Question

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.

Answer 1

Correct Answer: 200

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 \(\Omega\).