- #1
Peter G.
- 442
- 0
Hi,
I have a simple arrangement: Two 500 kΩ resistors connected in series. The teacher illustrates what it would be like to measure the p.d across one of these 500 kΩ resistors with a voltmeter with 50 kΩ resistance. To do so, he calculated the resistance in parallel of the voltmeter and one of the resistors. He got 45.5 kΩ. So far so good, but what he did next however is what I don't understand. He divides 45.5 kΩ by the total 545 kΩ and multiplies it by the e.m.f. Why would the voltmeter read (45.5 kΩ / 545 kΩ) x 12 V? I thought it would measure 6V out of the 12 of the source since both resistors had the same resistance. The voltmeter accounts for itself?
Thanks,
Peter G.
I have a simple arrangement: Two 500 kΩ resistors connected in series. The teacher illustrates what it would be like to measure the p.d across one of these 500 kΩ resistors with a voltmeter with 50 kΩ resistance. To do so, he calculated the resistance in parallel of the voltmeter and one of the resistors. He got 45.5 kΩ. So far so good, but what he did next however is what I don't understand. He divides 45.5 kΩ by the total 545 kΩ and multiplies it by the e.m.f. Why would the voltmeter read (45.5 kΩ / 545 kΩ) x 12 V? I thought it would measure 6V out of the 12 of the source since both resistors had the same resistance. The voltmeter accounts for itself?
Thanks,
Peter G.