Courses
Courses for Kids
Free study material
Offline Centres
More

When a current of $5mA$ is passed through a galvanometer having a coil of resistance$15\Omega $, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range $0 - 10V$ is:
(A) $4.005 \times {10^3}\Omega $
(B) $1.985 \times {10^3}\Omega $
(C) $2.045 \times {10^3}\Omega $
(D) $2.535 \times {10^3}\Omega $

seo-qna
Last updated date: 29th Feb 2024
Total views: 21k
Views today: 1.21k
Rank Predictor
Answer
VerifiedVerified
21k+ views
Hint We are given with the magnitude of the current through the galvanometer and the coil resistance is given and we are asked to calculate the resistance required to convert the galvanometer to a voltmeter of full scale deflection of$10V$. Thus, we will take into account the resistance used to convert a galvanometer to a voltmeter, we have to connect that resistance in series with the galvanometer.
Formula Used
$V = IR$
Where, $V$ is the potential difference across the circuit, $I$ is the current through the circuit and $R$ is the resistance of the circuit.

Complete Step By Step Solution



Here,
$V = 10V$
$I = {I_g} = 5mA = 5 \times {10^{ - 3}}A$
${R_g} = 15\Omega $
Now,
${R_g}$ and ${R_S}$ are in series.
Thus, the total resistance will be$({R_g} + {R_S})$.
Now,
$V = {I_g}({R_g} + {R_S})$
Putting in the values, we get
$10 = (5 \times {10^{ - 3}})(15 + {R_S})$
Further, we get
$15 + {R_S} = \dfrac{{10}}{{5 \times {{10}^{ - 3}}}}$
Also, we get
$15 + {R_S} = 2 \times {10^3}$
Again, after calculation, we get
${R_S} = 2000 - 15$
Finally, we get
${R_S} = 1985\Omega = 1.985 \times {10^3}\Omega $

Hence, the correct option is (B).

Additional Information
In the case of voltmeter, the resistance connected to the galvanometer should have a high resistance as the voltmeter should be such that the equipment pulls the minimal amount of current. On the other hand, the resistance connected to the galvanometer for converting into an ammeter, the resistance should be low as the targeted equipment should pull the maximum amount of current.

Note In this case, the resistance is connected in series and due to which the total resistance was the summation of the coil resistance and the extra resistance. But if the case was for ammeter, the resistance was to be connected in parallel with the galvanometer and thus the formula for the net resistance would change and the net answer will also change.