The internal resistance of the ammeter is RG = 200 Ω, and the full bias current is Ig = 500 μ a. the correct way to refit the ammeter into a voltmeter with a range of 1.0V is () A. A 0.1 Ω resistor shall be connected in series B. a 0.1 Ω resistor shall be connected in parallel C. A 1800 Ω resistor shall be connected in series D. a 1800 Ω resistor shall be connected in parallel

The internal resistance of the ammeter is RG = 200 Ω, and the full bias current is Ig = 500 μ a. the correct way to refit the ammeter into a voltmeter with a range of 1.0V is () A. A 0.1 Ω resistor shall be connected in series B. a 0.1 Ω resistor shall be connected in parallel C. A 1800 Ω resistor shall be connected in series D. a 1800 Ω resistor shall be connected in parallel

The resistance to be connected in series is: r = Uig − RG = 1.0500 × 10 − 6 − 200 = 1800 Ω. A series resistance is wrong, so a error B should not be connected in parallel. So B error C meets the requirements. So C is correct, D should not be connected in parallel. So D error: C

A sensitive ammeter with a full current of 200 microampere is refitted into an ohmmeter. When it is used to measure the resistance of 10K ohm, the needle just points at 100 microampere. When it is used to measure the other resistance R, the needle just points at 50 microampere

Let the internal resistance of ohmmeter be r and the electromotive force of internal power supply be e
Solutions e = 0.2R and E = 1 + 0.1R
E = 2V, r = 10kom
R=E/0.05-r=30kom