Edits to “Statement”, “Answer”
en/8.3.3.md
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| ### Statement | |||
| − | $8.3.3.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
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| − | ### 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>. | |||
| @@ -28,7 +22,3 @@Solution | |||
| #### Answer | |||
| $$\Delta\varphi = 1 \text{ kV}$$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### 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? | $8.3.3.$ 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 | ### Solution | ||
| The current at which the voltmeter's needle deflects to the maximum scale value is: | The current at which the voltmeter's needle deflects to the maximum scale value is: | ||
| $$I = \frac{U_{\text{v}}}{R_{\text{v}}}$$ | $$I = \frac{U_{\text{v}}}{R_{\text{v}}}$$ | ||
| where $U_{\text{v}}$ is the measurement limit of the voltmeter, and $R_{\text{v}}$ is its internal resistance. | 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 $R_{\text{v}} + R_{\text{add}}$. | 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 $I$: | The maximum potential difference $\Delta\varphi$ that such a system can measure is achieved at the same maximum current $I$: | ||
| $$\Delta\varphi = I \cdot (R_{\text{v}} + R_{\text{add}})$$ | $$\Delta\varphi = I \cdot (R_{\text{v}} + R_{\text{add}})$$ | ||
| Substitute the expression for the current into the potential difference formula: | Substitute the expression for the current into the potential difference formula: | ||
| $$\Delta\varphi = \frac{U_{\text{v}}}{R_{\text{v}}} \cdot (R_{\text{v}} + R_{\text{add}}) = U_{\text{v}} \left( 1 + \frac{R_{\text{add}}}{R_{\text{v}}} \right)$$ | $$\Delta\varphi = \frac{U_{\text{v}}}{R_{\text{v}}} \cdot (R_{\text{v}} + R_{\text{add}}) = U_{\text{v}} \left( 1 + \frac{R_{\text{add}}}{R_{\text{v}}} \right)$$ | ||
| Perform the calculation by substituting the known values ($U_{\text{v}} =$ 100 V, $R_{\text{v}} =$ 10 kOhm, $R_{\text{add}} =$ 90 kOhm): | Perform the calculation by substituting the known values ($U_{\text{v}} =$ 100 V, $R_{\text{v}} =$ 10 kOhm, $R_{\text{add}} =$ 90 kOhm): | ||
| $$\Delta\varphi = 100 \cdot \left( 1 + \frac{90}{10} \right) = 100 \cdot (1 + 9) = 1000 \text{ V}$$ | $$\Delta\varphi = 100 \cdot \left( 1 + \frac{90}{10} \right) = 100 \cdot (1 + 9) = 1000 \text{ V}$$ | ||
| Convert the result to kilovolts: <b>1000 V = 1 kV</b>. | Convert the result to kilovolts: <b>1000 V = 1 kV</b>. | ||
| @@ -28,7 +22,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $$\Delta\varphi = 1 \text{ kV}$$ | $$\Delta\varphi = 1 \text{ kV}$$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||