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How can calculate the approximat internal resistance of the multimeter?
Wiki User
∙ 12y ago
If the meter is digital you will be told in the instrument's specifications what the internal resistance is. Most digital multimeters have an input resistance, on all voltage ranges of 10 Mohms.
The resistance of most analogue multimeters depends on the voltage range, being given as "x kohm/volt", and is called the meter's sensitivity
The better analogue multimeters have a sensitivity of 20k/volt, so, for instance, on the 30V range of such an instrument, the resistance would be 30 x 20,000 = 600k.
There are, or were, also analogue instruments which had a constant 10 Mohm resistance on all voltage ranges. The Philips PM2505 being one example.
Wiki User
∙ 12y ago
Wiki User
∙ 9y agoSwitch the dial to "Ohms" (there may be several ranges of Ohm measurements, choose the appropriate range), place the probes across the object to me measured and read off where the needle pointer is on the dial. Remember to switch off the meter afterwards as Ohm readings drain the battery in the merer.
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If you have a circuit consisting of a battery and a wire What two factors determine the amount of current in the circuit?
The voltage of the battery, and the resistance of the circuit (including the resistance of the wire and the internal resistance of the battery).
As a battery ages and is used up what will the internal resistance be?
As it ages, its Kirchhoff virtual 'internal resistance' increases. Pictured this way, the practical effect is that the open-circuit voltage of a failing cell still measures good, but the voltage quickly sags when you try to draw any current from it
When a multimeter is set to measure resistance what reading would you expect when the probes aren't touching?
If the test leads are not in contact with anything, the needle or pointer of an analog meter will be resting at the left most position. This is represents an infinite amount of resistance, or an "open circuit"; it is also safe to say there is the no continuity, or path between the black and red probes.
Why the input impedence of op-Amp is very high?
An ideal amplifier will have a very high input resistance, and low output resistance. This is so it doesn't "load" the input circuitry, and can drive output circuitry. Say you hooked an op amp up to a microphone that kicked out 100mV unloaded. If you shorted out the microphone, the output would approach 0mV (very low resistance load). If you connected a very high resistance load instead, the output would remain close to 100mV. As the load becomes a smaller resistance/bigger load, the output voltage will decrease until it is no longer useful. This is because the microphone has an internal resistance that remains constant while you're varying the attached load. The voltage drop across the internal resistance will increase as current output increases, causing less of a signal to be delivered to the amplifier.
What happens to voltage when resistance increases?
Ohm's law states that "The current is directly proportional to the applied EMF (voltage) and inversely proportional to the resistance in the circuit." <<>> if resistor exists, resistance decreases according to ohm's law, current is directly proportional to voltage and current is inversely proportional to resistance it means as current increases, voltage increases. resistance increases, current decreases so as voltage if there is no resistor, there should be no resistance except internal resistance of voltmeter and ammeter
Can you measure the internal resistance of dry cell using multimeter?
Well, you do use a multimeter to measure it, but not directly. You need at least one other thing besides the multimeter and the dry cell. Here's a simple way to do it: -- Take a resistor. With the meter set to measure resistance, measure the resistor's value. Call that 'R'. -- With the meter set to measure voltage, measure the voltage of the dry cell. Call it 'V1'. -- Connect the resistor across the dry cell. Measure the dry cell's voltage again. Call it 'V2 ' this time . V2 will be less than V1 . The internal resistance of the dry cell is R times (V1 - V2) / V2
How do you measure a resistor using multimeter?
a multimeter probably has a Ohm meter as well. If yes, use it and make sure you have the correct scale adjusted. If there is no Ohm meter you should calculate the resistance using Ohms Law. So if you use your multimeter as an Amp meter (in series with the unknown resistor) and you apply a know voltage across you should be able to calculate the resistance by dividing the voltage you applied by the current you meassured (R=v/I) ANSWER: 1 make sure ther is no external power connected 2 short the leads together and adjust for zero reading on the meter. This step is necessary to eliminate the internal battery drifting with time. Unlike voltage and amps reading the scaling of the meter is not important to begin with since a peg meter or a overflow will not destroy the meter immediately but do change the scale to get maximum deflection or digital display.That it
Do meters usually have a high or low internal resistance?
It depends on the application. Voltmeters have a high internal resistance, while ammeters have a low internal resistance.
Is it desirable for a battery to have a high internal resistance?
No, it is desirable for a battery to have a low internal resistance.
Define Internal Resistance?
The internal resistance of a device is the resistance in ohms of that device. It is the resistance electrons need to overcome before electricity is said to flow.
How resistivity effects internal resistance of a cell experiment?
Resistivity effects internal resistance of a cell experiment by not allowing the cell to react. Internal resistance experiments are performed in advancing physics classes.
Do Dead batteries have high internal resistance?
yes, batteries have high internal resistance. The higher the resistance the lower power you get out of the batter. Therefore if you no power you have very high resistance.
What is internal resistance?
it is a type of resistance which inbuilt in voltage source
The no-load output voltage of a DC power supply is measured at 15V When a 600 ohms load is connected to the output the output drops to 13.7V Calculate the internal resistance of the power supply?
This question can be answered using voltage dividers. Assume the power supply consists of a voltage source and a resistor. With no load, all of the voltage source's voltage is dissipated by the internal resistor of 15V. When there is a load, there are two resistors in series. To calculate the internal resistance:1. I=V/R. You know the 600ohm resistor dissipated 13.7V. So that would mean a current of 13.7/600=22.8mA2. If the 600ohm resistor dropped 13.7, kirchoff's voltage law would tell us the internal resistor dropped 15-13.7=1.3V.3. R=V/I, Use the current to calculate the internal resistance. 1.3/22.8mA = 56.9ohmsCommentFurther to the above answer, a voltage-source's voltage is not 'dissipated by the internal resistance when on no load'. On no load, there is no current passing through the internal resistance, so no 'voltage dissipation' can takes plac -i.e. the non-load voltage is 15 V.
Actual internal resistance of 1.5 volt battery?
The value of internal resistance of 1.5 volt battery is 0.5 ohms.
What is resistance of an ideal ammeter?
ideal ammeter has zero internal resistance
What is the value of internal impedance of ideal current source?
internal resistance is always infinite in ideal current source .the internal resistance is in shunt with current source
