Question

A voltmeter has an internal resistance of $1000\Omega $ and gives full scale deflection when 2V is applied across the terminals. Now a resistance of$4000\Omega $ is connected in series with it. Then it gives full scale deflection with
A. 8V
B. 10V
C. 6V
D. 4V

Answer 1

Hint: In order to solve the given question you could consider it as two different parts. For the first part you could use Ohm's law to find the current passing through the voltmeter when it shows full scale deflection with the given internal resistance and voltage. Now apply ohm’s law for the second part and substitute the same current to get voltage across the voltmeter to show full scale deflection.

Formula used: Ohm’s law,
$V=IR$

Complete step by step answer:
In the question, we are given a voltmeter that has an internal resistance of $1000\Omega $ and shows full scale deflection when there is a potential difference of 2V across it. We are now connecting a resistance of $4000\Omega $ in series with the voltmeter and we are supposed to find the voltage for which the voltmeter gives full scale deflection.
In the initial situation when the only resistance was that provided by the internal resistance,
${{V}_{1}}=2V$
$r=1000\Omega $
But from ohm’s law, we have,
$V=IR$
$\Rightarrow I=\dfrac{{{V}_{1}}}{r}$
Substituting the values,
$I=\dfrac{2}{1000}A$ ………………………………. (1)
Now for the second situation, when a resistance of $4000\Omega $ is connected in series with the voltmeter, the current will be the same. So, by ohm’s law,
${{V}_{2}}=I{{R}_{eff}}$ ………………………… (2)
Now the effective resistance will be the sum internal resistance(r) and resistance R. So,
${{R}_{eff}}=4000+r=\left( 4000+1000 \right)\Omega $
$\Rightarrow {{R}_{eff}}=5000\Omega $ …………………………… (3)
Substituting (1) and (3) in (2), we get,
${{V}_{2}}=\dfrac{2}{1000}\left( 5000 \right)$
$\therefore {{V}_{2}}=10V$

So, the correct answer is “Option B”.

Note: The statement of Ohm’s law goes like this: the current through the conductor between any two points is directly proportional to the voltage across these points. The resistance R is actually introduced as the constant of proportionality in the relation. Also, the resistance is independent of the current and is actually the opposition offered to the current flow.