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Doing it with a single resistor is not a good idea because it can only be done with exactly the right amount of current.
If the current is 1 amp, for a voltage drop of 10.8 volts you need 10.8 ohms (volts/amps).
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What is the voltage drop running through resistor 1 resistor 1 equals 3 ohms?
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.
What is the 6 x 6 D matrix for an isotropic linear elastic body?
D=E/((1+v)(1-2v))*[1-v v v 0 0 0; v 1-v v 0 0 0; v v 1-v 0 0 0; 0 0 0 0.5(1-2v) 0 0; 0 0 0 0 0.5(1-2v); 0 0 0 0 0 0.5(1-2v)]
What resistor to use to drop 9volt's dc to 6 volt's dc?
To drop a 12 volt source to 6 volts with a resistor, you have to drop 6 volts. The value of the resistor you need would be 6 divided by the current the device pulls in amps. For example, if the device pulls a half an amp the resistor has to be 6/0.5 or 12 ohms. As this device runs on 6 volts and draws 1/2 amp, it's wattage is 3 watts (volts x Amps). Common practice is to double this, or the resistor will probably get too hot and may open. I'd use a 10 watt to resistor to maintain a good margin for safety, and they're readily available. Use a 12 ohm, 10 watt resistor.
Relationship between bulk modulus and poissons ratio?
K=E/(3*(1-2v)) K: Bulk modulus E: young modulus v: poison's ratio on the other hand: delta V/V=(1-2v)*delta L/L relative change in Volume equals to: (1-2v) * relative change in length.
What is the function of a cement resistor?
A cement resistor is typically used as a power resistor (a resistor whose power rating is greater than 1 W).
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 drop running through resistor 1 resistor 1 equals 3 ohms?
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.
What size resistor to drop 9 dc to 3 dc at 1 amp?
330 milli ohms
Can someone explain how to differentiate 1400 v plus 2v 7?
Do you mean 1400v + 2v^7 (is 2v to the power 7)? When you differentiate, e.g. 2v^7, you bring the 7 to the front and reduce the power by 1. If so the answer is 1400 + 7*2v^6 (where * means multiply) or 1400 + 14v^6
What is the 6 x 6 D matrix for an isotropic linear elastic body?
D=E/((1+v)(1-2v))*[1-v v v 0 0 0; v 1-v v 0 0 0; v v 1-v 0 0 0; 0 0 0 0.5(1-2v) 0 0; 0 0 0 0 0.5(1-2v); 0 0 0 0 0 0.5(1-2v)]
What resistor to use to drop 9volt's dc to 6 volt's dc?
To drop a 12 volt source to 6 volts with a resistor, you have to drop 6 volts. The value of the resistor you need would be 6 divided by the current the device pulls in amps. For example, if the device pulls a half an amp the resistor has to be 6/0.5 or 12 ohms. As this device runs on 6 volts and draws 1/2 amp, it's wattage is 3 watts (volts x Amps). Common practice is to double this, or the resistor will probably get too hot and may open. I'd use a 10 watt to resistor to maintain a good margin for safety, and they're readily available. Use a 12 ohm, 10 watt resistor.
What Is v in v plus 1 equals 2v plus 2?
v = -1
Why 250 ohm resistance using for transmitter calibration?
In this situation, to calibrate a transmitter you need a power circuit and communicator circuit. The Hart communicator used in the calibration process is connected to the power source circuit in parallel. The power source circuit is the one that has ammeter, 250 Ohm resistor, and power source all connected in series. As the transmitter sends output mA, it creates volt drop across the 250 Ohm resister. Let's say the volt drop across the resistor was 1 Volt. Now, back to the Hart communicator. It is a load, meaning there will be a volt drop across the Hart communicator. Since it is in parallel with the power circuit, it is also parallel with the resistor. So, the 1 volt drop across the 250 Ohm resistor will also make 1 volt drop across the Hart communicator. Technically speaking, the 1 volt drop across the Hart communicator is only true if its resistor is also 250 Ohm. However, it does NOT matter what voltage drop is in the Hart communcator. It only sees the "relative" voltage drop changes to measure the changes in transmitter outputs.
Relationship between bulk modulus and poissons ratio?
K=E/(3*(1-2v)) K: Bulk modulus E: young modulus v: poison's ratio on the other hand: delta V/V=(1-2v)*delta L/L relative change in Volume equals to: (1-2v) * relative change in length.
How much ohms resistor is needed to drop 12 volts to 5 volts?
The size of the resistor will depend on the load. Let's look at this a bit to see if we can make sense of it. You want to drop the applied voltage to a device from 12 volts AC to 11 volts AC. That means you want to drop 1/12th of the applied voltage (which is 1 volt) across the resistor so that the remaining 11/12ths of the applied voltage (which is 11 volts) will appear across the load. The only way this is possible is if the resistor has 1/11th of the resistance of the load. Here's some simple math. If you have an 11 ohm load and a 1 ohm resistor in series, you'll have 12 ohms total resistance ('cause they add). If 12 volts is applied, the 1 ohm resistor will drop 1 volt, and the 11 ohm load will drop the other 11 volts. A ratio is set up here in this example, and each ohm of resistance will drop a volt (will "feel" a volt) across it. See how that works? If the resistance of the load is 22 ohms and the resistance of the (series) resistor is 2 ohms, each ohm of resistance will drop 1/2 volt, or, if you prefer, each 2 ohms of resistance will drop 1 volt. The same thing will result, and the load will drop 11 volts and the series resistance will drop 1 volt. That's the math, but that's the way things work. You'll need to know something about the load to select a series resistance to drop 1/12th of the applied voltage (which is 1 volt) so your load can have the 11 volts you want it to have. There is one more bit of news, and it isn't good. If your load is a "dynamic" one, that is, if its resistance changes (it uses more or less power over the time that it is "on"), then a simple series resistor won't allow you to provide a constant 11 volts to that load. What is happening is that the effective resistance of the load in changing over time, and your resistor can't "keep up" with the changes. (The resistor, in point of fact, can't change its resistance at all.) You've got your work cut out for you figuring this one out.
What is the voltage drop through resistor 1?
There is insufficient information in the question to properly answer it. You don't state the value of resistor 1, nor do you state how it is connected in the circuit, nor do you state the value of the other components in the circuit. Please restate the question, giving more details.
What is the voltage drop through a resistor of 6 ohms and 2 amps?
12 volts...! The voltage drop across a 2 ohm resistor depends on the current flowing through it. As voltage (E) equals current (I) times resistance (R), if 1/2 amp is flowing through your 2 ohm resistor, 1/2 times 2 = 1 volt. If 1 amp is flowing through your 2 ohm resistor, 1 times 2 = 2 amps. Piece of cake. If the two ohm resistor is the only component in the circuit, it will drop whatever the applied voltage is. Put a 2 ohm resistor across a 6 volt battery, it drops 6 volts. If you put your 2 ohm resistor across a 9 volt battery, it drops 9 volts. Another way to say voltage drop may help. The voltage drop across a resistor is the voltage it "feels" when in a circuit. And that last couple of examples says that very well. In a circuit where a given resistor is the only component, it drops all the voltage in the circuit. It "feels" all the voltage in the circuit. In a circuit where there are 2 resistors of equal value in series, each one drops or "feels" half of the applied voltage. (The sum of the voltage drops equals the applied voltage.) As you work more with simple circuits using resistors in different arrangements with a given voltage source, try thinking of the voltage drop of a resistor as the voltage it "feels" when the circuit is energized.
