Solution Process

It depends on the motor. In the case of general brush motors, they operate at considerably lower voltages. In the case of a 12V DC motor, it rotates very slowly even if it is over 1V. However, cooling fans, etc., have semiconductors inside. So, it does not operate at about 1V like a normal DC motor, and it operates only when it is at least about 4~5V. It is difficult to confirm whether it will run at 3V without directly experimenting because the minimum operating voltage is different for each product. ​ The voltage written on the motor is considered to be the highest efficiency at that voltage. Even if it is lower than that, it rotates, and even if it is slightly higher than that, it operates without problems, but the efficiency is lowered and the life span may be reduced. (when voltage rises) ​

DC motors can operate even below the rated voltage. However, there is a risk of overheating and burnout due to insufficient rotational speed or power. The motor can be used with the highest efficiency at its rated voltage.

1. I have a question.. For example, I usually use a DC 12V 0.16A motor. Can I drive this motor with a 3v battery..?? .. The previous specification was for a cooling fan, but in reality it runs with a 1.5V battery... Each motor has its own specification. Hey.. Does this make sense... It's the best, but less than that is ok.. I wonder... 1) All products have product specifications (specification sheet = SPEC). 2) Example: Rated voltage, rated current, and rated current consumption are basic. Characteristics according to voltage and current depending on the product Changes etc.. And sometimes even taboos.. 3) In general, it is common sense that the voltage operates normally within the ±10% error range of the rated voltage. 4) However, in the case of a DC motor, it is rare for a 12V motor to operate even if a voltage of 1/4 of 3V is supplied. ** Of course it turns, but the number of revolutions (RPM). Torque etc are far below spec. 5) Therefore, when the load is light, the speed is often controlled by adjusting the voltage.

Answer 2 It can be used in parallel with the 220V power circuit. First, 6V power must be supplied to the control circuit, so make a circuit that can rectify the voltage down (you can use an adapter DC 6V) and install a 6V DC relay instead of a 6V lamp. And you can use 220V lamp by using 6V relay contact. The important thing here is that the relay capacity should not be large. So, if necessary, if the capacity of the LAMP or electric furnace is large, use the relay contact again to install a large MAGNET, etc., and you can use it as you want.

Answer 1 I use a 6V relay in the place of the lamp. Just connect a 220V light bulb to the relay contact. Relay contact and PIC circuit are separated. There are no problems with the operation of the PIC. For TR protection, connect a rectifier diode (1N4004) in parallel to the relay coil. (The direction of the diode is 6V as the cathode)

If the input signal is +, negative voltage is not needed, but if + and - are swinging around 0V, positive power is needed to amplify both sides. And in most cases of OPAMP, both power supply is basic. When performing OPAMP experiments, it is recommended to use a linear type power supply rather than the same as SMPS. Schools and companies often use power supplies that support both power sources. http://www.powertm.co.kr/html/pro01_dcpower_01_ope_q.htm If you like decorating circuits, try making your own. If you use 7809 and 7909 instead of 7815,7915 above, you get ±9V. Otherwise, buy two 9V adapters and connect one side with + and the other side like this (+)(-) -- 0V -- (+)(-) If it is set to 0V (COM), it becomes positive power.

If you connect a conductor to the + pole of a power source with a high potential, the conductor will also have the same potential as the + pole. In other words, the + pole of the power supply and the conductor become equipotential. So, the free electrons in the wire do not show any movement. In order for free electrons to move in a conductor, the potential on one side of the conductor must be high and the other side must be low.

it's okay. Corona discharge in technical terms ​In corona discharge, if any part of the electrode exceeds the limit, insulation breakdown of the air occurs and there is no flashover. It means that the insulation of the air is partially broken and it discharges with a low sound and a faint light. In other words, corona is a local discharge phenomenon just before a spark discharge. ​The minimum voltage at which corona occurs is called the corona threshold voltage. The corona threshold voltage varies depending on the surface condition, thickness, and distance between the wires. When it rains or snows, the threshold voltage rises. As it descends, the insulation of the air is partially broken and a sound is generated or a faint light is emitted and the discharge occurs. This phenomenon is called corona discharge. ​ In particular, when it rains or snows, the threshold voltage goes down and corona discharge occurs more easily. ​ Therefore, since it has been raining a lot recently, it is a sound that appears because of it, so you can just leave it as it is.

R L C all can be said to be a load, ​ The only load that consumes power is R.

I'll explain as far as I know. What is electronic imbalance? This is in line with the meaning of electro-magnetic imbalance. One circuit in the metal tube, for example, if it is a single-phase 2-wire type, all 2-wires are 3-phase, and if it is a 3-wire type, it means all three wires. Simply put, when electricity flows through a wire, it creates magnetism around it. The magnetism then affects the other wires. So if you put the two wires together In other words, if you put + and - wires together, they cancel each other out to maintain equilibrium. If not, it will be unbalanced. In case of 3-phase 3-wire system, if the voltage or current flowing through each wire is different, The equilibrium may be broken. We are also using a method called transposition as a measure to prevent imbalance of electric wires in pylons.

A blown fuse indicates that something is wrong internally. The role of the fuse is as a last resort to prevent secondary damage (fire or more serious failure) due to an abnormality occurring inside the product, causing an overcurrent to flow. ​ If you keep replacing the fuse even though the fuse is blown, what can be easily repaired becomes more and more serious. Therefore, if the fuse is blown, it is not necessary to replace the fuse alone, but to find the cause of the overcurrent flow, repair it, and then replace the fuse. And never change the fuse capacity. ​

Second Answer You don't need to get electricity by dangerously detonating hydrogen. The picture above is a hydrogen fuel cell made by Hyundai. It can produce about 95kW of electricity, so in terms of horsepower, it can produce about 127 horsepower, which is a level that can supply about 200 households of electricity. There is no need to create electricity by exploding hydrogen that is difficult and dangerous. It is a fuel cell structure using hydrogen and oxygen. ​

First answer A method of generating electricity by combining (reacting) hydrogen and oxygen is called a fuel cell. It is used in submarines and is also used as an emergency generator in large buildings. What is important here is the reaction rate, that is, whether humans can control the reaction rate. In general, the word "explosion" refers to a time when you can't control your reaction rate and react on its own, so your question needs to be modified a bit. Explosions cannot generate electricity. Any chemical or physical explosion can produce electricity (with certain devices added). It's just a matter of efficiency, so we don't actually produce it.

The power supply has zero impedance (=resistance), so try shorting the left Vin and GND. Then R1 and R2 are connected in parallel. Therefore, the resistance value seen from Vout becomes R1//R2 (// means parallel connection), so it becomes 50kohm.

Another interpretation. If high-pass filters with the same high-pass cutoff frequency are cascaded to two stages, the high-pass cutoff frequency becomes the same as when connected to one stage. So the answer is (4).

The formula for the total bandwidth for a two-stage amplifier with the same bandwidth is a bit complicated. Since it is two-stage above, if you put n = 2, you get 500k x 0.643 = 321.79kHz, so the answer is number 3.

another interpretation If you come across this problem a lot, you have to memorize it to some extent. 20dB at voltage and current = 10 times 3dB = 1.41x 6dB = 2x Memorize this much. The dB just adds up. So the total is 60dB. Since 20 dB is 10 times, 60 dB is 1000 times. So the output will be 1V.

The answer is number one.

second answer For resistors in series, R = R1 + R2 + R3 +... In parallel, 1/R = 1/R1 + 1/R2 +1/R3 + .... An inductor is like a resistor. ​ For capacitors in series, 1/C = 1/C1 + /1C2 + 1/C3 + .... In parallel, C = C1 + C2 + C3 + ..... ​ I think you're confused because the calculations for capacitors and resistors are opposite to each other.