If the arc AB is 1 / 3 of the circumference, what is the resistance between ab? If point a is fixed and point B is optional on the circumference, what is the maximum value of resistance between ab?

If the arc AB is 1 / 3 of the circumference, what is the resistance between ab? If point a is fixed and point B is optional on the circumference, what is the maximum value of resistance between ab?


The resistance between AB is r = 20 ohm
If point a is fixed and point B is optional on the circumference, the maximum resistance between AB is 22.5 Ω



Bend a uniform wire with a resistance of 18 Ω into a closed metal ring with a diameter of D = 0.80M as shown in the figure. The angle of AB arc on the ring to the center of the circle is 60 degrees. Fix the ring perpendicular to the direction of the magnetic induction line in a uniform magnetic field with a magnetic induction intensity of B = 0.50t, and the direction of the magnetic field is perpendicular to the paper surface. A straight wire PQ with a resistance of 1.25 Ω per meter slides to the left at a speed of V = 3.0m/s along the plane of the ring The direction is perpendicular to PQ), and the straight wire is in close contact with the ring during sliding (ignoring the resistance at the contact). When it passes through a and B positions on the ring, calculate: (1) the induced electromotive force generated by ab section of the straight wire, and indicate the direction of the current in this section of the straight wire; (2) the electric power of heat loss on the ring at this time


(1) Suppose the length of AB section of straight wire is l, the diameter of ring is D, and the induced electromotive force is e, then there is a geometric relationship, l = D2 = 0.4m & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; so: e = BLV = 0.6V & nbsp; & nbsp; & nbsp; & nbsp; from the right hand rule, the direction of induced current of straight wire is obtained from a to B. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; &At this time, the resistance of AB arc segment on the ring: Rab = 186 Ω = 3 Ω, the resistance of ACB arc segment: RACB = 18 × 56 Ω = 15 Ω, the total resistance of Rab and RACB in parallel: r outer = 2.5 Ω & nbsp; & nbsp; & nbsp; & nbsp; internal resistance of power supply: r = 1.25 × 0.40 = 0.50 Ω & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; The total current should be: I = ER + r = 0.20a & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; suppose the electric power of heating loss on the ring is p, then: P = I2R and = 0.10w A: (1) the induced electromotive force generated by ab section of straight wire is 0.6V, and the direction of current in this section of straight wire is from a to B; (2) at this time, the electric power of heating loss on the ring is 0.10w



As shown in the figure, the ring is made of wire with uniform thickness and resistance value R. a, B and C divide the ring into three equal parts (each equal part of resistance is 13R). If any two points are connected into the circuit, the resistance value of the circuit is______ .


The resistance wire with resistance value of R is bent into a ring, and a, B and C divide the ring into three equal parts, so the resistance of AB section radb = 13R, the resistance of ACB section RACB = 23R, ∵ two conductors are in parallel, ∵ 1R = 1radb + 1racb, that is, 1R = 1r3 + 12r3, and the solution is: r = 29R; so the answer is: 29R



A 10 m long copper conductor with a resistance of 10 ohm is pulled into a 20 m long copper conductor with a resistance of 10 ohm
To analyze


1, known: r = k * L / S is (length divided by cross-sectional area, and then multiplied by resistivity)
2. Original wire: 10 = 10 * k / S
3. According to the principle of constant volume: volume = s * l (bottom * height)
4, S1 * L1 = S2 * L2, L2 = 2 * L1. (1)
5,S1=2S2 .(2)
6,R2=K*L2/S2 .(3)
7, from the second part: 10 = 10 * k / S1. (4)
8. From the above four formulas, it is concluded that:
R2 = 4r1 = 40 (Euro)
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