A milliammeter of range 120 mA and resistance $9\Omega$  is joined in a circuit as shown. The meter gives full scale deflection for current I when A and B are used as its terminal. i.e., current enters at A and leaves at ‘B’  [C is left isolated] .T he value of I is ……. A.

 

$$\begin{gathered} Let say current I enters at A . \\ \Rightarrow i_g=120 mA goes into galvanometer and remaining (i - 120mA) \\ goes through 0.1 \Omega \end{gathered}$$

$R_1=0.1\Omega , R_2=0.9\Omega and R=9\Omega$

$$\begin{gathered} \Rightarrow Consider 9\Omega and 0.9\Omega in series. \\ \Rightarrow V_A-V_B=\left(i-i_g\right)R_1=i_g\left(R_2+R\right) \\ \Rightarrow Equating the potential drop across upper and lower branch we get \\ \Rightarrow \left(i-120\times {10}^{-3}\right)\times 0.1=\left(9+0.9\right)\times 120\times {10}^{-3} \\ \Rightarrow i-0.120 = 11.88 \\ \Rightarrow i= 12.00 A \end{gathered}$$