The Ammeter and the Voltmeter — Question 3

Question 3

An ammeter changes the current that it is intended to measure

An ideal 6.00 V source originally drives a 3.00 ohm resistor. Inserting an ammeter adds series resistance RA=.150 ohm.

1. Find the original current, indicated current and fractional disturbance relative to the original current.

2. Derive the largest RA that would limit this disturbance to one percent in the same circuit.

Question 3 - Solution

An ammeter changes the current that it is intended to measure

(a) Method and calculation

$$I_0=E/R=2.00\,\mathrm A,\quad I_A=E/(R+R_A)=40/21\,\mathrm A.$$$$\frac{I_0-I_A}{I_0}=\frac{R_A}{R+R_A}=1/21\simeq4.76\%.$$

(b) Interpretation and further calculation

$$\frac{R_A}{R+R_A}\le\epsilon\Rightarrow R_A\le\frac{\epsilon R}{1-\epsilon}.$$$$\epsilon=.0100:\quad R_A\le.030303\,\Omega\simeq30.3\,\mathrm{m\Omega}.$$

Physical check

A perfectly calibrated ammeter can still disturb the current because its resistance is nonzero.