The power of your generator is too low and the frequency is too low (your speed is 3000n/s Although you have increased the voltage, think about it
If the power is the same, the primary liter of 6v If it increases 33 times to 220V, then the 2 currents will be 1/33 of the 1 current Of course, the voltage you measured is zero, because it is equivalent to a no-current supply if it There is no current in a closed circuit, naturally there will be no voltage Although your idea is good, the equipment if the capacity is not good, you can add a. battery for floating charging Choose a small charging voltage and choose 6V the speed is enough, it should be fine (simple answer)
In a hydroelectric power plant, the water flow is 1m3/s. , and the drop is 5m, the generator efficiency is 50%, the output voltage is 350V, and the total resistance of the transmission line is 4Ω
The density of water is ρ, the output power of the generator: P out=
Voltage output U 0 = 240 V; transmission current is I, transmission power loss is P loss, P loss=
Get: I send=
The high voltage transmission voltage is U, U=
Then, therotation ratio of the primary and secondary coils of the step-up transformer is n 1:n 2 =240:5000=6:125; The voltage loss in the transmission line is the U loss. , U loss = I send r=10×30V=300V; The voltage reaching the step-down transformer is U drop, U drop=U send-U loss=5000-300V=4700V User voltage For U=220V, the ratio between the turns of the primary and secondary coils of the step-down transformer is n 3:n 4=4700:220=235:11; The power delivered to the user is the loss P=P -P; Suppose the power of the bulb is P. The maximum number of bulbs that can emit light normally is N, then N=
Answer: The ideal pitch rotation ratio -transformer up is 6:125; the transformation ratio of the step-down transformer is 235:11; The electric lamps used by users are all "220V, 100W", and up to 470 lamps of this specification can be used. |
(1) The output power of the generator is P＝
| mghη |
| t |
(2) The current flowing through the primary coil of the step-up transformer is
I1＝
| P < /td> |
| U1 |
According to the meaning of the question, the power lost on the transmission circuit is< /p>
P line=P×5%= I22R line
The current on the transmission line can be obtained as I2＝
Thus, the transformation ratio between the primary coil and the secondary coil of the step-up transformer is
(3) liter The output voltage of thevoltage transformer is U2=n2
The voltage at both ends of the primary coil of the step-down transformer is < p>U3=U2-I2R line= 1329VThe output voltage of the step-down transformer is known, that is to say the user voltage U4=220V, therefore the transformation ratio of the primary and secondary coil of the step-down transformer coil is
Answer: (1) The output power of the generator is 2.5 × 104 W; (2) The ratio of the primary and secondary coils of the step-up transformer is 1:4; (3) The ratio of the turns of the primary and secondary coils of the step-down transformer is of 6:1
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