Q139 / 160 · TS EAMCET 2015 · Physics
A voltmeter of $250 \mathrm{mV}$ range having a resistance of $10 \Omega$ is converted into an ammeter of $250 \mathrm{~mA}$ range. The value of necessary shunt is (nearly)
- A$2 \Omega$
- B$0.1 \Omega$
- Correct answerC$1 \Omega$
- D$10 \Omega$
Solution
A voltmeter of $250 \mathrm{mV}$ range having resistance of $10 \Omega$
$$
\begin{aligned}
& \therefore & V_g & =G I_g \\
\Rightarrow & & 250 \times 10^{-3} & =10 \times I_g \\
\Rightarrow & & I_g & =25 \times 10^{-3} \mathrm{~A}
\end{aligned}
$$
Shunt required to convert this voltmeter into ammeter.
$$
\begin{aligned}
S & =\frac{G I_g}{I-I_g} \\
& =\frac{10 \times\left(25 \times 10^{-3}\right)}{250 \times 10^{-3}-25 \times 10^{-3}}=\frac{10}{9}=1 \Omega
\end{aligned}
$$
Hence, the correct answer is option: 'C'.