Question

A circuit you’re building needs an ammeter that goes from 0 mA to a full-scale reading of 50.0 mA. Unfortunately, the only ammeter in the storeroom goes from 0 μA to 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 Figure 65 This lets a certain fraction of the current pass through the meter; knowing this value, you can deduce the total current. Assume that the ammeter is ideal.

a. What value of R must you use so that the meter will go to full scale when the current I 50.0 mA?

Hint: When I = 50.0 mA, the ammeter should be reading its maximum value.

---

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

FIGURE 65

Answer 1

(a) The resistance of the resistor to support the ammeter can be determined as follows.

The maximum permissible current through the ammeter is. To measure the total current as 50 mA, we construct the ammeter by using a resistor in parallel with it as shown in the following figure. Thus, the total current flows into two paths in which the maximum current of flows through the ammeter and the remaining current 0.0495A flows through the parallel resistor.

Draw the circuit contains ammeter along with resistor to measure the current.

As the ammeter and the resistor are in parallel, there exists same potential difference across the both; hence the voltage across the resistor is equal to voltage across the ammeter.

Since the resistance of the ammeter is, then the voltage across the ammeter is,

The voltage across the resistor is

Here current in the resistor is.

Rearrange the equation for.

Substitute forandfor.

Therefore, the resistance of the resistor used to see the full scale deflection in the ammeter is.

(b) The effective resistance of the ammeter can be determined as follows.

When the ammeterof resistance and resistorof resistance are in parallel, then the effective resistance is

Here resistance of the ammeter is.

Thus,

Therefore, the effective resistance of the ammeter is.