To calculate the power rating of a resistor required to drop the voltage from 5V to 4.5V using a 30 ohm resistor, we can use the formula:
Power = Voltage Drop x Current
The current flowing through the resistor can be calculated using Ohm's law:
Current = Voltage Drop / Resistance
Substituting the values we get:
Current = (5V - 4.5V) / 30 ohms = 0.0167 A (rounded to four significant figures)
Now we can calculate the power required:
Power = (5V - 4.5V) x 0.0167 A = 0.00835 watts
So a resistor with a power rating of at least 0.00835 watts (or 8.35 milliwatts) should be sufficient for this application. However, it is recommended to use a resistor with a slightly higher power rating to ensure that it can handle any temporary power surges or variations in current that may occur. A 0.25 watt resistor should be suitable for this purpose.
A resistor in parallel with a voltages source will not cause the voltage to drop, theoretically. To get a 20 volt drop you need a resistance in series, and the number of ohms is 20 divided by the current in amps. If the current is unknown or variable, the voltage can't be dropped by using a resistor.
Normally through the resistor's internal construction. It flows through any part of the resistor that has low resistance- be it anywere. And then there's this. It might be that one should consider that current flows through a resistor and voltage is dropped across a resistor. Perhaps this is where the question began. The former is fairly straight forward. The latter can be vexing. Voltage is said to be dropped across a resistor when current is flowing through it. The voltage drop may be also considered as the voltage measureable across that resistor or the voltage "felt" by that resistor. It's as if that resistor was in a circuit by itself and hooked up to a battery of that equivalent voltage.
Voltage division is a very complicated way to change the voltage supplied to a piece of equipment. You need to install a transformer, if the difference is significant, and transformers should only be installed by qualified professionals.
It does not matter. Kirchoff's Current Law states that the signed sum of the currents entering a node is zero. A consequence of that law is that the current in every part of a series circuit is the same. The only thing that resistor location affects is the potential voltage of the LED terminals with respect to the rest of the circuit. Certainly, if you are driving the LED with high voltage, such as 120VAC, you should consider the resistor location so as to reduce electrocution hazard but, the LED's performance is not affected by resistor location in the circuit.
You reduce voltage in a DC circuit with a resistor. If the power demands are high, however, you may need a switching voltage regulator.
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.
No, the meter used to make the measurement will be applying whatever voltage to the resistor it needs to make the measurement. Any additional voltage will disturb this, resulting in at least an incorrect reading and at worst damaging or destroying the meter.
Has nothing to do with the intensity of the LED, and all to do with the voltage/amperage of thediode, and the voltage of the system it is supposed to be used with.
There is a thing called the resistor box that drops the voltage through the switch leaving the full voltage to run on number 4 of the switch. Locate this resistor and replace, this should fix problem.hope this helps
The rf output voltage should be proportional to the signal voltage in AM. A change in the DC supply voltage should also cause a proportional change to the rf output voltage.
If you mean the HVAC fan, the resistor should be located within the fuse block.
the blower motor resistor should be behind the glove box. just unscrew the two screws and that should be it.