A galvanometer of resistance \( 20\,\Omega \) gives full-scale deflection for a current of 5 mA. It is converted into an ammeter of range 0 to 5 A by connecting a suitable shunt across it. When this ammeter reads 2 A, the potential difference across its terminals is nearly
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.04 V
In the ammeter, the shunt is joined in parallel with the galvanometer. So the two always have the same potential difference across them. At full scale the galvanometer carries its full-scale current \( I_g \), and this fixes the potential difference across the whole ammeter.
At the full-scale reading of 5 A:
\( \begin{aligned} V_{\mathrm{fs}} &= I_g\,G \\ &= 5\times10^{-3}\,\mathrm{A}\times20\,\Omega \\ &= 0.1\,\mathrm{V} \end{aligned} \)
So the resistance of the ammeter is:
\( \begin{aligned} R_A &= \frac{V_{\mathrm{fs}}}{5\,\mathrm{A}} \\ &= \frac{0.1\,\mathrm{V}}{5\,\mathrm{A}} \\ &= 0.02\,\Omega \end{aligned} \)
When the reading is 2 A, the potential difference across the ammeter is:
\( \begin{aligned} V &= I\,R_A \\ &= 2\,\mathrm{A}\times0.02\,\Omega \\ &= 0.04\,\mathrm{V} \end{aligned} \)