0
\$\begingroup\$

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.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ this question has been asked here a lot ... the rest of the voltage is dropped across the battery internal resistance ... |l------\/\/\-----\/\/\----- \$\endgroup\$ – jsotola Apr 2 at 21:50
  • 1
    \$\begingroup\$ Pay attention not to fry the 10 Ohm resistor when using a fresh battery. \$\endgroup\$ – tobalt Apr 3 at 5:44
  • \$\begingroup\$ @tobalt, I already saw it smoking ;-). How do I decide what current resistor can withstand? \$\endgroup\$ – user1700890 Apr 3 at 13:34
  • 1
    \$\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 Apr 3 at 14:17
3
\$\begingroup\$

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\$.

\$\endgroup\$
2
  • \$\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 Apr 2 at 21:59
  • 1
    \$\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 Pefhany Apr 2 at 22:00
3
\$\begingroup\$

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

\$\endgroup\$

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.