Question

A circuit you're building needs an ammeter that goes from 0 mA to a full-scale reading of 40.0 mA. Unfortunately, the only ammeter in the storeroom goes from 0 μAto a full-scale reading of only 500 μA. Fortunately, you can make this ammeter work by putting it in a measuring circuit, as shown in the figure. This lets acertain fraction of the current pass through the meter; knowing this value, you can deduce the total current. Assume that the ammeter is ideal.

1. What value of must you use so that the meter will go to full scale whenthe current is 40.0 ? Hint: When $I;=;$ 40.0 , the ammeter is reading its maximum value.

2. What is the equivalent resistance of your measuring circuit?

Answer 1

From ohms law -500x10-6 A)(50 Ω) 0.025 V The current going through R is, I40.0 mA -0.5 mA 39.5 mA From ohms law R=- 0.025 V 39.5 mA = 0.633 Ω It is a parallel circuit, (0.633 Ω)(50 Ω) (0.6330)-(50 Ω) 0.624 Ω eq