Q71 / 90 · JEE Main 2024 (05 Apr Shift 2) · Physics

A galvanometer of resistance $100 \Omega$ when connected in series with $400 \Omega$ measures a voltage of upto $10 \mathrm{~V}$. The value of resistance required to convert the galvanometer into ammeter to read upto $10 \mathrm{~A}$ is $x \times 10^{-2} \Omega$. The value of $x$ is :

  • A
    $2$
  • B
    $800$
  • Correct answer
    C
    $20$
  • D
    $200$

Solution

$\mathrm{i}_{\mathrm{g}}=\frac{10}{400+100}=20 \times 10^{-3} \mathrm{~A}$ For ammeter Let shunt resistance $=\mathrm{S}$ $\begin{aligned} & \mathrm{i}_{\mathrm{g}} \mathrm{R}=\left(\mathrm{i}-\mathrm{i}_{\mathrm{g}}\right) \mathrm{S} \\ & 20 \times 10^{-3} \times 100=10 \mathrm{~S} \\ & \mathrm{~S}=20 \times 10^{-2} \Omega \end{aligned}$

Hence, the correct answer is option: 'C'.

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