Question 180: Current Electricity

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 $\_\_\_\_$ .

Accepted Answer

For a voltmeter, to increase its range, we connect a
high resistance in series
. Let us calculate it.
The voltmeter has internal resistance $x \, \Omega$ and can measure up to $20 \, 	ext{V}$.
So, the full-scale current of the voltmeter is
$$ I = \frac{V}{R} = \frac{20}{x} $$
Now we want the new range to be $30 \, 	ext{V}$.
Let the additional resistance connected in series be $R$.
Then for the same full-scale current,
$$ 30 = I(x + R) $$
Substitute $I = \frac{20}{x}$:
$$ 30 = \frac{20}{x}(x + R) $$
$$ 30x = 20(x + R) $$
$$ 30x = 20x + 20R $$
$$ 10x = 20R $$
$$ R = \frac{x}{2} $$
So, we must
connect a resistor of $\frac{x}{2}\,\Omega$ in series with the voltmeter
.
Therefore, the correct option is:
$$ \boxed{	ext{Option A}} $$

Answer

connect resistor of $\frac{x}{2} \Omega$, in series with voltmeter.

Answer

connect resistor of $\frac{x}{2} \Omega$, in parallel to voltmeter.

Answer

connect a resistor of $x \Omega$ in series with voltmeter.

Answer

connect resistor of $2 x \Omega$ in parallel to voltmeter.

Current Electricity

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 \_\_\_\_ .

Choose the correct answer: