It is not 330 ohms it is a value that must be installed in series to insure a 20ma or so of current limiting. This value of 330 ohms can change greatly if the source voltage changes greatly. It also has to be noted that the LEDS are not created equally So each LED voltage drop can be added to the equation.
You need to use the 250 Ohm resistor in series with HART protocol communication because it acts as a shunt resistor.
To find the current through the 40 ohm resistor, first calculate the total resistance of the parallel circuit: 1/Rt = 1/120 + 1/60 + 1/40. Then, calculate the total current using Ohm's Law, I = V/Rt. Finally, use the current divider rule to find the current passing through the 40 ohm resistor.
You can only use a resistor to drop a voltage at a constant current. If you know the current, use Ohm's law to calculate the resistor value.
To calculate the current running through the 60 ohm resistor in a parallel circuit, you first need to find the total resistance of the circuit. For a parallel circuit, the reciprocal of the total resistance (1/RT) is equal to the sum of the reciprocals of the individual resistances (1/R1 + 1/R2 + 1/R3). Once you find the total resistance, you can use Ohm's Law (I = V/R) to calculate the current running through the 60 ohm resistor.
1). 6V battery, 1-ohm resistor, 2-ohm resistor, all in series:Total resistance = 3 ohms.Current in the loop = 6/3 = 2 amperesPower dissipated by the 2-ohm resistor - I2R = 8 watts.2). 4V battery, 12-ohm resistor, 2-ohm resistor, all in parallelThe 12-ohm resistor is irrelevant.4 volts across the 2-ohm resistor.Power dissipated by the 2-ohm resistor = E2/R = 8 watts.
To calculate the total current in the circuit, you first need to find the total resistance by using the formula for resistors in parallel: 1/Total Resistance = 1/120 + 1/60 + 1/40. Once you have the total resistance, you can use Ohm's Law: Current = Voltage / Total Resistance. Finally, to find the current running through the 40 ohm resistor, you divide the total current by the resistance of the 40 ohm resistor.
If they're in parallel, then each resistor acts as if it were the only one,and the presence of any others is irrelevant.The current through the 60-ohm resistor is I = E/R = (120/60) = 2 amperes.
To limit the overload current of an LED, you can use a current-limiting resistor in series with the LED. The resistor value can be calculated using Ohm's law, considering the LED's forward voltage and desired current. Alternatively, constant current drivers or LED driver circuits can be employed to automatically regulate the current flowing through the LED, providing more efficient and reliable operation.
Assuming that you're talking about 12V DC you would use a 4 ohm resistor. If you mean AC then you would need a step-down transformer with a 4:1 ratio.
You should use the same size resistor; a 47 kilo-ohm.CommentThe correct spelling is 'kilohm'.
if you have 12v dc then you can use a 4 ohm resistor
How can you see a 1 ohm resistor on a 100k scale ?