Two resistances of 100Ω and 200Ω are connected in series with a battery of 4V and negligible internal resistance. A voltmeter is used to measure voltage across the 100Ω resistance, which gives a reading of 1V. The resistance of the voltmeter must be _____ Ω.
Two resistances of 100Ω and 200Ω are connected in series with a battery of 4V and negligible internal resistance. A voltmeter is used to measure voltage across the 100Ω resistance, which gives a reading of 1V. The resistance of the voltmeter must be _____ Ω.
Correct Answer: 200
Solution and Explanation
The voltmeter \( R_v \) is connected in parallel with the 200\(\Omega\) resistor. The equivalent resistance of this parallel combination is:
\[ R_{\text{parallel}} = \frac{R_v \cdot 200}{R_v + 200}. \]
The total resistance of the circuit is:
\[ R_{\text{total}} = 100 + R_{\text{parallel}}. \]
Using the voltage division rule, the voltage across the 100\(\Omega\) resistor is given as:
\[ V_{100} = \frac{100}{R_{\text{total}}} \cdot V_{\text{total}}. \]
Substitute the given values:
\[ \frac{4}{3} = \frac{100}{100 + \frac{R_v \cdot 200}{R_v + 200}} \cdot 4. \]
Simplify by dividing through by 4:
\[ \frac{1}{3} = \frac{100}{100 + \frac{R_v \cdot 200}{R_v + 200}}. \]
Take the reciprocal:
\[ 3 = \frac{100 + \frac{R_v \cdot 200}{R_v + 200}}{100}. \]
Multiply through by 100:
\[ 300 = 100 + \frac{R_v \cdot 200}{R_v + 200}. \]
Rearrange:
\[ 200 = \frac{R_v \cdot 200}{R_v + 200}. \]
Simplify by cross-multiplying:
\[ 200(R_v + 200) = R_v \cdot 200. \]
Expand terms:
\[ 200R_v + 40000 = R_v \cdot 200. \]
Cancel \(200R_v\) on both sides:
\[ 40000 = 200R_v. \]
Solve for \(R_v\):
\[ R_v = 200\Omega. \]
Final Answer: The resistance of the voltmeter is:
200 \(\Omega\).
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