I wanted to measure the resistance in ohms of an apple, so I plugged my digital multimeter into either end of the apple and adjusted the settings up to 200k and it gave me a reading of about 68. Because the meter was set to 200k does that mean I have to do 68 x 200000 to get the number of ohms for the apple or something else? Sorry, I'm new to all this.
-
\$\begingroup\$ What does your multimeter manual have to say? Usually the range given is the maximum value that can be measured. \$\endgroup\$– user2943160Commented Aug 27, 2016 at 18:49
-
\$\begingroup\$ @user2943160 it's long gone. Maybe it could be on the internet. \$\endgroup\$– MrSchmuckCommented Aug 27, 2016 at 18:51
-
2\$\begingroup\$ Generally on the 200k range the multi-meter can read up to a maximum of 199. So your reading of 68 means 68k. \$\endgroup\$– Steve GCommented Aug 27, 2016 at 19:04
1 Answer
Despite the fact the manual for your particular meter is missing, there are some fundamental concepts to using multimeters which you should familiarize yourself with.
I would recommend checking out some of the numerous online tutorials, such as this one from LadyAda, about multimeter usage. (Measuring resistance, specifically.)
Alternatively, you may want to acquire or borrow a meter with a manual and read up on it.
Ranges on meters denote the maximum value that the range supports. So 200k
means you can measure from 0 to 200 kΩ. All of the readings shown while using that range should be in kΩ.
For example, if you instead set the range to 200 Ω, it would measure from 0 to 200 Ω, in which case the apple at 68kΩ would be out of range. The multimeter would display some sort of message to indicate out of range, but this differs by manufacturer/model.
-
\$\begingroup\$ Perfect. Explained everything perfectly! Thank you so much! \$\endgroup\$ Commented Aug 27, 2016 at 19:23
-
\$\begingroup\$ For a lot of multimeter, this denotes the maximum range for which the accuracy is guaranteed. They have some percent of over range that they still display, often 10 or 20% \$\endgroup\$– PlasmaHHCommented Aug 27, 2016 at 20:54