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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.
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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.
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 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.
Is a miliohm 1000 ohms?
No, a miliohm is 0.001 ohms. A kiloohm is 1000 ohms.
What resistor to drop 2V to 1 1.5V?
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).
What happens to the current flowing through a metal resistor when the voltage across is increased?
When the voltage is increased across a metal film resistor, the current flow will also increase. Ohm's law states that the current flowing through a resistor is directly proportional to the voltage across that resistor. I = V/R Let us assume an initial voltage drop across a 4.99K ohm metal film resistor is 5V. The current flow through the resistor is calculated to be: I = 5/4990 = 0.001 Amps or 1 mA If that voltage were to say double to 10V: I = 10/4990 = 0.002 Amps or 2 mA Using these values it is also possible to calculate the power dissipated by the resistor. P = I*V = 0.002 * 10 = 0.02 Watts This power calculation determines the minimum physical case size needed for the resistor to function within these conditions. Anything smaller, the resistor will fail.
How current passes through resistor.?
What is the current running through resistor four?1 amps..!What is the current running through resistor one? 3 amps...!What is the current running through resistor three? 2amps..!What is the current running through resistor five? 3 amps..!What is the voltage drop running through resistor five? 45 volts...!What is the equivalent resistance through the parallel portion of the circuit? 6 ohmsAnswerA resistor is a conductor, albeit one with a higher resistance than a length of wire, so current passes through it without any problem. The magnitude of the current will, of course, be somewhat lower because of the additional resistance.
What is the voltage drop through a resistor of 6 ohms(2amps)?
Assuming the these resistors are connected in series, the voltage drop across each resists can be calculated simply using V = IR. In a series circuit, the same current must flow through each resistor. So, 2*i + 2*i + 2*i = 6 volts. or i * 6 = 6, or i = 1. So, the voltage drop across all resistors i*r is 2 volts.
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 are the three great truths about circuitry?
1. The sum of the component voltage drops in a series circuit is equal to the voltage at the source.Vs=V1+V2+...+Vn (s=source, n=total number of voltage drops in the circuit)2. The greater the resistance imposed by a component, the greater the voltage drop across it.Larger resistor=larger voltage drops, Smaller resistor=smaller voltage drops3. In a series circuit, the percentage of resistance contributed by a component is equal to the percentage of voltage dropped by that component.
What is a potential divider and what does it do?
The most common type is resistive division. Say we have a source of V volts and we wish to apply a portion of the V volts (aV where a is between 0 and 1) to a load of resistance R. We can place a resistor R2 in series with load R to drop off the unwanted part of the voltage V. The math is a = R / (R + R2). This only works if the resistors are constant. For a changing load other methods are necessary. Say the load needs 5 volts and it draws 10 milliamperes at 5 volts, but the supply is 10 volts. The effective load resistance is 500 ohms. A 500 ohm resistor in series with the load would drop off the unwanted 5 volt excess. Suppose the load changes by 10%, but we wish to hold load voltage to 5%. We can place a 500 ohm resistor in parallel with the load and use a 250 ohm resistor to drop the voltage to 5 volts. The voltage change is now only 5% for a 10% load change.
How do I measure AC current using microcontroller.?
1. PIC16F877A 2. CT 3.difference amplifier DIFFERENCE AMPLIFIER: To measure this current with pic microcontroller,we have to use ADC module of PIC microcontroller.To use ADC module we will convert current into voltage form by using a .1 ohm shunt resistor across CT and we will measure this voltage drop across shunt resistor.Then this voltage drop can be easily converted into current again.For example voltage drop across .1 ohm shunt resistor =8v then current according to ohm law V=IR I=V/R I=8/.1=8A but the problem is ADC of pic microcontroller can never measure voltage greater than 5 volt. so to solve this problem we can use differenceamplifier. By adjusting gain of difference amplifier we can reduce voltage lower than 5 volts.
