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 :

<p>To convert a galvanometer into an ammeter to measure larger currents, a low resistance known as the <em>shunt resistance</em> ($R_{\text{sh}}$) is connected in parallel with the galvanometer. The value of this shunt resistance can be calculated using the principles of parallel circuits and the desired maximum current the ammeter should read.</p> <p>The original configuration of the galvanometer allows it to measure up to $10\,\text{V}$, and it has a resistance of $100\,\Omega$. When connected in series with a $400\,\Omega$ resistor, the total resistance in the circuit is $100\,\Omega + 400\,\Omega = 500\,\Omega$. Given this configuration measures up to $10\,\text{V}$, we can calculate the maximum current it is designed to measure using Ohm's law:</p> <p>$I = \frac{V}{R} = \frac{10\,\text{V}}{500\,\Omega} = 0.02\,\text{A}$</p> <p>Now, to recalibrate the device to measure up to $10\,\text{A}$, we require the calculation of the shunt resistor $R_{\text{sh}}$ that needs to be connected in parallel with the galvanometer. The total current $I$ will now be $10\,\text{A}$, and the part of this current flowing through the galvanometer ($I_g$) remains $0.02\,\text{A}$ (as before, to ensure we do not exceed the device's original maximum measuring capability), leaving the rest to flow through the shunt. Thus, $I - I_g$ flows through the shunt.</p> <p>Since the voltage drop across both the shunt and the galvanometer must be the same for parallel components, we use Ohm’s Law $V = IR$ for both and set up an equation to calculate $R_{\text{sh}}$:</p> <p>$I_g R_g = (I - I_g) R_{\text{sh}}$</p> <p>Substituting known values ($R_g = 100\,\Omega$, $I = 10\,\text{A}$, and $I_g = 0.02\,\text{A}$):</p> <p>$0.02\,\text{A} \times 100\,\Omega = (10\,\text{A} - 0.02\,\text{A}) R_{\text{sh}}$</p> <p>This simplifies to:</p> <p>$2\,\text{V} = 9.98\,\text{A} \times R_{\text{sh}}$</p> <p>Solving for $R_{\text{sh}}$ gives:</p> <p>$R_{\text{sh}} = \frac{2\,\text{V}}{9.98\,\text{A}} \approx 0.2004\,\Omega$</p> <p>Expressing this in terms of $\times 10^{-2} \Omega$ gives $R_{\text{sh}} \approx 20.04 \times 10^{-2} \Omega$. Therefore, the value of $x$ is approximately $20.04$, and rounding it according to the provided options leads to the closest value:</p> <p>Option B: 20</p>