A multirange current meter can be constructed by using a galvanometer circuit as shown in figure. We want a current meter that can measure `10mA`, `100mA` and `1A` using a galvanometer of resistance `10Omega` and that produces maximum deflection for current of `1mA`. Find `S_1, S_2` and `S_3` that have to be used.

Options

A. `18 K Omega `
B. `180 k Omega `
C. `1.8 k Omega `
D. `1800 k Omega `

Answer

A

Explanation

Here , `I_g = 1mA = 1 xx 10^(-3) A, G = 10 Omega`
` R_1 = ? , R_2 = ? , R_3 = ? `
Using `R + G = (V)/(i_g)`

At point P, `R_1 +10 = (2)/(10^(-3)) 2000 Omega`
`R_1 = 2000 -10 = 1990 Omega = 1.9 k Omega`
At point `Q, R_1 + R_2 +10 =(20)/(10^(-3)) = 20000 Omega`
`R_2 + 2000 = 20000`
`therefore R_2 = 18000 Omega =18 K Omega`
At point R,
`R_1 + R_2 + R_3 +10 =(200)/(10^(-3))= 200000 Omega`
` therefore R_3 = 200000 -1990 -18000-10`
`180000 Omega = 180 K Omega`