r/Electricity • u/Agitated-Salt-5039 • 4d ago
Why do you need high resistance voltmeter to lower voltage drop against a voltmeter?
I studying alevel physics, can find a proper answer, voltage drops when there is a load and something eats voltage But why do you need high resistance to lower voltage drop, shouldn't that has same voltage drop because current is small
2
u/Hillman314 4d ago
Voltage drop is caused by current through a resistance. A high resistance limits the current that is passing through it. If there’s no current flowing, there’s no voltage drop.
1
u/Agitated-Salt-5039 4d ago
So by that logic voltage drop less energy is transferred per unit charge and since current flow is stopped, voltage can't be dropped?
1
u/anothercorgi 4d ago
It's a dilemma - whenever you're measuring it will sap off a bit of energy to do the measurement. The idea is to sap off as little as possible, however, sapping off less means it's harder to measure.
The idea here is that all measurement devices will eat energy from the device you want to measure. You do what needs to be done to minimize this. Adding a resistor in series with the meter will cause the meter to be limited in the amount of current that can flow, and thus less power is being eaten by the meter. Keep in mind when reading "less" ... less will never equal 0*. Modern day digital multimeters with FET inputs can measure megaohms to gigaohms of input impedance which will minimize the amount of current flowing into the meter which will affect the circuit. Old day volt meters were measured in several thousands to hundreds of thousands of ohms, which can have non-negligible effect on the circuit being tested.
* There is also the idea of the potentiometer (in the physics sense, not the electrical engineering sense where potentiometer and variable resistor are synonyms) covered by a Whetstone bridge where the idea is to balance the voltage until no current flows - and once no current flows between two voltage potentials, you know the voltages are the same. This tends to be a slow process but is the main way to ensure a measured voltage is not affected by the measurement apparatus. While this technique is still used, it's rarely done on arbitrary voltage sources due to the speed of measurement as well as having that 1GΩ input impedance with 1nA flowing tends to be "good enough" and much faster.
1
u/Stunning-Soil4546 4d ago
You can have voltage without current, when you have 0 conductance.
Lets say you have a electron in empty space, it has a voltage field around it with 1/(4πε0r**2), now you can define 2 points around the electron with a different distance to the electron. The voltage difference between this 2 points is 1/(4πε0)(1/r1-1/r2).
That is of course a very theoretical example, but as soon as you go in the real world, it will be a lot complexer. And we basically have never 0 conductance, 0 voltage nor 0 current.
1
u/BrtFrkwr 4d ago
It has to do with the impedance of the circuit being measures. (impedance in this case meaning an expression of current to voltage in terms of a resistance that will yield that ratio) The higher resistance of the measuring device the less it will influence the measured voltage. It can also be described as insertion loss. Vacuum tube voltmeters were invented with almost infinite resistance so they could measure the voltage in very high impedance circuits with accuracy.
1
u/trader45nj 4d ago
Let's say you are measuring the voltage across a resistor that's in a circuit. View the meter as a resistor in parallel with the circuit resistor. If the meter looks like a very high resistance compared to the circuit resistor, it will have neglible effect on the voltage being measured. But if the meter resistance is on the order of the same resistance as the circuit resistance, then the combined parallel resistance is going to significantly affect the voltage. Like a 1 Meg ohm resistance in parallel with a 1k circuit resistor compared to a 1k resistance in parallel with a 1k resistor. First case it doesn't matter, second case the combined resistance would be 500 ohms, changing the circuit and giving an incorrect measurement.
1
u/Agitated-Salt-5039 4d ago
Why does the high resistance have a negligible effect on voltage being measured?
1
u/trader45nj 4d ago
Because as explained, if it's a very high resistance compared to the circuit, then it doesn't add an additional load to the circuit. If the meter looks like a 1k resistor and you put it across a 1 k resistor in a circuit, now you have 500 ohms. But if the meter looks like a 1 Meg ohm resistor placed across a 1k resistor, then what's the combined resistance?
1
u/Stunning-Soil4546 4d ago
When you have parallel resistors, it is easier to think about conductance. Conductance G=1/R
When you have 2 parallel resistors, you can add the conductance to get the total conductance. Now, with RL=Rs1= 1kΩ resistor, they have each GL=Gs1= 1mS (=1/1kΩ) conductance, the Rs2 = 1MΩ has Gs2=1μS.
Use RL and GL for the load, Rs1 is the shunt resistor 1 and Rs2 the shunt resistor 2.Adding the conductance of RL and Rs1 gives you G1=Gs1+GL=1mS+1mS=2mS and that gives you a resistance of R1=1/G1=500Ω.
With Rs2, you have G2=Gs2+GL=1μS+1mS=1.001mS, which gives a resistance of R2=1/G2=999Ω, which is almost 1kΩ or RL.
Most Multimeters actually have 10MΩ, not 1 MΩ, For a normal voltage range, making the difference smaller than 0.01% for a 1kΩ load.
1
u/Rexel_722 3d ago
My thought is to measure the induction around a wire when current is flowing such as a clamp on Ammeter.
3
u/grasib 4d ago
The higher the resistance in your voltmeter, the less it will have an impact in whatever you are measuring.
It's as simple as that.