Question

A voltmeter of resistance $2000\,Omega$ reads $1\,volt/div$ . It is made to read $10\,volt/div$ . We should connect a resistance.
(A) $18\,\Omega \,$ in series
(B) $1800\,\Omega $ in series
(C) $1800\,\omega $ in parallel
(D) $18000\,\Omega $ in series

Answer 1

Hint: Substitute the value of the resistance and the corresponding potential difference given in the question in the ohm’s law. Multiply the whole equation obtained by $10$ , to find the value of the resistance to be conn$V = IR$ected for the $\,10\,volt/div$ .

Formula used:
The ohm’s law is given by
$V = IR$
Where $V$ is the potential difference developed across the circuit of the voltmeter, $I$ is the current flowing through it and $R$ is the resistance in the circuit of the voltmeter, while current flowing through it.

Complete step by step solution
It is given that the
Resistance of the voltmeter, $R = 2000\,Omega$
Volt per division on the voltmeter is $10$
Let us consider that the voltmeter reading, ${V_0} = 1\,Vdi{v^{ - 1}}$ and the ${V_1} = 10\,Vdi{v^{ - 1}}$
By using the ohm’s law,
Substituting the values for the voltmeter reading of $1\,Vdi{v^{ - 1}}$
$1V = I\left( {2000} \right)$
Multiplying both sides by $10$ to get the resistance for the values of the voltmeter reading of ${V_1} = 10\,Vdi{v^{ - 1}}$ .
$1 \times 10\,V = 2000 \times 10\,\Omega $
By simplifying the above step,
$10\,V = 20000\,\Omega $
Hence for the $10\,volt/div$ resistance of the $20000\,\Omega $ is required. Already the $2000\,\Omega $ resistance is connected in the voltmeter, so additionally, the $18000\,\Omega $ is connected in series and it is required for the maintenance of the resistance in the voltmeter.

Thus the option (D) is correct.

Note: In this solution, the $18000\,\Omega $ is made to connect in series , this is because in series the total resistance across the circuit is the sum of the individual resistance. Hence already there is $2000\,\Omega $ and along with it when the $18000\,\Omega $ is connected in series, the total resistance will be $20000\,\Omega $ .