$8.3.3.$ A voltmeter with a scale of 100 V has an internal resistance of 10 kOhm. What is the greatest potential difference that can be measured with this device if an additional resistance of 90 kOhm is connected to it?
Solution
The current at which the voltmeter's needle deflects to the maximum scale value is: $$I = \frac{U_{\text{v}}}{R_{\text{v}}}$$ where $U_{\text{v}}$ is the measurement limit of the voltmeter, and $R_{\text{v}}$ is its internal resistance.
If an additional resistance $R_{\text{add}}$ is connected in series with the voltmeter, the total resistance of the circuit becomes $R_{\text{v}} + R_{\text{add}}$. The maximum potential difference $\Delta\varphi$ that such a system can measure is achieved at the same maximum current $I$: $$\Delta\varphi = I \cdot (R_{\text{v}} + R_{\text{add}})$$
Substitute the expression for the current into the potential difference formula: $$\Delta\varphi = \frac{U_{\text{v}}}{R_{\text{v}}} \cdot (R_{\text{v}} + R_{\text{add}}) = U_{\text{v}} \left( 1 + \frac{R_{\text{add}}}{R_{\text{v}}} \right)$$
Perform the calculation by substituting the known values ($U_{\text{v}} =$ 100 V, $R_{\text{v}} =$ 10 kOhm, $R_{\text{add}} =$ 90 kOhm): $$\Delta\varphi = 100 \cdot \left( 1 + \frac{90}{10} \right) = 100 \cdot (1 + 9) = 1000 \text{ V}$$