
A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per mill ampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohm needs to be connected in series, with the coil will be:
A) ${10^3}$
B) ${10^5}$
C) $99995$
D) $9995$
Answer
123.3k+ views
Hint: A galvanometer is a device employed to detect current in a circuit. It is an electromagnetic device that has a very high sensitivity that can measure very low currents of the order of microamperes. It is based on the phenomenon of electromagnetic induction.
Complete step by step solution:
No. of divisions in the galvanometer $ = 150$
Current sensitivity $ = 10$ div per milliampere
Voltage sensitivity $ = 2$ div per millivolt
The maximum current that is used in galvanometer for a full scale deflection $ = {I_g} = \dfrac{{150}}{{10}} = 15mA$ ___ $(1)$
For the same full scale deflection, voltage required $ = {V_g} = \dfrac{{150}}{2}75mV$ ___ $(2)$
Using $(1)\& (2)$ , resistance of the galvanometer $ = {R_g} = \dfrac{{{V_g}}}{{{I_g}}}\dfrac{{75}}{{15}} = 5\Omega $
Therefore, the resistance required in series to get converted into voltmeter of range V is
$\Rightarrow 150 \times 1 = 150V$ is
$\Rightarrow R = \dfrac{V}{{{I_g}}} - {R_g}$
$\therefore R = \dfrac{{150}}{{15 \times {{10}^{ - 3}}}} - 5 = 10000 - 5 = 9995\Omega $
Final Answer, Resistance is $9995\Omega $
Hence (D) is correct.
Note: Moving coil galvanometer works on the principle of electromagnetic induction which states that when a current carrying piece of wire is placed in a magnetic field, it experiences a force. In the moving coil galvanometer, the magnetic field is radial.
Complete step by step solution:
No. of divisions in the galvanometer $ = 150$
Current sensitivity $ = 10$ div per milliampere
Voltage sensitivity $ = 2$ div per millivolt
The maximum current that is used in galvanometer for a full scale deflection $ = {I_g} = \dfrac{{150}}{{10}} = 15mA$ ___ $(1)$
For the same full scale deflection, voltage required $ = {V_g} = \dfrac{{150}}{2}75mV$ ___ $(2)$
Using $(1)\& (2)$ , resistance of the galvanometer $ = {R_g} = \dfrac{{{V_g}}}{{{I_g}}}\dfrac{{75}}{{15}} = 5\Omega $
Therefore, the resistance required in series to get converted into voltmeter of range V is
$\Rightarrow 150 \times 1 = 150V$ is
$\Rightarrow R = \dfrac{V}{{{I_g}}} - {R_g}$
$\therefore R = \dfrac{{150}}{{15 \times {{10}^{ - 3}}}} - 5 = 10000 - 5 = 9995\Omega $
Final Answer, Resistance is $9995\Omega $
Hence (D) is correct.
Note: Moving coil galvanometer works on the principle of electromagnetic induction which states that when a current carrying piece of wire is placed in a magnetic field, it experiences a force. In the moving coil galvanometer, the magnetic field is radial.
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