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Assuming it's running on a 9V battery like most multimeters are, and the setting is on MegaOhms.

How much current will be drawn from the internal battery?

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  • \$\begingroup\$ Your question is a little ambiguous - are you asking about how much current is drawn from the internal battery in order to test resistance? \$\endgroup\$ – David Oct 19 '14 at 7:42
  • \$\begingroup\$ yeah ok, can u answer that? \$\endgroup\$ – minusatwelfth Oct 19 '14 at 8:08
  • \$\begingroup\$ Why are you asking? \$\endgroup\$ – starblue Oct 20 '14 at 14:36
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Technically, voltage cannot be drawn because its a potential energy between 2 points like gravitational pull. Thus, when the multimeter probes is placed on the resistor, it would draw the full voltage of the battery, the thing that is drawn would be current which causes the galvanometer on the multimeter to deflect.

*edit You could probably use another multimeter to measure the current draw by connecting it in series with the element you are measuring.

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    \$\begingroup\$ It looks like the OP is asking about how much power of its own supply the DMM dissipates in the act of measuring a circuit, not of the circuit's supply. \$\endgroup\$ – Shamtam Oct 19 '14 at 7:31
  • \$\begingroup\$ Sorry I wrote full power when i meant full voltage. Still, when measuring resistance, 9v is passed through the element under test, resistance measured depends on the resulting current. \$\endgroup\$ – kenneth Oct 19 '14 at 8:02
  • \$\begingroup\$ @user1564795 sorry I can't comment on your post, only mine. Anyway, the amount of current depends on the resistive element you are measuring. Quoting from wikipedia, "To measure resistance, a small battery within the instrument passes a current through the device under test and the meter coil. Since the current available depends on the state of charge of the battery, a multimeter usually has an adjustment for the ohms scale to zero it. In the usual circuit found in analog multimeters, the meter deflection is inversely proportional to the resistance " \$\endgroup\$ – kenneth Oct 19 '14 at 9:56
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Multimeters nearly always inject a low current from a constant current generator into the resistor under test. This means that the voltage devloped across the resistor is directly proportional to the resistance being measured. On different ranges, the constant current injected may be bigger (in order to get better resolution on smaller values of resistance) and it may be smaller for the Mohm range.

Regarding the actual current drawn from the battery this entirely depends on the measurement system's efficiency which is entirely dependent on the circuit of the meter (which we don't have). For the Mohm range, 1 micro amp injected into the resistor-to-be-measured will produce 1 volt across 1 Mohm so this current is likely trivial compared to the current taken from the battery by the measurement system and LCD - if it's an analogue meter this may take more current than an LCD.

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  • \$\begingroup\$ On the two handheld multimeters I have it is 1µA on the single-digit megaohm range (Fluke 175 and Extech 430). The Fluke goes up to 1mA, the Extech to 0.1mA on lower ranges. \$\endgroup\$ – starblue Oct 20 '14 at 15:07

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