Q152 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Physics
A moving coil galvanometer of resistance $100 \Omega$ is used as an ammeter using a resistance $0.1 \Omega$. The maximum deflection current in the galvanometer is $100 \mu \mathrm{A}$. Find the minimum current in the circuit, so that ammeter shows maximum deflection?
- Correct answerA100.1 mA
- B1000.1 mA
- C10.01 mA
- D1.01 mA
Solution
We are given with following ammeter
Now, $\begin{aligned} & V_{A B}=I_g G=\left(I-I_g\right) S \\ & \quad\left(100 \times 10^{-6}\right) 100=\left(I-100 \times 10^{-6}\right) 0.1 \\ & \quad I-100 \times 10^{-6}=10^{-1} \\ & I=0.1+100 \times 10^{-6} \\ & \quad=0.1+0.0001=0.1001 \mathrm{~A}=100.1 \mathrm{~mA} \end{aligned}$ So, circuit current is $100.1 \mathrm{~mA}$.
Now, $\begin{aligned} & V_{A B}=I_g G=\left(I-I_g\right) S \\ & \quad\left(100 \times 10^{-6}\right) 100=\left(I-100 \times 10^{-6}\right) 0.1 \\ & \quad I-100 \times 10^{-6}=10^{-1} \\ & I=0.1+100 \times 10^{-6} \\ & \quad=0.1+0.0001=0.1001 \mathrm{~A}=100.1 \mathrm{~mA} \end{aligned}$ So, circuit current is $100.1 \mathrm{~mA}$.
Hence, the correct answer is option: 'A'.