The maximum permissible current through a galvanometer G of internal resistance 10Ω Ω is 0.05A. A resistance R is used to convert G into a voltmeter with a ...

Answer Details

The question is asking for the value of the resistor R that needs to be connected to a galvanometer G of internal resistance 10 ohms, to convert it into a voltmeter with a maximum reading of 100V. The maximum current that can flow through the galvanometer is given as 0.05A. To convert the galvanometer into a voltmeter, a resistor needs to be connected in series with it. The value of this resistor can be calculated using the formula: R = (Vmax/Ig) - Rg Where R is the value of the resistor to be connected, Vmax is the maximum voltage that the voltmeter can read (in this case, 100V), Ig is the maximum current that can flow through the galvanometer (0.05A), and Rg is the internal resistance of the galvanometer (10 ohms). Substituting the given values, we get: R = (100/0.05) - 10 R = 1990 ohms Therefore, the value of the resistor R that needs to be connected in series with the galvanometer G is 1990 ohms. It should be connected in series with the galvanometer to convert it into a voltmeter. So, the correct answer is "1990 ohms in series".