"Suppose that you measure the resistance of a resistor sing a voltmeter, a current meter, and a battery. Suppose that the voltmeter reads 0.2% high and the current meter reads 0.3% low. What will be the percent error in your measurement of resistance?"

I'm confused on how to solve this problem. If someone could walk me through this, I'd really appreciate it.

Question

"Suppose that you measure the resistance of a resistor sing a voltmeter, a current meter, and a battery. Suppose that the voltmeter reads 0.2% high and the current meter reads 0.3% low. What will be the percent error in your measurement of resistance?"

I'm confused on how to solve this problem. If someone could walk me through this, I'd really appreciate it.

jack1 · Answer

just a guess/start: but Ohms law to begin with?
V= IR   so R  = V/I

anonymous · Answer

Okay, that's a good place to start . . . I understand Ohm's law, it's just figuring out this percent difference that's really confusing me.

jack1 · Answer

i guess (V * 1.02)/(i * 0.97)
1.02/0.97
=1 and 5/97  or 1.0516...ish

jack1 · Answer

does that make sense?

anonymous · Answer

I think it does, actually! I'm going to try it. Thank you so much :D

jack1 · Answer

all good, hope it works man  =)