Question

Find the resistance (in ohms) that must be placed in parallel with a 25.0-Ω galvanometer (50.0-μA sensitivity) to allow it to be used as an ammeter with a 300-mA full-scale reading.

(a) 83.3 Ω
(b) 90.0 Ω
(c) 96.7 Ω
(d) 100.0 Ω

Answer 1

The resistance that should be placed in parallel with the 25.0-Ω galvanometer to convert it into a 300-mA full-scale ammeter is approximately 83.3 Ω,

after applying the formula for parallel resistances and using the given sensitivities and desired full-scale deflection.

Step-by-step explanation:

To find the resistance that must be placed in parallel with a 25.0-Ω galvanometer having a 50.0-μA sensitivity to convert it into an ammeter with a 300-mA full-scale reading, we can use the formula for parallel resistances in a circuit:
V = Ig * Rg = Ishunt * Rshunt

Where:

• V is the voltage across the galvanometer and the shunt resistor
• Ig is the current through the galvanometer
• Rg is the resistance of the galvanometer
• Ishunt is the current through the shunt resistor
• Rshunt is the resistance of the shunt resistor

Since we want the full scale deflection to be 300 mA, and the sensitivity of the galvanometer is 50 μA, it means that 299.95 mA should go through the shunt resistor when 50 μA goes through the galvanometer.

Solving for Rshunt:

Rshunt = Rg * (Ig / Ishunt)

Rshunt = 25.0 Ω * (50.0 μA / 299.95 mA)

Rshunt = 25.0 Ω * (50.0 × 10-6 A / 299.95 × 10-3 A)

Rshunt ≈ 4.17 mΩ

So, the resistance that must be placed in parallel with the 25.0-Ω galvanometer is approximately 83.3 Ω to be used as an ammeter with a 300-mA full-scale reading.