A galvanometer is to be turned into a voltmeter that deflects full scale for a potential difference of \(100.0 \mathrm{V}\) What size resistor should be placed in series with the galvanometer if it has an internal resistance of $75 \Omega\( and deflects full scale for a current of \)2.0 \mathrm{mA} ?$

The full-scale deflection current, \(2.0 \mathrm{mA}\), must be converted into amperes (A). To do this, multiply the value by \(10^{-3}:\) \(2.0 \mathrm{mA} = 2.0 \times 10^{-3} \mathrm{A}.\)

We can use Ohm's law to find the potential difference across the internal resistance of the galvanometer. The formula is given by \(V_{\mathrm{internal}} = I_{\mathrm{internal}} \times R_{\mathrm{internal}}\), where \(V_{\mathrm{internal}}\) is the potential difference across the internal resistance, \(I_{\mathrm{internal}}\) is the current for full-scale deflection, and \(R_{\mathrm{internal}}\) is the internal resistance of the galvanometer. Plugging in the given values, we get: \(V_{\mathrm{internal}} = (2.0\times 10^{-3} \mathrm{A})(75 \Omega) = 0.15 \mathrm{V}\)

To find the potential difference across the external resistor (\(V_{\mathrm{external}}\)), subtract the potential difference across the internal resistance from the total potential difference: \(V_{\mathrm{external}} = V_{\mathrm{total}} - V_{\mathrm{internal}} = 100.0 \mathrm{V} - 0.15 \mathrm{V} = 99.85 \mathrm{V}\)

Now we need to find the size of the external resistor (\(R_{\mathrm{external}}\)) that we must add in series with the galvanometer to make it read a full-scale deflection of \(100.0 \mathrm{V}\). We can use Ohm's Law, \(V_{\mathrm{external}} = I_{\mathrm{internal}} \times R_{\mathrm{external}}\), where \(V_{\mathrm{external}}\) is the potential difference across the external resistor, \(I_{\mathrm{internal}}\) is the current for full-scale deflection, and \(R_{\mathrm{external}}\) is the external resistance. Rearranging the formula to solve for \(R_{\mathrm{external}}\), we get: \(R_{\mathrm{external}} = \frac{V_{\mathrm{external}}}{I_{\mathrm{internal}}} = \frac{99.85 \mathrm{V}}{2.0\times 10^{-3}\mathrm{A}} = 49,925 \Omega\)

Since we have three significant figures in all given values, the size of the external resistor should also have three significant figures: \(R_{\mathrm{external}} = 49,900 \Omega\) Hence, a \(49,900\ Ohm\) resistor should be placed in series with the galvanometer to turn it into a voltmeter that deflects full scale for a potential difference of \(100.0 \mathrm{V}\).