0
\$\begingroup\$

I want to measure the current that can be supplied from a lead acid battery I have. (To know it's remaining current supplying capacity). To measure current I need to connect a load and an ammeter (multimeter, with knob to direct current) in series with the battery. Can I use another multimeter (I have two of them) with a larger internal resistance (knob towards dc voltage), as the series load, if I don't have any spare loads with me?

\$\endgroup\$
3
  • 1
    \$\begingroup\$ No, it's not a suitable series load. By design, it draws as little current as possible, making for a poor variable load. Buy a suitable power resistor. \$\endgroup\$ – nanofarad Mar 2 at 17:25
  • \$\begingroup\$ What power range do you need? Meters can’t handle much <1W \$\endgroup\$ – Tony Stewart EE75 Mar 2 at 17:44
  • \$\begingroup\$ @nanofarad Thanks! \$\endgroup\$ – Namboo3 Mar 2 at 17:46
1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Test circuit.

  • Voltmeters are typically designed to have very high impedance (resistance) and 1 MΩ and 10 MΩ is typical. This is so they draw as little current as possible from the circuit under test. As a result a voltmeter would be useless for loading a battery.
  • Ammeters are typically designed to have very low resistance so they don't cause any loss of voltage in the circuit. As a result if you were to connect one in place of VM1 a very high current would flow and probably destroy the meter or blow the fuse - if it has one (and it should).

The correct way is to use a load and monitor the current and voltage simultaneously. This way you can calculate the power at any time and you can stop the test before you over-discharge the battery. If yours is a 12 V lead-acid battery then car bulbs are a handy test load and you can get 6 W tail lamps, 21 W brake lamps and 55 W headlamps and mix and match. These are readily available but have the disadvantage, like all incandescent bulbs, that their resistance increases with temperature so as the battery voltage falls so will the lamp resistance.

\$\endgroup\$
0
\$\begingroup\$

Yes I suppose you can but chances are quite good that you will ruin your multimeter. These usually have a 10A maximum shunt in them for measuring current.

The usual process for measuring large currents, like from your lead-acid battery are:

  1. Use a high-current shunt in series with the load and you measure the voltage across this shunt.
  2. Use a "clamp" style ammeter.
\$\endgroup\$
2
  • \$\begingroup\$ Thanks for the answer!, Can you please explain me how that would ruin the multimeter? If I put the knob to a setting such that it would have a reasonably high resistance, will it not limit the current through the circuit? Pardon me, I'm a beginner. \$\endgroup\$ – Namboo3 Mar 2 at 17:42
  • \$\begingroup\$ I assumed you were planning to measure the current with the multimeter. These have relatively small current-sensing shunts which are sometimes fused but other times not. If you allow too much current to flow it will burn your shunt and it will no longer work. \$\endgroup\$ – jwh20 Mar 2 at 17:47

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.