2
\$\begingroup\$

I know how to do it with ammeter, but how can I find internal resistence without ammeter?

\$\endgroup\$
  • \$\begingroup\$ An ammeter is simply a combination of a resistor and a voltmeter. \$\endgroup\$ – Finbarr Oct 30 '17 at 14:42
  • \$\begingroup\$ If it is not ideally calibrated, you cannot do that without some other measurement device. \$\endgroup\$ – Eugene Sh. Oct 30 '17 at 14:43
  • \$\begingroup\$ Please edit your question to explain what type of meter - digital or analog - and provide a part number, if possible. Usually it is marked on them and is certainly in the specification or user manual. \$\endgroup\$ – Transistor Oct 30 '17 at 14:43
  • \$\begingroup\$ ... or more commonly, a voltmeter is the combination of an ammeter and a resistor. Do you have an ohmmeter? \$\endgroup\$ – Dave Tweed Oct 30 '17 at 14:44
5
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. DC voltmeter impedance measurement technique.

For a DC meter such as a typical digital multimeter you can try the scheme of Figure 1.

  • Use a very stable reference voltage. (OK use a PP3 9 V battery.)
  • Take a voltage reading from your reference voltage with the meter in question (Figure 1a).
  • Use a precision reference resistor. (OK grab a 1M resistor out of your bag of resistors.)
  • Take a second voltage reading from your reference (Figure 1b). If R1 is significant relative to the meter the reading will drop.

e.g. If the meter reading falls to half when using a 1M resistor then the resistor must have an input impedance of 1M.

The easiest way to do it is to keep adding or subtracting resistance until you get a reading of half of the battery voltage. That resistance is equal to that of the meter.


Update

VM2/9V = (R1+Ri)/R1 Is this correct?

No. \$ \frac {meter \ voltage}{total \ voltage} = \frac {meter \ R}{total \ R} \$. So, \$ \frac {VM2}{9} = \frac {Ri}{R1 + Ri} \$.

\$\endgroup\$
  • \$\begingroup\$ You don't have to keep trying resistors. As long as the resistor is reasonably close to the voltmeter resistance, a simple calculation using the voltage divider formula will yield a good value for the resistance. \$\endgroup\$ – Barry Oct 30 '17 at 14:57
  • 2
    \$\begingroup\$ Correct but I hadn't the energy to explain that to the OP. Add it as a separate answer if you want. You can copy my diagrams. \$\endgroup\$ – Transistor Oct 30 '17 at 15:02
  • \$\begingroup\$ @Barry Can you write that formula? \$\endgroup\$ – Ninalol Oct 31 '17 at 10:36
  • \$\begingroup\$ @Ninalol: Why don't you write the formula in the question and we'll tell you if it's right. It will have three knows: VM1, VM2 and R1 and one unknown - Rmeter. It's just a ratio calculation. \$\endgroup\$ – Transistor Oct 31 '17 at 16:25
  • \$\begingroup\$ VM2/9V = (R1+Ri)/R1 Is this correct? \$\endgroup\$ – Ninalol Nov 1 '17 at 10:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.