Question 47: Current Electricity

Question

A voltmeter of resistance $$1000 \Omega .0 .5 \mathrm{~V} /$$ div is to be converted into a voltmeter to make it to read $$=1 \mathrm{~V} / \mathrm{div}$$. The value of high resistance to be connected in series with it is

Accepted Answer

To convert a voltmeter of resistance $$1000 \Omega$$ with a sensitivity of $$0.5 \mathrm{~V} / 	ext{div}$$ into a voltmeter with a sensitivity of $$1 \mathrm{~V} / 	ext{div}$$, we need to connect an additional high resistance in series with the voltmeter.
The original voltmeter reads $$0.5 \mathrm{~V} / 	ext{div}$$ which means for every $$0.5 \mathrm{~V}$$, the internal resistance is $$1000 \Omega$$. To make it read $$1 \mathrm{~V} / 	ext{div}$$, we effectively need to double the voltage range it can measure for the same current passing through the meter.
Let the required high resistance to be connected in series be $$R_{s}$$. The total resistance of the voltmeter circuit for $$1 \mathrm{~V} / 	ext{div}$$ would then be given by:
$$ R_{total} = R_{v} + R_{s} $$
Given the original voltmeter resistance ($$R_{v}$$) is $$1000 \Omega$$ and considering we need to double the effective voltage being measured, the total resistance should also be doubled:
$$ R_{total} = 2 	imes R_{v} $$
Substituting $$R_{v}$$:
$$ 2 	imes 1000 \Omega = 2000 \Omega $$
This gives the equation:
$$ R_{total} = 1000 \Omega + R_{s} = 2000 \Omega $$
Solving for $$R_{s}$$:
$$ R_{s} = 2000 \Omega - 1000 \Omega = 1000 \Omega $$
Thus, the value of the high resistance to be connected in series with it is:
Option D: $$1000 \Omega$$

Answer

$$6000 \Omega$$

Answer

$$5000 \Omega$$

Answer

$$4000 \Omega$$

Answer

$$1000 \Omega$$

Current Electricity

A voltmeter of resistance $1000 \Omega .0 .5 \mathrm{~V} / div is to be converted into a voltmeter to make it to read =1 \mathrm{~V} / \mathrm{div}$. The value of high resistance to be connected in series with it is

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