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What is the result in voltage of applying a 2.7 ohm resistor to a 12 volt DC circuit?
Wiki User
∙ 14y ago
The question has just stated clearly that the applied voltage is 12 volts DC.
Provided that the power supply is capable of maintaining its output voltage while supplying some current ... i.e. that the effective internal resistance of the power supply is small ... and that the 2.7 ohm resistor is the only external element connected to the power supply's output, the voltage across the resistor is exactly 12 volts DC.
The current through the resistor ... supplied by the 12 volt DC supply ... is 12/2.7 = 4.44 Amperes (rounded).
The power dissipated by the resistor ... supplied by the DC supply ... is 122 / 2.7 = 53.23 watts !
Wiki User
∙ 14y ago
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If the resistance in the circuit is increased what will happen to the current and voltage?
* resistance increases voltage. Adding more resistance to a circuit will alter the circuit pathway(s) and that change will force a change in voltage, current or both. Adding resistance will affect circuit voltage and current differently depending on whether that resistance is added in series or parallel. (In the question asked, it was not specified.) For a series circuit with one or more resistors, adding resistance in series will reduce total current and will reduce the voltage drop across each existing resistor. (Less current through a resistor means less voltage drop across it.) Total voltage in the circuit will remain the same. (The rule being that the total applied voltage is said to be dropped or felt across the circuit as a whole.) And the sum of the voltage drops in a series circuit is equal to the applied voltage, of course. If resistance is added in parallel to a circuit with one existing circuit resistor, total current in the circuit will increase, and the voltage across the added resistor will be the same as it for the one existing resistor and will be equal to the applied voltage. (The rule being that if only one resistor is in a circuit, hooking another resistor in parallel will have no effect on the voltage drop across or current flow through that single original resistor.) Hooking another resistor across one resistor in a series circuit that has two or more existing resistors will result in an increase in total current in the circuit, an increase in the voltage drop across the other resistors in the circuit, and a decrease in the voltage drop across the resistor across which the newly added resistor has been connected. The newly added resistor will, of course, have the same voltage drop as the resistor across which it is connected.
What happens to the current in a circuit as a capacitor charges?
What happens to the current in a circuit as a capacitor charges depends on the circuit. As a capacitor charges, the voltage drop across it increases. In a typical circuit with a constant voltage source and a resistor charging the capacitor, then the current in the circuit will decrease logarithmically over time as the capacitor charges, with the end result that the current is zero, and the voltage across the capacitor is the same as the voltage source.
If the resistance of an electric circuit is 12 ohms and the voltage in the circuit is 60 volts the current flowing through the circuit?
V = I times R where V = voltage, I = current and R = resistance. Further, I = V / R.As I = V / R, I = 60 /12 = 5 amps.V=IR , where V=60 volts R=12 ohms so I = V/R = 60/12 = 5 Amp.
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 minimum voltage necessary to operate a light emitting diode?
That will vary with the type (i.e. "color") of the LED: IR LEDs operate at the lowest voltage (1.2V), red LEDs operate at low voltages (1.8V), green LEDs operate at medium voltages (2.5V), blue LEDs operate at high voltages (3.3V), and UV LEDs operate at the highest voltages (4V). Intermediate color LEDs operate at correspondingly intermediate voltage between those given above. The reason an LED cannot produce light below these voltages is it takes more voltage drop to get the energy to produce higher energy photons and the different types of binary semiconductors needed to produce each color/energy of photon result in different junction forward bias voltages.However LEDs are really current operated devices, not voltage operated devices, so they need a series resistor or a current source to limit the current through them. Simply applying a voltage source with the necessary "minimum operating voltage" across an LED will generally destroy it instead of lighting it.
If the resistance in the circuit is increased what will happen to the current and voltage?
* resistance increases voltage. Adding more resistance to a circuit will alter the circuit pathway(s) and that change will force a change in voltage, current or both. Adding resistance will affect circuit voltage and current differently depending on whether that resistance is added in series or parallel. (In the question asked, it was not specified.) For a series circuit with one or more resistors, adding resistance in series will reduce total current and will reduce the voltage drop across each existing resistor. (Less current through a resistor means less voltage drop across it.) Total voltage in the circuit will remain the same. (The rule being that the total applied voltage is said to be dropped or felt across the circuit as a whole.) And the sum of the voltage drops in a series circuit is equal to the applied voltage, of course. If resistance is added in parallel to a circuit with one existing circuit resistor, total current in the circuit will increase, and the voltage across the added resistor will be the same as it for the one existing resistor and will be equal to the applied voltage. (The rule being that if only one resistor is in a circuit, hooking another resistor in parallel will have no effect on the voltage drop across or current flow through that single original resistor.) Hooking another resistor across one resistor in a series circuit that has two or more existing resistors will result in an increase in total current in the circuit, an increase in the voltage drop across the other resistors in the circuit, and a decrease in the voltage drop across the resistor across which the newly added resistor has been connected. The newly added resistor will, of course, have the same voltage drop as the resistor across which it is connected.
What happens to the current in a circuit as a capacitor charges?
What happens to the current in a circuit as a capacitor charges depends on the circuit. As a capacitor charges, the voltage drop across it increases. In a typical circuit with a constant voltage source and a resistor charging the capacitor, then the current in the circuit will decrease logarithmically over time as the capacitor charges, with the end result that the current is zero, and the voltage across the capacitor is the same as the voltage source.
Why is series resistor in all bias experiment?
biasing resistor is important because the voltage passing through it will limit the current and derive the next device, i.e. transistor etc. when a signal is applied to this circuit, biasing resistor helps to signify that signal and as a result we can examine our output.
How do you measure current without an amp meter?
Depends on the current. Put a resistor in-line with the current, then measure the voltage across the resistor. V=RI. So, divide the measured voltage by resistor value. Be careful with the size of the resistor, as Power dissipated in a resistor is R*I^2 or V^2/2. So, a 1-Amp current into a 1 Ohm resistor will result in a 1Watt power dissipated in the resistor. If it's too small, it'll burn. Also, notice that if you do that, you haven't measured the current in the original circuit. You've measured the current when an extra resistor is installed in the original circuit, and that's different.
What is a passive element in a circuit?
A passive element is an element of the electrical circuit that does not create power, like a capacitor, an inductance, a resistor or a memristor.
What arrangement of three cells would result in the greatest electric current?
I haven't studied this for awhile, but... I assume by cell, you mean a voltage supply, like a battery. It depends on what else is in the circuit. If your circuit has a typical amount of resistance, then connecting the cells in series (as opposed to in parallel) will result in the largest voltage. Higher voltage means greater current across a resistor. However, if the resistance of the circuit is very low (like in a short circuit), then your batteries' own internal resistance may be the most significant factor, and batteries arranged in parallel may be able to sustain a higher current.
Why you use resistor in a circuit?
I=V/R, this is a most basic equation to use to understand what a resistor is and what it does, from the equation it can be seen that an increase of V, will result in an increase of I, also an increase of R will result in a decrease of I, because of that, resistors are used to reduce I so as to control the level of V in a circuit.
How do you increase resistance?
Since resistance is the ratio of voltage to current, if the voltage is constant then increasing the resistance will result in a reduction in current.
How to do the megger test in electric motor?
Briefly, a megger test is testing an electric circuit at load. The circuit might prove good at normal resistance measuring but fails when applying high voltage. Read the megger result correctly, every electric circuit will fail if stressed high enough. Disconnect any servo drive or frequency controller before testing.
What happens when you halve the resistance in a circuit when voltage remain constant?
Since resistance is the ratio of voltage to current, we can say that halving the resistance will result in twice the current.
If current drop does it effect the voltage?
Generally, yes. It would depend on the device you are talking about. In a resistor a change in current will result in an instant change in voltage. Inductors and capacitors do not change instantaneously but will over time.
What has a larger current the parallel circuit with 1 resistor or the parallel circuit with 2 resistors?
By Ohm's law, current is voltage divided by resistance.In order to determine which circuit draws more current, you need to look at voltage and resistance. Assuming similar voltage, then, less resistance would result in more current.Now you need to know the values of the resistors in both cases. You did not state those important pieces of information. The net resistance of two resistors, R1 and R2, in parallel is R1 time R2 divided by (R1 plus R2). All you need to do is calculate for the two cases.If the two resistors were the same, you can generalize the answer by saying that, with contant voltage, two similar resistors in parallel will pull more current than one similar resistor - specifically, two will pull twice the current of one.
