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Moving Charges and Magnetism - Moving Coil Galvanometer

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Moving Coil Galvanometer (MCG) is an instrument used for detecting and measuring small electric currents.

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The principle of an MCG is that a current-carrying coil placed in a uniform magnetic field experiences a torque, which causes it to rotate.

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In a radial magnetic field, the plane of the coil is always parallel to the magnetic field, ensuring that the angle between the area vector and magnetic field θ=90∘\theta = 90^\circ at all times. This makes the deflecting torque τ=NIAB\tau = NIAB (independent of the angle of rotation).

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Equilibrium condition: The deflecting torque τd=NIAB\tau_d = NIAB is balanced by the restoring torque τr=kϕ\tau_r = k\phi provided by the spring, where kk is the torsional constant and ϕ\phi is the angular deflection.

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Current Sensitivity (SiS_i): It is defined as the deflection per unit current, given by ϕI\frac{\phi}{I}. Increasing NN, AA, or BB, or decreasing kk, increases the sensitivity.

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Voltage Sensitivity (SvS_v): It is defined as the deflection per unit voltage, given by ϕV\frac{\phi}{V}. Note that increasing the number of turns NN does not necessarily increase SvS_v if the resistance RR also increases proportionally.

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Conversion to Ammeter: A galvanometer is converted into an ammeter by connecting a low resistance called a 'shunt' (SS) in parallel with the galvanometer coil.

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Conversion to Voltmeter: A galvanometer is converted into a voltmeter by connecting a high resistance (RR) in series with the galvanometer coil.

📐Formulae

τ=NIABsin⁡θ\tau = NIAB \sin \theta

ϕ=(NABk)I\phi = \left( \frac{NAB}{k} \right) I

Si=ϕI=NABkS_i = \frac{\phi}{I} = \frac{NAB}{k}

Sv=ϕV=NABkRS_v = \frac{\phi}{V} = \frac{NAB}{kR}

G=1Si (Figure of Merit)G = \frac{1}{S_i} \text{ (Figure of Merit)}

S=IgGI−Ig (Shunt for Ammeter conversion)S = \frac{I_g G}{I - I_g} \text{ (Shunt for Ammeter conversion)}

R=VIg−G (Series resistance for Voltmeter conversion)R = \frac{V}{I_g} - G \text{ (Series resistance for Voltmeter conversion)}

💡Examples

Problem 1:

A galvanometer coil has a resistance of 15Ω15 \Omega and the meter shows full scale deflection for a current of 4mA4 mA. How will you convert the meter into an ammeter of range 00 to 6A6 A?

Solution:

Given: G=15ΩG = 15 \Omega, Ig=4mA=4×10−3AI_g = 4 mA = 4 \times 10^{-3} A, I=6AI = 6 A. To convert to an ammeter, a shunt resistance SS is connected in parallel: S=IgGI−IgS = \frac{I_g G}{I - I_g} S=4×10−3×156−0.004=0.065.996≈0.01ΩS = \frac{4 \times 10^{-3} \times 15}{6 - 0.004} = \frac{0.06}{5.996} \approx 0.01 \Omega.

Explanation:

To measure a higher current, most of the current must bypass the galvanometer through the shunt. A shunt of approximately 0.01Ω0.01 \Omega must be connected in parallel.

Problem 2:

In a galvanometer, the deflection falls from 5050 divisions to 1010 divisions when a shunt of 12Ω12 \Omega is connected across it. Calculate the galvanometer resistance GG.

Solution:

Let the initial current be II. Initial deflection ϕ1=50∝I\phi_1 = 50 \propto I. When shunt SS is connected, the current through galvanometer becomes IgI_g. Deflection ϕ2=10∝Ig\phi_2 = 10 \propto I_g. Using the relation Ig=I(SG+S)I_g = I \left( \frac{S}{G+S} \right): 1050=12G+12\frac{10}{50} = \frac{12}{G + 12} 15=12G+12\frac{1}{5} = \frac{12}{G + 12} G+12=60  ⟹  G=48ΩG + 12 = 60 \implies G = 48 \Omega.

Explanation:

The deflection is proportional to the current flowing through the galvanometer. By using the current division rule between the galvanometer and the shunt, we solve for the unknown resistance GG.