I have to measure the current of a battery of 3.7 volt(used in HP IPAQ VM) and I am connection a resistor of 10 ohms in series with multimeter and the battery. Since the actual current of battery is 1.26 amps (1260 mA) but the meter shows only 0.37 amps.

How to measure actual current of source using multimeter?


1 Answer 1


Your meter is correct. With a 3.7 V battery and a 10 ohm resistor, Ohm's law (\$V = I\cdot R\$) tells you that the current should be 0.37 A.

The 1260 mA you mentioned is probably 1260 mAh (milliamp hours). This tells you the capacity of the battery, not the current draw. For example, if you drew 100 mA from the battery continuously, it would last 1260 mAh / 100 mA = 12.6 hours.

  • 1
    \$\begingroup\$ Note that the listed mAh rating is at a specific current draw only (given in the datasheet). \$\endgroup\$
    – tyblu
    Jan 7, 2012 at 22:11
  • 1
    \$\begingroup\$ And at a specific temperature, phase of moon, and species of dead fish waved over the battery during the measurment. \$\endgroup\$ Jan 7, 2012 at 22:51
  • 3
    \$\begingroup\$ Olin's just slightly less helpful than could be comment means that the stated capacity in milliamp-hours mAh is an approximation only as there are many parameters that affect the result. Rate of discharge, temperature, past history and past number of cycles and more will all have some effect on what is seen. The important thin is that mAh = milliamp-hours is a measure of current x time capacity whereas millamps is actual instantaneous current draw. \$\endgroup\$
    – Russell McMahon
    Jan 7, 2012 at 23:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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