Current Electricity and Electrical InstrumentMedium
Question
A voltmeter with internal resistance of $x\Omega$ can be used to measure upto 20 V . In order to increase its measuring range to 30 V , the required modification is to $\_\_\_\_$ .
Options
A.connect resistor of $\frac{x}{2}\Omega$, in series with voltmeter.
B.connect resistor of $\frac{X}{2}\Omega$, in parallel to voltmeter.
C.connect resistor of $x\Omega$, in series with voltmeter.
D.connect resistor of $2 \times \Omega$, in parallel to voltmeter.
Solution
Sol. [Key Concept: Full-scale current is unchanged: $\frac{20}{x} = \frac{30}{x + R}$; solve $R$ and note it goes in series.]
Since maximum current through voltmeter remains same
So, $\frac{20}{x} = \frac{30}{x + R}$
$$\begin{matrix} & 20x + 20R = 30x \\ & 20R = 10x \\ & R = \frac{x}{2} \end{matrix}$$
This extra resistance must be connected in series.
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