An electrical meter of internal resistance` 20 Omega `gives a through it. What is the maximum current, that can be measured by connecting three resistors each or resistance` 12 Omega `in parallel with the meter?

Options

A. `6 mA`
B. `8 mA`
C. `10 mA`
D. `4 mA`

Answer

A

Explanation

When three resistance are joined I parallel, their effective resistance will be `40 Omega`i.e. `S=4 Omega and G=20 Omega` The maximum current through the meter=`1mA=i_g therefore i_g/I=S/(S+G)therefore(10^(-3))/i=4/(4+20)=1/6therefore i=6 xx 10^(-3)A=6mA`.