance, that is, the DC resistance of the stator coil, not the insulation resistance. The manufacturer said that they tested it in the semi-finished product. Do I have to dismantle the motor to test it?

It is estimated that the motor will be disassembled and the test will be on time. You'd better contact the manufacturer!

The coil resistance of DC motor is very small, and the starting current is very large, which will produce adverse consequences. In order to reduce the starting current, it is necessary to connect a starting resistance R in series, and then gradually reduce R after the motor starts. If the power supply voltage is 220 V and the coil resistance of the motor is 2.0 Ω, then, (1) how much is the starting current when the starting resistance is not connected in series? (2) how much should R be in order to reduce the starting current to 20 a?

（1）I1=U/r=220/2=110A
（2）I2=U/(R+r)
R+r=U/I2=220/20=11Ω
So r = 11-r = 9 Ω

A small bulb marked with "2.5V & nbsp; & nbsp; 0.5W" must be connected in series with a large resistor in order to connect it to a 6V power supply for normal lighting?