A milliammeter of range 120 mA and resistance 9 Ω  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.

Let say current I enters at A . ⇒ig=120 mA goes into galvanometer and remaining (i - 120mA) goes through 0.1 Ω ⇒R1=0.1Ω, R2=0.9Ω and R=9Ω  ⇒Consider 9Ω and 0.9Ω in series. ⇒VA-VB=i-igR1=igR2+R ⇒Equating the potential drop across upper and lower branch we get ⇒i−120 × 10−3 × 0.1 = 9+0.9 × 120 × 10−3 ⇒i-0.120 = 11.88 ⇒i= 12.00 ≃12.00