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

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 $\Omega$.