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Not enough information. But if you know the current, you can use Ohm's Law (just multiply the resistance times the current).
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What is a maximum voltage across a resistor?
The voltage across the resistor is whatever voltage is applied. The only maximum here would be a voltage that would damage the resistor. If you think this might happen, you'll have to look up such a voltage from the data sheets.
What is the voltage across the first ohm resistor?
if R4 is the only resistor (the load), then the drop would be the same as the energy source
How would you connect a resistor in order to reduce an LED's applied voltage?
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.
When resistors are connected in series in a circuit. what are the relationships between the voltage drops across the resistor and the currents through the resistors?
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
Three identical bulbs are connected in parallel across a 12-V battery If one of the bulbs were removed from its socket then the potential difference measured across the empty socket would be?
12V. Every resistor in a parallel circuit shares the same voltage. It is the current that gets divided.
What is a maximum voltage across a resistor?
The voltage across the resistor is whatever voltage is applied. The only maximum here would be a voltage that would damage the resistor. If you think this might happen, you'll have to look up such a voltage from the data sheets.
The voltage drop across a resistor is 1.0 V for a current of 3.0 A in the resistor what is the current that will produce a voltage drop of 9.0 V across the resistor?
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.
What is the voltage across the first ohm resistor?
if R4 is the only resistor (the load), then the drop would be the same as the energy source
Why do resistor voltage decrease while capacitor discharges?
The reason why resistor voltage decreases while a capacitor discharges is because the resistor acts like a source of electrical energy. As the capacitor discharges, it draws energy from the resistor, which causes the voltage across the resistor to decrease. This is because the capacitor is acting like a drain, and is taking energy out of the resistor, thus causing the voltage across the resistor to decrease. The resistor and capacitor work together in order to create a discharge circuit. This is done by connecting the capacitor to the resistor, and then to a voltage source. The voltage source supplies the energy to the resistor, and then the resistor transfers this energy to the capacitor. As the capacitor discharges, it takes energy from the resistor, which causes the voltage across the resistor to decrease. In order to understand this process better, it is important to understand the basics of Ohm's Law. Ohm's Law states that the voltage across a resistor is equal to the current through the resistor multiplied by the resistance. As the capacitor discharges, it takes energy from the resistor, which means that the current through the resistor decreases, and therefore the voltage across the resistor will also decrease.
How would you connect a resistor in order to reduce an LED's applied voltage?
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 the volts going in equals 5 and there is a resistance that equals 6 what would be the volts coming out?
The question is a bit ambiguous, but I will try to address it. If the 6 ohm resistance is in series with another resistance then some of the 5 volts would be dropped across the 6 ohm resistance and the remainder of the voltage would be dropped across the other resistance. To calculate the voltage, use the 'resistor voltage divider equation' (Google it). If the 5 volts is applied across only a 6 ohm resistance, then the top of the resistor is at 5 volts and the bottom of the resistor would be at 0 volts. The resistor would drop all of the voltage.
What is the difference between an inductive low pass filter and a capacitive low pass filter?
Where you are measuring. A simple filter will be two elements - a capacitor or inductor and a resistor. A capacitor will tend to "trap" low frequencies. In the case of a lowpass filter made of a capacitor and resistor, the output voltage will be measured across the capacitor. Inductors are the opposite, so the output would be across the resistor.
When resistors are connected in series in a circuit. what are the relationships between the voltage drops across the resistor and the currents through the resistors?
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
Three identical bulbs are connected in parallel across a 12-V battery If one of the bulbs were removed from its socket then the potential difference measured across the empty socket would be?
12V. Every resistor in a parallel circuit shares the same voltage. It is the current that gets divided.
How do you connect and measure voltage using a voltmeter?
with a voltomiter
What is the voltage drop per led in a 25 led series circuit with 120 volt supply and leds are 3.0 volts 20 ma with a resistor of 5000 ohms?
There is not enough information to answer this question correctly. As the current increases through a diode, the voltage dropped across it increases. This correlation is not a linear function such as a resistance, and a datasheet for the diode or a curve tracer would be necessary to obtain the correct function. To explain further, if there were 25 series LEDs all with a 3 volt drop across them, a 75 volt potential would be measured across the diodes, and 45 volts would be measured across the 5000 ohm resistor. Ohms law will show that 45 volts applied to 5K ohms will equal 9mA. This violates the 20mA criteria in the question.
What size of resistor is required to operate the LED from a 9V battery?
In order to determine what size of resistor is required to operate an LED from a 9V battery, first start by knowing the current and voltage required for the LED. That information is available in the LED's specifications. For discussion purposes, lets assume a typical LED at 2.5V and 50mW. The translates to a forward current of 20mA. Build a simple series circuit containing a 9V battery, a resistor of an as yet unknown value, and the LED. By Kirchoff's current law, the current in the LED is the same as the current in the resistor, which is also the same as the current in the battery. This is 20ma. By Kirchoff's voltage law, the voltage across the LED plus the voltage across the resistor equals the voltage across the battery. This is 6.5V. (9 - 2.5) By Ohm's law, resistance is voltage divided by current, so the resistor is 6.5 / 0.02, or 325 Ohms. The nearest standard value to that is 330 Ohms. Cross check the power through the resistor. Power is voltage times current, or 6.5V times 0.02A, or 0.13W. A half watt resistor is more than adequate for this job.
