As shown in the figure, the length of each side of square wire frame ABCD is l, the mass of AB side is m, and the mass distribution is uniform, and the mass of other sides is not included. The total resistance of wire frame is R. the CD side is connected with smooth fixed rotating shaft OO ', and the wire frame can rotate freely around the OO' axis. The whole device is in a uniform magnetic field with magnetic induction intensity of B and vertical downward direction The results show that: (1) the quantity of electricity Q passing through the cross section of the wire in time t; (2) the magnitude and direction of Ampere force on the AB side at the lowest point; (3) the tensile force on the AB side at the lowest point; (4) the heat generated in the wire frame in time t

(1) From △ Φ △ t = BL2 △ T, we know the current I = ER in the loop, the charge q = I △ t = bl2r (2) AB through the coil is subjected to the horizontal right ampere force, the velocity is v = ω L, the instantaneous electromotive force is Ee1 = BLV, the current I1 = BLVR, the ampere force F1 = bi1l = b2l3 ω R (3) at the lowest point f-mg = mv2lab side is subjected to CD, the tensile force F2 = Mg + m ω 2l2 (4) from the conservation of energy, we know the heat generated by MGL = 12mv2 + Q Q = mgl-12mv2 = mgl-12m ω 2l2 A: (1) the electric quantity Q passing through the cross section of the conductor in time t is bl2r; (2) at the lowest point, the ampere force on the AB side is b2l3 ω r horizontally to the right; (3) at the lowest point, the tensile force on the AB side is F2 = Mg + m ω 2l2; (4) the heat generated in the wire frame in time t is mgl-12m ω 2l2
When the resistance of a circuit increases by 4 ohm, the current becomes 1 / 5 of the original. What is the original resistance of this circuit?
It's better to solve the problem with the knowledge of electricity in grade two. It doesn't matter~
U = 1V, because if the resistance is 1 ohm and the current is 5 A, then the voltage is 5 v. if the resistance plus 4 ohm is equal to 5 ohm, then the voltage 5 divided by 5 is equal to 1, which is just one fifth of the original
It's purely personal thinking. If you have any mistakes, please point them out. I'm also a sophomore. Let's discuss them together
In fact, the method upstairs is the same as that in my draft. I'm afraid you don't understand it, so I'll use more intuitive
1 ohm, let the voltage be u and the resistance be X. from the current = voltage / resistance, we can get u / (x + 4) = (1 / 5) * (U / x), and solve the equation to get x = 1
In fact, it's the same as that simply according to Ohm's Law: I = u / R
If u remains the same, then I becomes 1 / 5 of the original, then the resistance must become 5 times of the original, so it is very easy to know that the original resistance is 1 ohm
How much current can a square meter of copper wire carry?
The safe current carrying capacity of 2.5 square millimeter copper power line is - 28a
The safe current carrying capacity of 4mm2 copper power cord is 35A
The safe current carrying capacity of 6mm2 copper power line is 48A
The safe current carrying capacity of 10 square millimeter copper power line is - 65A
95 square millimeter energy band about 230 current
As shown in the figure, the mass of the square wire frame ABCD is m, the side length is l, and the total resistance of the wire frame is R. the wire frame falls freely from the top of the horizontal bounded uniform magnetic field from the vertical paper surface to the inside. During the falling process, the wire frame is always in the vertical plane perpendicular to the magnetic field, and the side of CD remains horizontal. The magnetic induction intensity of the magnetic field is B, and the direction is perpendicular to the paper surface to the inside The horizontal distance between the upper and lower interfaces of the field is L. when the CD edge just enters the magnetic field, the wire frame just moves at a uniform speed. The gravitational acceleration is g (1) (2) please prove that the work done by the wire frame to overcome the ampere force is equal to the electric power consumed by the wire frame at any moment when the CD edge of the wire frame moves in the magnetic field. (3) the work done by the wire frame to overcome the ampere force from the moment when the CD edge of the wire frame just enters the magnetic field to the moment when the AB edge just leaves the magnetic field
(1) Let the velocity of the wire frame CD just entering the magnetic field be V, then the induced electromotive force E = BLV generated when the wire frame CD enters the magnetic field. According to Ohm's law of closed circuit, the induced current through the wire frame is I = BLVR, the ampere force F A = bil = b2l2vr, and the wire frame operates at a uniform speed when the wire frame CD just enters the magnetic field
Voltage divided by current and resistance, the unit of resistance should be ohm or K ohm. Voltage divided by resistance equals the unit of current. For example, 5V divided by 3.23a equals 1.54798761. Is this 1.5 ohm or 1.5 kiloohm? Is the current unit of voltage divided by resistance ampere or milliampere,
1.5 ohm
Ohm, ampere
Unit correspondence: voltage (volt), current (ampere), resistance (Ohm)
If voltage (kV), current (a) and resistance (K Ω), from the formula, the units in physics are also involved in the calculation
Si ohm ampere
How much current can 25 square aluminum wires carry
Below 150A
As shown in the figure, the two ends of the parallel metal guide rail AB and CD with the distance L are respectively connected with the resistance R1 and R2 with the resistance R. the uniform magnetic field perpendicular to the plane of the guide rail has the magnetic induction intensity of B. now, pull the straight conductor rod with the resistance R through the magnetic field area with the speed V at a uniform speed. If the width of the magnetic field area is D, the friction and the resistance of the guide rail are ignored, then the work done by the external force in this process is___ The amount of charge passing through R1 is___ .
The induced electromotive force E = BLV produced by the conductor bar Mn, the induced current through Mn is I = ER + R2 = 2blv3rmn, and the ampere force is f a = bil = 2b2l2v3r. Because the bar Mn moves at a uniform speed, the external force F = f a = 2b2l2v3r. Therefore, the work done by the external force w = FS = 2b2l2vd3r; the electric quantity through Mn q = it = 2blv3r · DV = 2bld
If the resistor is R1, 10 ohm, R2, 20 ohm and they are connected in series for 6 volts, how many amperes is the voltage at both ends of R1 and how much is the current through R2?,
The current of R1 and R2 is the same, I = u / r = 6V / (10 Ω + 20 Ω) = 0.2A
The voltage at both ends of R1 is U1 = I * R1 = 0.2A * 10 Ω = 2V
The voltage at both ends of R2 is U2 = I * R1 = 0.2A * 20 Ω = 4V
How much power can a 2.5 square copper wire withstand without danger?
It is generally believed that the safe carrying current of 1 square millimeter copper wire is 7a, so the maximum safe current of 2.5 square millimeter copper wire is 17.5a. As for power, it depends on the voltage of the electrical appliances you connect. I think what you ask is the mains power is 220 V, and the power is 3850 watts
Make a square metal frame with a straight wire of 80cm in length and 8 ohm in resistance, and place it in a uniform magnetic field with B = 5T,
There is a "0.1W, 1V" small light bulb connected with the metal frame, which can rotate around an edge perpendicular to the magnetic induction intensity B. the normal light emission of the small light bulb is the rotation angular velocity of the metal frame? (B is perpendicular to the paper surface inward, and the metal frame rotates counterclockwise)
Because the small bulb is connected in parallel with a 2 ohm resistor_
I figured out 12.7
The total resistance is 18 ohm, and 1.8V total voltage is needed for the small bulb to light normally
The metal frame can rotate around an edge perpendicular to the magnetic induction intensity B, which means that sinusoidal voltage is formed when rotating at a constant speed, so the maximum value of the 2 sinusoidal voltage needs to be 1.8 * 1.414 = 2.545
5*0.2*0.2*w=2.545
The result is w = 12.7
But the answer is 32.5 rad / s. I don't know why
I figured out 12.7
The total resistance is 18 ohm, and 1.8V total voltage is needed for the small bulb to light normally
The metal frame can rotate around an edge perpendicular to the magnetic induction intensity B, which means that sinusoidal voltage is formed when rotating at a constant speed, so the maximum value of the 2 sinusoidal voltage needs to be 1.8 * 1.414 = 2.545
5*0.2*0.2*w=2.545
The result is w = 12.7
But the answer is 32.5rad/s, I don't know why
The answers from upstairs didn't seem right,
It seems that the third floor is just making mistakes
The confused rabbit answered correctly...

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