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

  • \$\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


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.


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} \$.

  • \$\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

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