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 $x imes 10^{-2} \Omega$. The value of $x$ is :

Question

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 $x 	imes 10^{-2} \Omega$. The value of $x$ is :

Aicharya · Accepted Answer

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

A galvanometer of resistance $100 \Omega$ when connected in series with $400 \Omega$ measures a voltage of…

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