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Below is a picture of my very simple circuit.

photo of circuit built on breadboard

There are three resistors 10 ohm, 100 ohm, and 10 kilohm. I connect them one at a time in serial fashion and measure voltage drop across them. When I measure battery voltage I get 6.5 Volt and this is what I expect as voltage drop on all of my measurements, but here is what I get:

  1. When I measure voltage drop across 10 ohm resistor I get 0.2 volt.
  2. When I measure voltage drop across 100 ohm resistor I get 1.0 volt.
  3. When I measure voltage drop across 10 kilohm resistor I get 5.8 volt.

Why do I get all these different readings and not something around 6.5 Volt?

P.S. On the image the circuit is not closed, but this is how I took picture, circuit is closed on all measurements.

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    \$\begingroup\$ this question has been asked here a lot ... the rest of the voltage is dropped across the battery internal resistance ... |l------\/\/\-----\/\/\----- \$\endgroup\$
    – jsotola
    Commented Apr 2, 2021 at 21:50
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    \$\begingroup\$ Pay attention not to fry the 10 Ohm resistor when using a fresh battery. \$\endgroup\$
    – tobalt
    Commented Apr 3, 2021 at 5:44
  • \$\begingroup\$ @tobalt, I already saw it smoking ;-). How do I decide what current resistor can withstand? \$\endgroup\$
    – user1700890
    Commented Apr 3, 2021 at 13:34
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    \$\begingroup\$ In this simple circuit, the worst case power in the resistor will be 81/R Watts. The resistor are probably rated for 0.25 W... The real power will be less because of internal Resistance in the battery but don't go grossly over this limit. \$\endgroup\$
    – tobalt
    Commented Apr 3, 2021 at 14:17

2 Answers 2

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Your battery is very, very dead. It has an internal resistance of around 500 ohms.

A fresh Panasonic alkaline battery (consumer grade, not industrial) has an internal resistance of less than 4\$\Omega\$.

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  • \$\begingroup\$ Thank you so much, I will try to do the same with new battery. But why does voltage drop changes with battery conditions? I did only a couple of measurements on this dead battery and they were fairly consistent. \$\endgroup\$
    – user1700890
    Commented Apr 2, 2021 at 21:59
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    \$\begingroup\$ Do the calculations of the voltage you expect assuming a (say) 500 ohm internal resistance in the battery and using the voltage divider equation. The internal resistance will vary a bit during measurements and will increase as current is drawn. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Apr 2, 2021 at 22:00
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9V in 10 ohms needs a current of 9V/10 ohms= 900mA. 9V in 100 ohms needs a current of 90mA. A little 9V Name Brand alkaline battery cannot produce 900ma but can produce 90mA if it is brand new but for only a few minutes.

Your battery has no name brand so it probably is garbage.

9V Name Brand batteries

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