Question

Find the resistance (in kΩ) that must be placed in series with a 25.0 Ω galvanometer having a 50.0 µA sensitivity (the same as the one discussed in the text) to allow it to be used as a voltmeter with a 0.215 V full-scale reading.

Answer 1

Answer:

4.28 kΩ

Explanation:

A galvanometer is used for measuring small currents. A galvanometer can be converted to a voltmeter by connecting a high resistance or multiplier in series with the galvanometer. In the given question:

The sensitivity of the galvanometer is 50×10-6A. The voltage across the galvanometer (V1) is 1.25×10-3 V as shown in the solution attached. This is then subtracted from the expected full scale reading of the voltmeter to obtain the potential difference across the multiplier as shown. This voltage across the multiplier is then used to obtain the multiplier resistance as shown.

Answer 2

Answer:

R = 4.275 kΩ

Explanation:

Let the required resistance be R

Obeying Ohm's law,

Total voltage in the setup = (max current sensitivity) × (total resistance of the circuit)

Total voltage in the setup = 0.215 V

Current sensitivity = 50.0 µA = 0.00005 A

Total resistance of the setup = (R + 25) Ω

0.215 = 0.00005 × (R + 25)

(R + 25) = (0.215/0.00005)

R + 25 = 4300

R = 4300 - 25 = 4275 Ω

R = 4.275 kΩ

Hope this Helps!!!