A milliammeter of range 0-30 mA has internal resistance of ` 20 Omega.` The resistance to be connected in series to convert it into a voltmeter of maximum reading 3V is :

Options

A. ` 49 Omega`
B. ` 80 Omega`
C. ` 40 Omega`
D. ` 50 Omega`

Answer

B

Explanation

` R = (V)/(I_g) - G `
` = (3)/(30 xx 10^(-3) ) - 20 = 10 ^(2) - 20`
` = 80 Omega`