The resistance of a motor coil is 1 ohm, When the voltage applied at both ends of the coil is 2V, the current is 0.8A, and the motor works normally. Calculate the electric energy consumed by the motor in normal operation for 1min The second question is how much mechanical energy is converted by the motor working for one minute?

First calculate the power P = UI = 2V * 0.8A = 1.6W
According to w = Pt, the consumed electric energy w = 1.6W * 60s = 96j
You didn't tell me the mechanical efficiency, but if you didn't, it should be all converted into mechanical energy, which is 96j
A small DC motor, the voltage at both ends of the motor is 12V, and the coil resistance of the motor is 0.5 ohm
When a small DC motor works, the voltage at both ends of the motor is 12V, the coil resistance of the motor is 0.5 ohm, and the current passing through the motor is 2A during normal operation. What is the efficiency of the motor at this time?
P total = UI = 24W, P heat = I & sup2; r = 2W, P machine = P total - P heat = 22W
Efficiency = P machine / P total = 11 / 12
Can 12V100AH battery be reversed to 48V for three wheel charging
If you can charge while riding, it would be better! What I'm talking about is how long can I charge the battery continuously for three rounds when I stop to charge. I'm satisfied with the bonus points
First, let's talk about the equipment to be used
The original 12V storage battery, because it is DC, so it needs to increase the voltage (AC) through the inverter, and then use the rectifier to change the AC into DC 48V or higher
You need to add two kinds of equipment, one is inverter, the other is rectifier
This will charge the 48V battery
How long does the battery last for three rounds?
12V100AH batteries charge 48V and how many ah batteries? During charging, the loss of inverter and rectifier should be considered. Therefore, charging 48V batteries with large capacity (if it is also 100Ah) is not appropriate and not sufficient
A six volt six watt bulb is connected to the two ends of the three volt power supply to calculate the resistance of the lamp, the current passing through the bulb and the actual power of the lamp
The resistance is 6 ohm by V2 / P, the current is naturally 0.5A, and P = u × I = 1.5W
It should be. Pro
Copper resistivity
How much resistance does 100 meters of pure copper wire with a diameter of 0.45 mm have?
Resistivity is related to temperature
The resistivity of copper at 20 ℃ is 0.0172 (& micro; Ω &; m)
According to the formula of resistivity: ρ = RS / L
ρ is resistivity, commonly used in Ω· M
S is the cross-sectional area, commonly used in M2
R is the resistance value commonly used in Ω
L is the length of the conductor - commonly used unit: M
We already know the diameter and length. Let LZ do it by itself, 1000000 & micro; Ω = 1 Ω
Or to help you find it, but to give the points ha, get good pain, specifically to ask students for a calculator: the following are basic units
Area s = (0.45 / 2) ^ 2 * 10 ^ - 6 * PI L = 100 ρ = 1.72 * 10 ^ - 9, so the result is: 10.82 Ω
The power input of household appliances is 220 V 140 W. what is the conversion current? What is the formula? [3]
This problem should be considered from two aspects. For pure resistive circuit, P = UI is 0.64a. For inductive circuit, P = UI * power factor (for household power, the value is between 0.75 and 0.95), so its current will change. Generally, if the value is 0.85, then the problem will be 0.64/0.85 = 0.75a
The rated power of a small bulb is 2 watts, and it is connected in series with a 4 ohm resistor in a circuit with a supply voltage of 6 v. what is the voltage of the small bulb
Please tell the rated voltage of the bulb =?
What is the electrical conductivity of 0.71 Ω· mm2 · M-1, S / M
Conductivity = 1 / resistivity = 1 / 0.71 Ω· mm2 · M-1 = 1 / 0.71
=1.40845070422535S/0.001m*0.001m*m-1
=1408450.70422535S/m
After connecting the two bulbs marked with "220 V & nbsp; & nbsp; 200 W" and "220 V & nbsp; & nbsp; 40 W" in series into the 380 V circuit, what will be burned is______ .
The resistance of bulb 1 is: R1 = u2p = (220V) 2200W = 242 Ω, the resistance of bulb 2 is: R2 = u ′ 2p ′ = (220V) 240W = 1210 Ω, the two bulbs are in series, r = R1 + R2 = 242 Ω + 1210 Ω = 1452 Ω, the current in the circuit is: I = ur = 380v1452 Ω = 0.26A, the voltage at both ends of bulb 1 is: U1 = IR1 = 0.26A × 242 Ω
When the bulb L is connected in series with the resistor R, the power supply voltage is 6 V and the resistor R is 5 ohm. The heat generated by the current in R per minute when the bulb L is connected
When the bulb L is connected in series with the resistor R, the power supply voltage is 6 V and the resistor R is 5 Ω. When s is switched on, the heat generated by the current at R is 12 joules per minute, and the rated power of the bulb is 0.25 watt. If the resistance of the bulb remains unchanged, the rated voltage of the bulb is
2、5 V
t=1min=60s
According to the statement that the heat generated by the current at R per minute is 12 joules, we can use the formula P resistance = q / T = 12j / 60s = 0.2W
According to P = I square * r, I = root P resistance / R resistance = root 0.2w/5 ohm = 0.2A
Because the circuit is in series, the current is equal everywhere, and the resistance voltage U1 = IR = 0.2A * 5 ohm = 1V
Then the bulb voltage U2 = u-u1 = 6v-1v = 5V, calculate the bulb resistance R = U2 / I = 5V / 0.2A = 25 Ohm,
Next, we start to calculate the rated voltage of the bulb. According to P = usquare / R, we deduce u = root P = root 0.25W * 25 Ohm = 2.5V

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