You'll see a voltage drop across a resistor if current is flowing through it. It only
has to be a part of a complete circuit, i.e. one in which current is flowing.
You could use the voltage divider rule to reduce the voltage. Using two resistors in series, the input voltage will drop across each resistor by an amount that is proportionate to the values of the resistors. If the 1st resistor is 10K and the 2nd resistor is 100K, the voltage drop across the 10K will be 10 times LESS than that of the 100K resistor. The total voltage drop across both resistors will be equal to the supplied input voltage. Work out the ratio of voltage you need from the total input voltage and use 2 resistors will that give you the same ratio. Connect the LEDs in parallel with the resistor the gives you the voltage you want. Use a MM to measure the voltage across the resistor before wiring LEDs.
if R4 is the only resistor (the load), then the drop would be the same as the energy source
Kirchhoff's Voltage and Current Laws apply to circuits: series, parallel, series-parallel, and complex.If your circuit comprises just a single resistor, then they still apply. For example, the voltage drop across a single resistor will be equal and opposite the applied voltage (Kirchhoff's Voltage Law), and the current entering the resistor will be equal to the current leaving it (Kirchhoff's Current Law).
When resistors are connected in series in a circuit . the voltage drop across each resistor will be equal to its resistance, as V=IR, V is direct proportional to R. An A: The relationship is that the current will divide for each paths in a parallel circuit and the voltage drop across each will be the source voltage. In a series circuit the current will remain the same for each component but the voltage will divide to reflect each different component value. And the sum of all of the voltage drops will add to the voltage source
There is insufficient information in the question to answer it. You need to provide either the voltage across the resistor, or the power dissipated by the resistor. please restate the question.
It doesn't. In a series circuit, the largest voltage drop occurs across the largest resistor; the smallest voltage drop occurs across the smallest resistor.
The correct question is what is the voltage drop across a resistor or the current flowing through the resistor using Ohm's Law where Voltage = Current x Resistance
What is the amount of current flowing through the resistor? Voltage drop is dependent on the current. Ohm x Amps = Voltage drop
The resistor is 1/3 of an ohm. A 9 volt drop across the resistor would cause a draw of 27 amps through the resistor. The wattage you would need for that resistor is at least a 243 watts.
Any part of a circuit that has a voltage drop across it is a resistor.
when a resistor is connected in a circuit it drop some voltage across it.when a circuit have large input voltage then by using a resistor of suitable value we get the desired voltage.
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.
A resistor drops both voltage and current, however the term "drop" is generally used to indicate a voltage or current drop across the device, so it is more correctly stated that a resistor drops voltage, by allowing the current in the circuit to decrease.
Current flows in loops, voltage drops across elements. With relation to current, what flows in, must flow out, so no, current is not dropped across a resistor, it flows through a resistor and voltage is dropped across the resistor.
A: That resistor is there to limit the current to the LED it can be any value if the voltage is decreased or increased or no resistor if the voltage across the led is equal to the forward voltage drop.
You could use the voltage divider rule to reduce the voltage. Using two resistors in series, the input voltage will drop across each resistor by an amount that is proportionate to the values of the resistors. If the 1st resistor is 10K and the 2nd resistor is 100K, the voltage drop across the 10K will be 10 times LESS than that of the 100K resistor. The total voltage drop across both resistors will be equal to the supplied input voltage. Work out the ratio of voltage you need from the total input voltage and use 2 resistors will that give you the same ratio. Connect the LEDs in parallel with the resistor the gives you the voltage you want. Use a MM to measure the voltage across the resistor before wiring LEDs.
if R4 is the only resistor (the load), then the drop would be the same as the energy source