Physics question about voltage dividers?
If you are referring to a variable resistor like a potentiometer then just think of ohm's law where current = voltage/resistance. If resistance is decreased then the voltage drop will decrease which means that current will increase. Otherwise, if the variable resistor decreases to 0, then you will have all of the current going through that branch and a short circuit will occur because electricity follows the path of least resistance. If the short circuit occurs, and you have an element, like a light-bulb, then the light bulb will not work because it needs to have current, and current only exists when there is a difference in potential caused by a voltage drop which is only caused by a resistor. Possibly, if this is a parallel branch diagram, all the current will flow through the lower resistor and none will flow through the higher resistor. Again this possibility is if this is a parallel and not a series branch.
Voltage divider problem?
I get a series string of two resistors: 180 ohms and 60 ohms, where the load is placed across the 60 ohms. Unloaded, the voltage across the 60 ohm resistor will be 6 V with a current is 100 mA [ No load current = 24 V divided by (180 + 60) ohms]. With a 1k ohm load, the current will increase to 101 mA, giving about 5.7 volts. 6V out with 24 V in leaves 18 V across the remainder of the voltage divider. 100 mA unloaded implies 6V/100mA = 60 ohms. 18V/100 mA = 180 ohms.
Design of Voltage Divider Circuit?
Engineering Problem Solving-Design of Voltage Divider Circuit In a series resistance circuit, the voltage divider theorem states that the ratio of the voltage across any resistor to the source voltage is equal to the ratio of the resistance of that particular resistor to the total resistance of the circuit. a. Derive the voltage divider theorem from basic relations. b. Resistors with a 5% tolerance are to be used to make a precision voltage divider to provide 1.00, 2.00, 3.00, 4.00 and 5.00 steps from a 5.00 V voltage source. You are provided six resistors of nominal value 1200 Ω, which have the following values : 1.135 kΩ, 1.177 kΩ, 1.185 kΩ, 1.129 kΩ, 1.144 kΩ, and 1.156 kΩ. Using five of these resistors, design the “best” voltage divider that you can design. (How do you define “best” in this case?) You have to apply the Engineering Problem Solving steps to solve this problem.
How do I use a Voltage Divider circuit?
Connect the divider from the battery to ground. That should be as stable as the battery is. If you are concerned with minimum current, make the resistors in the divider as large as possible, but not too large that the reference input will load down the reference voltage. "connect a divider in parallel with the circuit" is meaningless without more detail. Where did you get a 2.3 volt battery? edit: always list all the info, that way you get a correct answer. Your problem seems to be in your "2.3 volt battery". "using a breadboard to provide 2.3 volts" the latter statement is better, as it explains a little more, but still not complete. Unless you are using a voltage regulator, your 2.3 volt source has problems, it's voltage changes with load. did you think of measuring your 2.3 volts? If you do, I think you will find it changes a lot. Get a better 2.3 volt source and your problems will be gone. But why 2.3 volts? why not 3 volts, ie, two 1.5 volt batteries? Then you can just use a couple of cheap batteries. .
What is loaded or unloaded voltage divider?
Unloaded voltage divider -- make of minimum two resistors,no other resistor ( load ) attaches in parallel to the divider output. At this stage,divider output has the highest designed voltage. Loaded voltage divider -- A resistor ( load ) attaches to the divider output, more current passing through voltage divider network causing more voltage drop at the divider output. At this stage,voltage at the divider output has lower voltage compare to no load stage.
Tough Voltage-Divider Design Question?
In this problem we need to choose 10% resistors to make a voltage divider that meets a given specification. Vin=30v and we need to provide an open-circuit output voltage of Vout equivalent to 6v . requirement is that the Thevenin resistance as seen from the output terminals is between 10kiloomh and 30kiloohm come up with resistors R1 and R2 such that the division ratio Vout/Vin is within 10% of the requirement CIRCUIT DIAGRAM LINK http://i42.tinypic.com/174zed.gif FIND : R1 (in Ohms): R2 (in Ohms): Vmax (in Volts): Vmin (in Volts) Additional requirements Resistor restricted set E12={10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82} Thus, you can buy 10% resistors with a nominal resistance of 330 or 33k , but not 350 . The "tolerance" means that if you buy a 10% 390 resistor you can be sure that its resistance is between 351 and 429 . Please help me here,I just have Vout/Vin equals to R2/(R1+R2) could someone please proved the answers and an explanation of the procedure to them . much appreciated
Should I use a voltage divider or a buck converter to get 6V from 12V?
It depends what the 6v is needed for.If you just need a 6v voltage source, with no significant current, then a voltage divider might be OK. This type of arrangement is common in circuits to set DC bias levels for example.If you need to draw significant current, or are powering something that uses variable amounts of current, a voltage divider is inadequate and inefficient.Example - you need 6v at 1 amp, but the 1 amp doesn’t vary. You can achieve that by inserting a 6Ω resistor in series with the 12v supply. The circuit being powered forms the ‘lower’ end of the voltage divider. 1A through 6Ω is 6v, so you have achieved the final voltage of 6v. But the 6Ω resistor is now dissipating 6W, which is a substantial amount of wasted power. In addition, that wasted power will manifest as heat, so the resistor will get very hot and need to be rated for that power.If the current varies, say from 50mA up to 1A, then when it’s drawing 50mA, the series resistor will only drop 0.3v, so the circuit will get powered by a voltage of 11.7v. You have lost all regulation of the supply voltage.A solution to the voltage problem is a linear regulator, which adjusts the conductance of a pass transistor to keep the output voltage constant whatever the current. A simple linear regulator like a 7805(6) will do this, but it’s no more efficient than the series resistor. When passing 1A, it will still be dissipating 6W, a lot of heat (though when the current is 50mA, the power dissipation will reduce to 15mW, which may be acceptable).A DC-DC converter avoids the efficiency problem by only transferring enough charge to maintain the required output current and voltage. Its efficiency can be very high, and does not waste power as heat to ‘get rid’ of the excess voltage. So this is the right choice where the current can vary, or is substantial, and you don’t have power to waste, like in a battery-powered device.
Why use a regulator when a simple voltage divider can do the job for stepping down voltage?
A voltage divider is fine for producing an intermediate voltage from a given higher supply voltage except:if the supply voltage changes it will change the output voltage proportionatelyif the load is more than infinity then the voltage divider ratio changes and the output voltage changes To regulate the output voltage even poor specs of 10% then you need to have 10 times the divider current running than load current so its very inefficient.These are known as line regulation and load regulation, respectively.A good voltage regulator establishes a reference voltage that has high line regulation then buffers it so that it also has good load regulation. Typically these would be much better than 1%, i.e. a change of either load or line of say 10% would result in 1% of that, or .1% change in output voltage.The technical answer is that the output impedance of the voltage divider is very high - its the parallel combination of the two resistors that make up the divider. An ideal regulator has zero ohms. To make a divider be near to ideal requires divider resistors in the range of an ohm or so... in doing that the divider will consume large amounts of power so you have a very inefficient system. You can either have efficient regulators int he 60-90% range with excellent 1% or better regulation, or voltage dividers with something like less than 10% efficiency and worse than 10% regulation.
What is the procedure for finding voltages using current divider rule and voltage divider rule?
In simple language voltage divider is used to obtain a desired fraction of input voltage as output, commonly used in electronic devices. Current division means dividing the current between the branches of divider in such a way to minimize the total energy expended. For mathematical expression and more explanations visit the link electrical science