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+### Statement
+
+$8.3.3.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$8.3.3.$ A voltmeter with a scale of <b>100 V</b> has an internal resistance of <b>10 kOhm</b>. What is the greatest potential difference that can be measured with this device if an additional resistance of <b>90 kOhm</b> 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):
+$$\Delta\varphi = 100 \cdot \left( 1 + \frac{90}{10} \right) = 100 \cdot (1 + 9) = 1000 \text{ V}$$
+
+Convert the result to kilovolts: <b>1000 V = 1 kV</b>.
+
+#### Answer
+$$\Delta\varphi = 1 \text{ kV}$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]