Question 61: Current Electricity

Question

The current sensitivity of a galvanometer having 20 divisions is $$10 \mu \mathrm{A} /$$ div. If the resistance of the galvanometer is $$100 \Omega$$ then the value of the resistance to be used to convert this galvanometer in to an voltmeter to read up to $$1 \mathrm{~V}$$ is :

Accepted Answer

To convert a galvanometer into a voltmeter, a high resistance needs to be added in series with the galvanometer. This is because a voltmeter must have a high resistance to ensure that it does not draw significant current from the circuit being measured, which can alter the voltage being measured. Therefore, out of the options given, we can immediately disregard the options mentioning a resistance in parallel with the galvanometer or resistances that are too low to make a sensible voltmeter.
The current sensitivity of the galvanometer is $$10 \mu \mathrm{A} / 	ext{div}$$, and it has 20 divisions. This means that the full-scale deflection current of the galvanometer is:
$20 	imes 10 \mu \mathrm{A} = 200 \mu \mathrm{A} = 200 	imes 10^{-6} \mathrm{A}$
To read up to $$1 \mathrm{V}$$, we require this full-scale current of $$200 \mu \mathrm{A}$$ to flow when a potential difference of $$1 \mathrm{V}$$ is applied across the modified galvanometer (now a voltmeter).
The internal resistance of the galvanometer, $R_g$, is $$100 \Omega$$. The voltage $V$ across the galvanometer and the series resistance $R_s$ together is given by Ohm's law, $V = I_t 	imes (R_g + R_s)$, where $I_t$ is the total current for full-scale deflection, which is $$200 \mu \mathrm{A} = 200 	imes 10^{-6} \mathrm{A}\). Set this equal to 1 V to find $R_s$.
$1 = (200 	imes 10^{-6}) 	imes (100 + R_s)$
$1 = (200 	imes 10^{-6} 	imes 100) + (200 	imes 10^{-6} 	imes R_s)$
$1 = 0.02 + (200 	imes 10^{-6} 	imes R_s)$
Subtracting 0.02 from both sides yields:
$0.98 = 200 	imes 10^{-6} 	imes R_s$
Solving for $R_s$ gives:
$R_s = \frac{0.98}{200 	imes 10^{-6}} = \frac{0.98}{0.2 	imes 10^{-3}} = 4900 \Omega $
Hence, the correct answer is a $$4900 \Omega$$ resistor in series with the galvanometer to convert it into a voltmeter that can read up to $$1 \mathrm{V}$$. This matches
Option D: $4900 \Omega$ in series with the galvanometer
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Answer

$$4 	imes 10^{-6} \Omega$$ in series with the galvanometer.

Answer

$$4900 \Omega$$ in parallel with the galvanometer.

Answer

$$4 	imes 10^{-3} \Omega$$ in series with the galvanometer.

Answer

$$4900 \Omega$$ in series with the galvanometer.

Current Electricity

The current sensitivity of a galvanometer having 20 divisions is $10 \mu \mathrm{A} / div. If the resistance of the galvanometer is 100 \Omega then the value of the resistance to be used to convert this galvanometer in to an voltmeter to read up to 1 \mathrm{~V}$ is :

Choose the correct answer: