How a 5 mA, 100 Ω galvanometer is converted into a 20 V voltmeter? (Answer: `\R_H` = 1900 N) 



Given:

Current through the galvanometer = `\I_g` = 5 mA = 5 ｘ 10⁻³ A

Resistance  = `R_g` = 100 Ω

Total Voltage = V = 20 V


To Find:

High Resistance = `\R_h` = ?


Solution:

The formula for shunt Resistant `\R_h` is


`\R_h` = `\frac {V}{I_g}`- `\I_g`

by putting the corresponding values


`\R_h` = `\frac {20 V}{5 ｘ 10⁻³ A}`- 100 A

`\R_h` = 4000 Ω - 100 Ω

`\R_h` = 3900 Ω ---------------------Ans. 




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