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# 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) $2.535 \times {10^3}\Omega$

Last updated date: 29th Feb 2024
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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).