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, denoted as \( R_v \), is connected in parallel with the 200\(\Omega\) resistor. The equivalent resistance of this parallel configuration is calculated as:

\[R_{\text{parallel}} = \frac{R_v \cdot 200}{R_v + 200}.\]

The total resistance of the circuit is determined by adding this parallel resistance to the 100\(\Omega\) resistor:

\[R_{\text{total}} = 100 + R_{\text{parallel}}.\]

According to the voltage division rule, the voltage across the 100\(\Omega\) resistor is expressed as:

\[V_{100} = \frac{100}{R_{\text{total}}} \cdot V_{\text{total}}.\]

Substituting the given values into the equation yields:

\[\frac{4}{3} = \frac{100}{100 + \frac{R_v \cdot 200}{R_v + 200}} \cdot 4.\]

Dividing both sides by 4 simplifies the equation to:

\[\frac{1}{3} = \frac{100}{100 + \frac{R_v \cdot 200}{R_v + 200}}.\]

Taking the reciprocal of both sides results in:

\[3 = \frac{100 + \frac{R_v \cdot 200}{R_v + 200}}{100}.\]

Multiplying both sides by 100 gives:

\[300 = 100 + \frac{R_v \cdot 200}{R_v + 200}.\]

Rearranging the equation to isolate the parallel resistance term:

\[200 = \frac{R_v \cdot 200}{R_v + 200}.\]

Cross-multiplying to eliminate the denominator yields:

\[200(R_v + 200) = R_v \cdot 200.\]

Expanding the terms on the left side gives:

\[200R_v + 40000 = R_v \cdot 200.\]

Subtracting \(200R_v\) from both sides cancels this term, leaving:

\[40000 = 200R_v.\]

Solving for \(R_v\) yields the voltmeter's resistance:

\[R_v = 200\Omega.\]

Final Answer: The resistance of the voltmeter is:

200 \(\Omega\).