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How to connect a resistance wire with a resistance of 5 and 10 ohm in the same circuit with the same power on time and voltage? Why?
Parallel connection. With the same voltage, the electric power is the sum of the electric power of the two resistors, and the work done is also the sum
Observe the following formulas: 152 = 1 × (1 + 1) × 100 + 52; 252 = 2 × (2 + 1) × 100 + 52; 352 = 3 × (3 + 1) × 100 + 52 According to this rule, please write the nth equation (n is a positive integer)______ ．
According to the meaning of the question: the nth equation is: (10N + 5) 2 = 100N (n + 1) + 52. So the answer is: (10N + 5) 2 = 100N (n + 1) + 52
In a parallel circuit, there are two constant resistors, R1, 10 ohm, R2 and 30 ohm to calculate the total resistance
R = (R1 * R2) / (R1 + R2) = 7.25 Ω
What are the rules for the arrangement of numbers in the pyramid of natural numbers
1. The number on the rightmost side of each row is the square of the number of rows in the row;
2. The number at the beginning of the line forms a difference of 1, 3, 5, 7, 9 (tolerance 2);
3. The number of numbers in the third line is odd;
4. The composition difference of the middle number on the line is 2, 4, 6, 8, 10, 12 Arithmetic sequence with tolerance 2
In the circuit shown in the figure, l is marked with "6V & nbsp; 3W "small bulb, the internal resistance of the power supply R = 2 Ω, the total resistance of the sliding rheostat r = 48 Ω, when the sliding plate P contact of the sliding rheostat is at the midpoint of the rheostat, the bulb l can light normally, at this time, the reading of the voltmeter is 27V, calculate: (1) the value of resistance RX; (2) the electromotive force of the power source E; (3) the electric power consumed by the sliding rheostat
(1) According to P = UI, the rated current of the bulb is I1 = 36 = 0.5A; the voltage of the parallel part is 6V; the voltage of the parallel part of the sliding rheostat is 6V, the current is I2 = 624 = 0.25A; the total current is I = I1 + I2 = 0.5 + 0.25 = 0.75a; the total voltage of the other half of the sliding rheostat and both ends of Rx is u = 27-6 = 21V; the total resistance is R '= RX + R2 = 210.75 = 28 Ω; the solution is that RX = 28-24 = 4 Ω (2) is determined by the ohm of the closed circuit According to the law, e = u + IR = 27 + 0.75 × 2 = 28.5V; (3) the total power of sliding rheostat is p = I21 × R2 + i2r2 = 0.52 × 24 + 0.752 × 24 = 19.5w; (1) the resistance of Rx is 4 Ω; (2) the electromotive force is 28.5V; (3) the total power consumed by sliding rheostat is 19.5w
Simplification ratio: 1 / 3 hour: 10 / 3 hour: 5 / 6 hour, 1 / 25 hour: 8 hour
1 / 3 hour: 45 minutes = 20 minutes: 45 minutes = 4:9
10 / 3:5 / 6 = 10 / 3x6 / 5 = 60 / 15 = 4:1
1,25:8:1 = 1.25:0.125 = 10:1
The electromotive force E of the power supply is 12V. The internal resistance R is 1 ohm, the resistance R1 is 9 ohm, and the resistance R2 is 15 ohm. The indication of ammeter A is 0.4A. Calculate the resistance value of resistance R3 and the electric power it consumes
(1) When the current expression number is 0.4A, U2 = i2r2 = 6V, I = 0.6ar3 = U3 / I3 = U2 / (i-i2) = 30 Ω (2) the electric power consumed by R3: P3 = ei-i2 (R1 + R) - [E-I (R1 + R)] ^ 2 / r2: 15p3 = - 250i2 + 420i-144 when I = 0.84a, P3 is the largest, U2 '= E-I (R1 + R)] = 3.6
The solution of the equation x & # 178; = 16
The answer is ± 4
How to convert the current of 200 ohm two-phase heating tube
I = u / r = 220 V / 200 = 1.1a = 1.1 A
(1 / 2-2x-1 / 4) divided by 2x & # 178; - 8 / x-3
You have solved the following problems, the above problems, the general problems, and then all the problems can be solved into the product of the equation. This problem is very simple. I'll calculate it carefully
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