When a power supply is connected with 8 ohm resistor, the current through the power supply is 0.15 When a power supply is connected with 8 ohm resistance, the current passing through is 0.15 a, and when it is connected with 13 ohm resistance, the current passing through is 0.10 A. calculate the electromotive force and internal resistance of the power supply!

When a power supply is connected with 8 ohm resistor, the current through the power supply is 0.15 When a power supply is connected with 8 ohm resistance, the current passing through is 0.15 a, and when it is connected with 13 ohm resistance, the current passing through is 0.10 A. calculate the electromotive force and internal resistance of the power supply!

1.5 V, 2 Ω
Solution of simultaneous equations
V =0.15(8+R)
V =0.1(13+R)

There is a semicircle with diameter d made of uniform resistance wire, where o is the center of the circle. A is the midpoint of the semicircle. If the resistance between two points of OA is r, what is the resistance of BC (the two ends of radius)?

This problem seems complicated, but it is actually a formula to find the arc length of a semicircle
BCr=pi*2R*/2=3.14R
BCr:B , C resistance
2R: resistance of diameter D

A ring with diameter D is made of 12 ohm uniform resistance wire and connected to the circuit as shown in the figure. The P end of wire a in the figure can move along the ring and keep good contact. It is known that R1 is 15 ohm and the power supply voltage is 9 v? Find out the maximum current intensity im; (2) when P moves to where is the maximum total resistance in the circuit? Try to prove. Find out the power output at this time

(1) It can be seen from the circuit diagram that when the slide P of the circuit turns to point a, there is a short circuit to the ring resistance; in the circuit, only R1 is connected, and the current is the maximum. According to Ohm's law, Im = ur1 = 9v15 Ω = 0.6A (2) Let P turn to a certain point, the two arc resistances are R1 and R2, and R1 + R2 = 12 Ω. The total resistance of the circuit is: rtall = r1r2r1 + R2 = R1 (12 Ω− R1) 12 Ω. When R1 = 12 Ω - R1, that is, R1 = 6 Ω, point P and point a are in the diameter of the ring, and Rmax = 6 Ω & nbsp; × (12 Ω− 6 Ω & nbsp; ）12 Ω = 3 Ω, then the power output P = u2r, total = u2r1 + R max = (9V) 215 Ω + 3 Ω = 4.5w. Answer: (1) when P moves to a, the maximum current intensity in the circuit is 0.6A; (2) when P moves to the middle, the total resistance in the circuit is the largest. At this time, the power output is 4.5w

In the circuit shown in the figure, the ring with diameter D is made of ultra-fine uniform resistance wire, and its resistance value is R. in the figure, a, B When the key K is closed and the slider P slides clockwise along the ring, the number of meters in the figure will change. Students a and B make experiments according to the circuit diagram, and record the number of meters when the slider P is in some positions, such as It is shown in Table 1 and table 2. Table 1 (experimental data measured by student a): position of slide P position B position C position D position e position f position x position a the indication of ammeter a (a) is 0.250.20 0.18 0.17 0.18 0.19 the indication of voltmeter V1 (V) is 3.75 3.00 2.70 2.55 2.70 2.85 the indication of voltmeter V2 (V) is 2.25 3.00 3.30 3.453.303.15 Table 2 (experimental data measured by Student B): slide P position B position C position D position e position f position G position x indication of ammeter a (a) 0.240.170.130.110.090.08 indication of voltmeter V1 (V) 3.602.551.951.651.351.20 indication of voltmeter V2 (V) 2.403.454.054.35 4.65 4.80 According to the above experimental data, answer the following questions: (1) according to the experimental data in Table 1, please complete the space under position a in Table 1 through calculation and analysis, and mark the position x (approximate position) on the graph. (2) according to the experimental data in Table 1 and table 2, please explain the reason why the experimental data measured by students a and B are different through comparison and analysis (please draw the electric diagram) (3) according to the analysis in (2), please think: when the slide P slides to the position X between GA, the possible data (i.e. the indication of each meter) of the space under position X in Table 2 are______ A. 0.05 a, 0.75 V, 5.25 v. & nbsp; & nbsp; & nbsp; b.0.07 a, 1.05 V, 4.95 v. c.0.16 a, 2.40 V, 3.60 v. & nbsp; & nbsp; & nbsp; d.0.25 a, 3.75 V, 2.25 v. e.0.30 a, 4.50 V, 1.50 v. & nbsp; & nbsp; & nbsp; f.0.35 a, 5.25 V, 0.75 v

According to the data in the table, the power supply voltage U = U1 + U2 = 3.75V + 2.25V = 6V, R resistance R1 = UI = 3.6v0.24a = 15 Ω; when the slide is connected to a, the circular resistance is short circuited, then only R1 is connected in the circuit; according to Ohm's law, the current I '= ur1 = 6v15 Ω = 0.4A; At this time, the voltmeter V1 measures the power supply voltage, so the indication of V1 is 6V; the voltmeter V2 is parallel connected at both ends of the wire, so the indication of V2 is 0; when the slide is connected to point E, the circular resistance in the circuit is divided into the same two parts, the resistance of each part is R2, and the two parts are in parallel; then the resistance of the parallel part R and = R4; It can be seen from the data in figure a that V2 is 3.45v and I ′ = 0.17a, then R and = u2i ′ & nbsp; can be obtained from Ohm's law; =45v0.17a ≈ 20 Ω; then the total resistance R = 4 × R and = 4 × 20 Ω = 80 Ω; because the contact plate divides the circular wire into two parts, according to the resistance law and circular symmetry of the parallel circuit, the current and voltage situation of the slide plate connecting g should be the same as that of the slide plate connecting C, that is, the current value is 0.2A; and the current gradually decreases from e to a; because the current of the X point is between 0.18a and 0.20a, it can be seen that , point x should be between F and G; so the answer is as follows: (2) comparing the two tables, we can see that the current value in table B is always a little smaller than that in table a, which means that the resistance in student B's experiment is larger than that in student a's experiment when it is connected in the same way; because the total resistance will be reduced after the resistance is paralleled, it means that student B can only connect a section of wire when he is doing the experiment, so it means that When student B did the experiment, it was supposed that a part of the wire had broken circuit; because there was a number indication from a to g, it indicated that only a certain point between G and a had broken circuit fault; the experimental circuit of the two students was equivalent as follows: (3) according to the analysis of 2, if there was a circuit break between G and a, it should be considered that point X was before or after the breakpoint, If the current is equal to 0.07a, the indication of voltmeter V1 is U1 ′ = 0.07a × 15 Ω = 1.05v; the indication of voltmeter V2 is 6v-1.05v = 4.95v; therefore, B is correct; if the contact is after the breakpoint, the part between the contact and a is connected If the current is 0.25A, the voltage U1 = 0.25A × 15 Ω = 3.75V; U2 = u-u1 = 1.25V; so D is correct; if the current is 0.30a, the voltage can be 4.5V and 1.5V respectively, so e is correct; if the current is 0.35A, the voltage can be 5.25V and 0.75V respectively, so F is correct; answer: (1) position The values under a are: 0.40a; 6V; 0; point x as shown in the figure; (2) the circuit between G and a is open circuit, and the formed circuit diagram is shown in figure (b); (3) B, D, e, F

A nickel chromium wire of uniform thickness has a cross-section diameter of D and a resistance of R. after it is drawn into a uniform wire with a diameter of D / 10, its resistance becomes a.r/100
A nickel chromium wire of uniform thickness has a cross-section diameter of D and a resistance of R. after it is drawn into a uniform wire with a diameter of D / 10, its resistance will become zero
A．R／1000 B．R／100
C．100R D．10000R
Can you make it easier to understand? I found a few that I didn't understand very well. Why should I choose D? I think the square of 10 is 100r. Why is it wrong?

1. When the diameter becomes 1 / 10, the area becomes 1 / 100, S2 = S1 / 100
2. If the volume remains unchanged and the cross-sectional area is 1 / 100 of the original, the length becomes 100 times of the original, L2 = 100l1
3. The resistance of a conductor is directly proportional to its length and inversely proportional to its cross-sectional area
Original resistance: R1 = ρ * L1 / S1 = R
Resistance after stretching = ρ * L2 / S2 = ρ * 100 * L1 / (L1 / 100) = 10000 * ρ * L1 / S1 = 10000r1 = 10000r
Note: stretch, not cut it to 1 / 10 of the original diameter without changing the length

A section of nickel chromium wire with uniform thickness has a diameter of D and a resistance of R. after drawing it into a uniform wire with a diameter of d10, its resistance becomes ()
A. 10000 RB. R10000C. 100 RD. R100

The formula s = 14 π D2 shows that the cross-sectional area of the nickel chromium wire is 1100; the length of the nickel chromium wire is 100 times of the original; the resistance law r = ρ LS shows that the resistance is 100 × 100 = 104 times of the original, that is, 10000r; therefore, a

Which one of the wires is the live wire? Which one is the ground wire? Which one is the zero wire? What color are they?
What is the role of each?

In general construction, the color of the live wire and the zero wire will be the same. In the electric pen test, the live wire is bright, and the zero wire is the same color as the live wire. As for the difference between the zero wire and the ground wire

Is n a live wire or a zero wire

Live wire is l, and neutral wire is n
There is also the ground wire is generally yellow and green, the symbol is a vertical three horizontal
I hope my answer can help you!

How do the live wire, zero wire, ground wire in the circuit correspond to the color of the wire and the letters n, l, e?
How do the colors of live wire, zero wire, ground wire and wire in the circuit correspond to the symbol letters n, l and E?

In general, in single-phase alternating current, the live wire is generally represented by L, and the color is mostly red and brown. The zero wire is represented by N, and the color is black and blue. The ground wire is represented by E, which is generally two-color wire, but the DC is different. L is the negative wire, and the color is generally black. N is the positive wire, and the color is brown. The ground wire is the same

What is the reason for the live line voltage of household 3-phase socket to be 220 V, the zero line voltage to be 110 V and the ground line voltage to be 36 V?
I use electronic measuring pen to check, live line inspection: live line 220, zero line 110, ground line 12
Power off inspection: live line 110, zero line 110, ground line 36
It is a single switch. The three-phase sockets connected to other switches are normal (live line 220, zero line and ground line 0V)

It is suggested that you use a multimeter to measure, your measurement results are very problematic
1. The ground to ground potential is 0. What benchmark do you use to test the ground voltage?
2. If your measurement result is correct, live line 110, zero line 110, then the voltage between live line and zero line is 0
In addition to inductive current, you are most likely to be grounded. It is recommended that you short the zero line and ground line to ground