Question

A voltmeter reads 5.0 volt at full scale deflection and is graded according to its resistance per volt at full-scale deflection as 5000$\Omega {V^{ - 1}}$. How will you convert it into a voltmeter that reads 20 V at full-scale deflection? Will it still be graded as 5000$\Omega {V^{ - 1}}$? Will you prefer this voltmeter to one that is graded 2000$\Omega {V^{ - 1}}$?

Answer 1

Hint: In this question use the concept that if we want to change the full scale deflection with same grading we have to add an external resistance later on in the solution using the concept that higher current system is not preferable compared to lower current system for same full scale deflection.

Complete Step-by-Step solution:
Given data:
Resistance per volt = 5000$\Omega {V^{ - 1}}$
So the potential current $\left( {{I_g}} \right)$ for 1 volt = $\dfrac{1}{{5000}}$ Amp.
Now, reading of voltmeter = 5 V at full scale deflection.
So the resistance (R) of that system is
$ \Rightarrow R = 5000 \times 5 = 25000$$\Omega $.
Now we have to convert it into a voltmeter that reads 20 V at full scale deflection.
So we have to add an external resistance (r), now the potential current remains same as there is no breaking of branch or any other connection, therefore
$ \Rightarrow 20 = {I_g}\left( {R + r} \right)$
Now substitute the values we have,
$ \Rightarrow 20 = \dfrac{1}{{5000}}\left( {25000 + r} \right)$
$ \Rightarrow 100000 = \left( {25000 + r} \right)$
$ \Rightarrow r = 100000 - 25000 = 75000$$\Omega $.
So the total resistance of the voltmeter become
$ \Rightarrow R + r = 25000 + 75000 + 100000$$\Omega $.
So the grading of the voltmeter that reads 20 V at full scale deflection is
Grading = $\dfrac{{100000}}{{20}} = 5000$$\Omega {V^{ - 1}}$.
So we will still graded as 5000$\Omega {V^{ - 1}}$.
Now the other grading is given as 2000$\Omega {V^{ - 1}}$.
So the potential current ${\left( {{I_g}} \right)_1}$ for 1 volt = $\dfrac{1}{{2000}}$ Amp.
As we see that $\dfrac{1}{{5000}} < \dfrac{1}{{2000}}$
$ \Rightarrow \left( {{I_g}} \right) < {\left( {{I_g}} \right)_1}$
The current drawn by 2000$\Omega {V^{ - 1}}$ grading system is more than 5000$\Omega {V^{ - 1}}$ grading system.
Which is not preferable as higher current systems will heat up quickly and there is a chance of catching fire.
So we will not prefer the 2000$\Omega {V^{ - 1}}$ grading system.
So this is the required answer.

Note – Whenever we face such types of question the key concept is same current as grading is same for two different full scale deflection and for higher we have to add an external resistance so calculate this resistance value as above, then for same full scale deflection if we prefer lower grading system than the current drawn by the meter is higher so that there is a chance of burning the meter or damage of the meter which is not preferable.