When you connect 230V ac supply directly to a diode with the help of resistor of the order kilo ohms, the diode will conduct on alternate half cycles. Forward bias current will be 230ma for 1KOhm, or 23ma for 10KOhm. Power across the resitor will be 25W for 1KOhm, or 2.5W for 10KOhm. Bottom line - the resistor will get very hot - and if not rated correctly, will be destroyed.
The protecting resistor is put in series with the LED so that you have a voltage divider - the supply voltage is split across the LED ( max 0.6v) and the remainder across the protecting resistor. So if your supply is 6volts, 5.4v will be across the resistor,
By using a voltage divider, that is two resistors of the same value in series across the DC supply. Half of the supply voltage will be at the point where the two resistors is connected. But how much wattage of those resistors is also an issue.
Usually you'll need 100ohm resistor for each led. In this kind of setting, you will get the LED intensity nonuniform because of its differences due tolerance value. the best thing is to use LED driver IC. Another benefit of using LED driver is, you can mix red led with other color. Now, if we wanted to connect a red LED with a forward voltage of 1.8V and a current of 20mA directly to the transformer output then use the following equation: R=V/I = 9 / 0.02 = 450 Ohms. You could use 470 ohms from the E12 resistor series.
If there's nothing else between the ends of the resistor and the power supply, then the voltage across the resistor is 24 volts, and the current through it is 2 amperes.
Theoretically, you could connect three identical lamps in star (wye) and connect this to the three line conductors. But as you asked how to connect a (single) 230-V lamp to the supply, then the answer is by using a step-down transformer.
You get a violent explosion as the diode vaporizes.
First you connect a trnsformer to the main supply and then a rectfier .youmay connect a zener diode or a resistor or a capacitor as a filter circuit
The fans usually connect to the motherboard, but they can connect directly to the power supply unit.
you cannot directly interface a relay to a microcontroller, you need a buffer circuit in between the two, you can use a NPN transistor BC 547, connect the controller pin directly to the base of the transistor (since there is an internal 10k pull up resistor inside the controller no need of any base resistor). Connect the transistors emitter to the gnd & the collector to the relay coil, the another end of the relay coil goes to the supply, check this link for complete information http://www.dnatechindia.com/index.php/Tutorials/8051-Tutorial/Relay-Interfacing.html
The protecting resistor is put in series with the LED so that you have a voltage divider - the supply voltage is split across the LED ( max 0.6v) and the remainder across the protecting resistor. So if your supply is 6volts, 5.4v will be across the resistor,
If by power supply you mean a voltage source, it really won't matter that the resistor is removed. The voltage source will provide infinite current, instantly charging the capacitor so that the capacitor's voltage is equal to the source.Alternative AnswerIf you are referring to an a.c. circuit, then a load current will continue to flow with its value being determined by the capacitive reactance of the circuit, and the resulting phase angle will lead the supply voltage be very close to 90 degrees.
The motherboard does not connect the power supply to the peripherals. The power is supplied directly to most peripherals such as HDDs and CD drives by cables from the power supply. Some special expansion cards, such as network cards, get enough power from the motherboard through the PCI slot to function.
The electric heater is basically a resistor, designed to have the right resistance to draw the required current. So a 2 kW heater designed for a 230 v supply is really a resistor of 28.8 ohms, so when it's connected the current is 8 amps and the power is 2 kW.
You would not connect a current transformer to a 230 v supply. To get 5.6 v 12 mA you could get a 230 to 6 volt transformer, then drop the supply from 6 to 5.6 using a 33-ohm resistor.
Then the voltage in will equal the voltage out. The purpose of a resistor is to reduce the amount of electrical flow of current. You 'short out' the supply and blow a fuse/circuit breaker.
If there is nothing else in the circuit, then the voltage drop across the resistor will be the full supply voltage of 5 volts. The size of the resistor does not matter in this case - it will always be 5 volts.
Need more information - a typical regulated power supply would include dozens of resistors, each for one of several different reasons. Which resistor are you asking about?