NCERT Based Test: The Moving Coil Galvanometer - NEET MCQ

Since magnetic torque on the coil,
τ = NIAB
This torque is balanced by counter torque
∴ kϕ = 
where k is torsional constant. It is a scalar quantity having dimension of torque or energy i.e. [ML²T⁻²]

For meter M₁, R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10^–3 m², B₁ = 0.25 T, k₁ = k 
For meter M₂, R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 × 10^–3 m², B₂ = 0.50 T, k₂ = k 
As, current sensitivity Is = NBA/k.

Voltage sensitivity V_s = 
So, 

= 
= 
= 1

Current sensitivity of galvanometer is deflection per unit current i.e

Similarly voltage sensitivity is deflection per unit voltage
i.e. 
= 
From (i) and (ii)
Voltage sensitivity = current sensitivity × 1/resistance
Now if current sensitivity is doubled, then the resistance in the circuit will also be doubled since it is proportional to the length of the wire, then voltage sensitivity
= 
= (current sensitivity) × 1 / (resistance)
Hence, voltage sensitivity will remain unchanged

To utilise a galvanometer (G) as an ammeter, there is the difficulty in measurement of current due to the sensitivity of galvanometer and also the connection of galvanometer with a very large resistance in series that may change the value of current in the circuit.
To overcome these difficulties one attaches a small resistance r_s called shunt resistance in parallel with the galvanometer coil as shown in the figure.

Here, R_G = 70Ω, r_s = 0.03Ω
∴ R = 
= 0.02998 
= 0.03Ω

= 0.21A

I_G = 50/5 = 10mA; R^G = 40Ω, R_s = 2Ω
Maximum current,
I = 
= 
= 210 mA

The equivalent circuit of given circuit is given by
Here,
= 
∴ RS = 0.0499Ω = 0.05Ω
The total resistance in the circuit
R = R_S + 5Ω = 0.05 + 5 = 5.05Ω
The current measured by the device

= 0.99 A

= 
= 
= 
= 
∵ S = 
∴ l  = 
= 3.0m

Galvanometer is a very sensitive instrument, therefore it can not measure high potential difference.
In order to convert a Galvanometer into voltmeter, a very high resistance known as "series resistance" is connected in series with the galvanometer.

Putting V = 2V,
I_g = 2mA
= 2 × 10⁻³A,
G = 12 Ω,
∴ R= 2 / 2 × 10⁻³ −12 = 1000 − 12

= 988Ω

R_eq = R_g +3 = 50 + 3 = 53Ω

The galvanometer can be transformed into a voltmeter by applying a high resistance in series. If the series resistance is R, then total resistance in circuit  = (15 + R)Ω
Now 
= 
= 4.5 × 10³ 
∴ R  = 4500−15
= 4485Ω

Total initial resistance
= G + R = 50Ω + 2950Ω = 3000Ω
Current,I = 

= 1 × 10⁻³ A = 1mA
If the deflection has to be reduced to 20 divisions, then current

Let x be the effective resistance of the circuit,
3V = 3000Ω × 1mA =  
or x = 
∴  Resistance to be added =(4500Ω − 50Ω) = 4450Ω