The resistance of the transmission line is 0.5 ohm, the transmission power is 1000000kw, and the transmission voltage is 1000000v. What is the current in the transmission line? Why can't I = u / R be used instead of I = P / u?

The resistance of the transmission line is 0.5 ohm, the transmission power is 1000000kw, and the transmission voltage is 1000000v. What is the current in the transmission line? Why can't I = u / R be used instead of I = P / u?

The purpose of this question is to check the understanding of the formulas I = u / R and P = UI
I = u / R is the most suitable in "closed loop". Although the title gives the wire resistance, it is not necessary because it does not provide "closed loop" condition
But p = UI can solve this problem

If f (x) = x & # 179; + X & # 178; F ↓ (1), then f ↓ (1) =?

F (x) = x & # 179; + X & # 178; f '(1)
For the above equation, the derivation on both sides of the equation is as follows:
F '(x) = 3x & # 178; + 2x * f' (1) substituting x = 1 has
f'(1)=3+2f'(1)
f'(1)=3

Why is the power consumption 0.50 (kW. H) / 24h marked on the refrigerator nameplate,

Economic power consumption is the energy consumption measured under normal temperature, that is, the power consumption in 24 hours is 0.5 kwh, also known as average power consumption. Generally, it is lower than this value in winter, and much higher in summer, especially in high temperature season. The average power consumption is about 15 kwh per month

2.38 + 0.99 =? Simple operation

2.38＋0.99
＝2.38+（0.99＋0.01）
＝2.38+1-0.01
＝1.38+0.01
＝1.37

Calculation process of connecting a 3V led to a 220V circuit

The above figure shows the circuit diagram of lighting led by 220 V AC. IN4007 diode is a protective function, and the 3V voltage drop of LED can be ignored. & nbsp; generally, the current of common light-emitting diode is 10 Ma, which is the brightest. If the current of LED is 10 Ma, then the value of resistance is 220 / 10 Ma = 22 K, & nbsp; but if 22 K resistance is selected, The power of the 22K resistor is (10mA) ^ 2 * 22K = 2.2W. If the power is too high, the resistor should be more than 3.5W. You can't buy a 22K resistor with that high power in the market. So you can choose a resistor of 2W, 100k or 80K. In this way, the current of the LED is 2.2 or 2.8ma, and the brightness of the lamp is OK, Choose 80K is 0.62w. Buy 2W resistance is absolutely safe

How to calculate the five questions: 45 / 8 + (- 3.257) + (- 4.625); - 45 -- (- 15) - 12 - (- 31); 25-16 - (- 21) - 32?
Please write the process clearly when you write!
Thank you~~

45÷8+（-3.257）+（-4.625）
=5.625-3.257-4.625
=(5.625-4.625)-3.257
=1-3.257
=-2.257
-45--(-15)-12-(-31)
=-45-15-12+31
=-72+31
=-41
25-16-(-21)-32
=25-16+21-32
=25+21-16-32
=46-16-32
=30-32
=-2

As shown in the figure, R1 = 5 ohm, R2 = 20 ohm. It is known that the current I1 passing through R1 is 0.4A. Calculate the voltage at both ends of the circuit and the current passing through resistance R2

u=2V,I2=0.1A

0.25 times 40.4 + 0.125 times 10.4

0.25 times 40.4 + 0.125 times 10.8
=0.25×4×10.1+0.125×8×1.35
=10.1+1.35
=11.45

The full current Ig of the ammeter is 1mA, and the internal group is 500ohm. To refit it into a voltage meter with a measuring range of 3V, it should be installed in the

3V / 1mA = 3000 ohm, 3000 ohm - 500 ohm = 2500 ohm, 2500 ohm resistor should be connected in series in the ammeter to make it full bias at 3V

Calculate (- 4) ^ 2 ^ (- 8) - (- 1) ^ 10 - 5 ^ 3 × (- 2 / 5) ^ 2 ^ (- 0.4) ^ 3
Write 5 ^ =, 3 ^ 5 =, (- 4) ^ 3 =, - 4 ^ 3 =, (- 5) ^ 4 =, - 5 ^ 4 =,
(- 2 / 3) ^ 3 =, - 2 / 3 =

(- 4) ^ 2 ^ (- 8) - (- 1) ^ 10 = - 16 ^ - 8-1 = - 2-1 = - 3 - 5 ^ 3 × (- 2 / 5) ^ 2 ^ (- 0.4) ^ 3 = - 125 × (2 / 5) ^ 2 ^ [- (2 / 5) ^ 3] = 125 ^ - 2 / 5 = 125 × 5 / 2 = 625 / 2, 5 ^ 2 = 25, 3 ^ 5 = 243, (- 4) ^ 3 = - 64, - 4 ^ 3 = - 64, (- 5) ^ 4 = 625, -