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# A range of galvanometer is $V$, when $50\;\Omega$ resistance is connected in series, its range gets doubled when $500\;\Omega$ resistance is connected in series. Galvanometer resistance is?A. $100\;\Omega$B. $200\;\Omega$C. $300\;\Omega$D. $400\;\Omega$

Last updated date: 12th Sep 2024
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Hint:-In this question, the concept of the galvanometer is used that is the galvanometer is designed to detect feeble current only as the current passes through it deflection depicts the presence of current and deflection is proportional to current and then first write the expression for the voltage of the galvanometer in terms of current, internal and external resistance, then obtain the separate equations for the given cases and calculate them to obtain the internal resistance of the galvanometer.

Complete step-by-step solution:-
In this question, it is given that the voltage range of the galvanometer is $V$ if the magnitude of the resistance is $50\;\Omega$ and if the resistance increases to $500\;\Omega$, the voltage range also increases to $2V$. In this question, the internal resistance of the galvanometer needs to be calculated.
The galvanometer is designed to detect feeble current only as the current passes through it; deflection depicts the presence of current and deflection is proportional to current. To convert galvanometer into ammeter a shunt resistance is connected to it in parallel.
In this question, there are two values of resistance therefore we will calculate the potential at these two given values of resistance that is, for $R = 50\;\Omega$
$V = IG(G + R)$
Here, the current passing through the galvanometer is ${I_G}$, the internal resistance of the galvanometer is $G$. Now, substitute the given values in the above expression as,
$V = IG\left( {G + 50} \right)......\left( 1 \right)$
Similarly, for $R = 500\;\Omega$, the value of the range of the galvanometer gets doubled as mentioned in the question. Hence, $V = 2V$
Now, on substituting the values in the above given equation we get,
$2V = IG\left( {G + 500} \right)......\left( 2 \right)$

Now, we divide equation (2) by equation (1) as,
$\dfrac{{2V}}{V} = \dfrac{{G + 500}}{{G + 50}}$
We apply the cross multiplication for the above equation and obtain,
$2\left( {G + 50} \right) = G + 500$
By simplifying the above expression, we get,
$\therefore G = 400\;\Omega$
Hence, the value of the resistance of the galvanometer is $400\;\Omega$.
Therefore, the correct option is D.

Note:- Make sure that the internal resistance and the external resistance are in the series so, the resultant resistance will be obtained by adding them and as the external resistance increases the voltage range of the galvanometer will also increase.