A 30 volt 90 watt lamp has 3 amps going through it. The series resistor also has 3 amps going through it, by Kirchoff's current law. The voltage across the resistor is 90 volts. With 3 amps, that is 30 ohms. (By the way... The resistor must be rated to carry 270 watts. That is a lot of power for a resistor.)
A: Since you know the current flow you need the voltage drop for this particular led at 20ma. All LED require a current and voltage to operate properly assuming a voltage drop of 2v then for a 12v source it becomes 12-2=10v 10v/.02=500 ohms in series to limit the current is required
An inductor will supply better current source.
It depends on the required output current, load rejection factor, and ripple. Also, efficiency enters into the picture.
When you connect a load (motor, light bulb, etc.) in series with a power supply, current will flow, whether AC or DC depends on the power supply you are using.
In a combination circuit, which includes both series and parallel components, the total supply voltage is distributed among the components. In series sections, the voltage is divided based on the resistance of each component, while in parallel sections, the voltage across each branch remains equal to the supply voltage. The total supply voltage remains constant throughout the circuit, but the voltage across individual components can vary depending on their configuration and resistance.
24 milli omps
a meter uses its own supply to measure resistance. you dont have to remove the resistor from the circuit but you must isolate it from the circuit supply to get an accurate reading.
It is required to supply chemical energy, organic molecules, and essential nutrients
+5v of supply is required
a resister is to RESIST current flow....if the LED gets the full effect of the power supply, the LED will immediately blow out.
Good question on a linear power supply a load is not really required since it is nothing but a low band power amplifier. ON a switching type power supply a minimum load is always required so the PWM has something to control to. Also the soft start circuit may blowup waiting for feedback .Critical circuit since it operates for a very short time. however the load is in parallel not in series.
To properly install a faucet water supply line, first turn off the water supply. Connect the line to the faucet and the shut-off valve using a wrench. Make sure the connections are tight to prevent leaks. Turn the water supply back on and check for any leaks. If there are no leaks, your faucet water supply line is properly installed.
A compression fitting is typically required to connect a refrigerator water line to the water supply.
No, the 1A power supply does not provide enough current (3200mA required) for the device to operate properly. You need a power supply that can deliver at least 3200mA to meet the device's power requirements. Using an underpowered supply may result in the device not functioning correctly or potential damage.
A stop switch is wired in series with the power supply and the load.
There are no example metric series in the question on which to supply a possible answer.
NO