A voltmeter has a switch that enables voltages to be measured with a maximum of \(25.0 \mathrm{V}\) or \(10.0 \mathrm{V}\). For a range of voltages to $25.0 \mathrm{V},\( the switch connects a resistor of magnitude \)9850 \Omega$ in series with the galvanometer; for a range of voltages to \(10.0 \mathrm{V}\), the switch connects a resistor of magnitude \(3850 \Omega\) in series with the galvanometer. Find the coil resistance of the galvanometer and the galvanometer current that causes a full-scale deflection. [Hint: There are two unknowns, so you will need to solve two equations simultaneously.]

In order to understand the situation, we will start by finding an expression for the total resistance for both configurations of the voltmeter. For the 25.0 V configuration, the total resistance is given by the sum of the 9,850 Ω resistor and the resistance of the galvanometer (which we will denote as R). So, we have: \(R_{total1} = 9850 + R\) For the 10.0 V configuration, the total resistance is given by the sum of the 3,850 Ω resistor and the resistance of the galvanometer. Thus, we have: \(R_{total2} = 3850 + R\)

Now that we have the total resistance for both configurations, we will now use Ohm's Law (\(V = IR\)) to find an expression for the current in each case. Let's denote the full-scale deflection current of the galvanometer as \(I_G\). For the 25.0 V configuration, we can apply Ohm's Law: \(I_{G1} = \frac{V_1}{R_{total1}}\) For the 10.0 V configuration: \(I_{G2} = \frac{V_2}{R_{total2}}\) We know the voltages are 25.0 V and 10.0 V in the respective configurations. This means: \(I_{G1} = \frac{25.0}{9850 + R}\) \(I_{G2} = \frac{10.0}{3850 + R}\)

Now we have two equations with two unknowns: the coil resistance \(R\) and the galvanometer current \(I_G\). By equating the values of \(I_{G1}\) and \(I_{G2}\), we can solve for \(R\) and \(I_G\). \(\frac{25.0}{9850 + R} = \frac{10.0}{3850 + R}\) Now, we can cross-multiply and simplify the equation: \(25.0(3850 + R) = 10.0(9850 + R)\) \(96250 + 25R = 98500 + 10R\) \(15R = 2250\) \(R = 150 \Omega\) Now that we have the coil resistance, we can substitute it back into one of the current equations to find the galvanometer current: \(I_G = \frac{10.0}{3850 + 150}\) \(I_G = \frac{10.0}{4000} = 0.0025 \mathrm{A}\) So the coil resistance of the galvanometer is 150 Ω, and the galvanometer current that causes a full-scale deflection is 0.0025 A.