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As per Ohms law , V = IR OR I = V/ R

So if I set R to 0 , then V should be equal to I ?

However when I connect 9 Volt battery poles to each other with multimeter in between the reading says 0.2 A.

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  • \$\begingroup\$ your 9v battery is flat. \$\endgroup\$ – Jasen Jul 30 at 8:00
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    \$\begingroup\$ Mathematics 101. \$\endgroup\$ – Andy aka Jul 30 at 8:04
  • \$\begingroup\$ @jasen what is flat ? \$\endgroup\$ – Mr Coder Jul 30 at 8:05
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    \$\begingroup\$ Your math is wrong. I = 9/0 which is "infinity" or cannot be computed. However, resistance is never zero. And there is a concept you did not address, which is the internal resistance of a battery. \$\endgroup\$ – mkeith Jul 30 at 8:09
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    \$\begingroup\$ @MrCoder: "101" from a common numbering system for college/university level courses. 100 is the base level, usually first year. 101 is typically the first year introductory course for a given subject. "Math 101" is therefore just a fancy way of saying "basic math." \$\endgroup\$ – JRE Jul 30 at 8:12
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Fortunately for you, batteries have internal resistance. If they hadn't then you would have either blown your meter's current limit fuse or destroyed your meter.

Never short-circuit a battery or power supply with an ammeter. Always connect it in series with the load.

You can model your battery as an "ideal" 9 V battery with a series resistance. From you measurement you can calculate that the battery's internal resistance. \$ R = \frac {V}{I} = \frac {9}{0.2} = 45 \ \Omega \$.


From the comments:

So does it mean if I short circuit 220 V (in house voltage), then reading should be 220 amps, since now there is no battery hence no internal resistance?

If there is no internal resistance on your 220 V supply then \$ I = \frac {V}{R} = \frac {220}{0} = \infty \ \text A \$.

Never short out a supply like that. Explosive amounts of energy can be released. (Look up "arc flash" on YouTube.) If you try it with your multimeter the results could be lethal.

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  • \$\begingroup\$ so does it mean if I shortcircuit 220V (in house voltage) , then reading should be 220 Amps , since now there is no battery hence no internal resistance ? \$\endgroup\$ – Mr Coder Jul 30 at 8:34
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    \$\begingroup\$ Sigh 2.0........ \$\endgroup\$ – Bart Jul 30 at 8:43
  • \$\begingroup\$ @MrCoder See mkeith's comment above. Your math is incorrect. \$\endgroup\$ – HandyHowie Jul 30 at 8:43
  • \$\begingroup\$ @HandyHowie thanks I made an error in calculations. \$\endgroup\$ – Mr Coder Jul 30 at 9:19
  • \$\begingroup\$ @MrCoder: See the update. \$\endgroup\$ – Transistor Jul 30 at 9:24
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First, as already said- don’t ever attempt to deliberately short any power source unless you are trying to obtain a UL rating for a device- you run a high risk of fire or (perhaps minor) explosion! I once worked for a power supply manufacturer and was assigned to do this sort of work one week- the supply was behind a blast shield for this reason.

Second, the ammeter itself has resistance in addition to the wires connecting it as well as the battery itself. There’s no way in a practical circuit to achieve zero resistance however close you might get and your current reading would only be 9A if the total resistance was exactly 1 ohm.

For internal resistance of typical battery types, see this article (the table therein shows a 9V zinc carbon battery is typically 35 ohms):

http://www.learningaboutelectronics.com/Articles/Battery-internal-resistance

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