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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

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

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.

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).

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.

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.

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.

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.

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

📐Formulae

τ=NIABsinθ\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=IgGIIg (Shunt for Ammeter conversion)S = \frac{I_g G}{I - I_g} \text{ (Shunt for Ammeter conversion)}

R=VIgG (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=4×103AI_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=IgGIIgS = \frac{I_g G}{I - I_g} S=4×103×1560.004=0.065.9960.01ΩS = \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=50I\phi_1 = 50 \propto I. When shunt SS is connected, the current through galvanometer becomes IgI_g. Deflection ϕ2=10Ig\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.

Moving Coil Galvanometer - Revision Notes & Key Formulas | CBSE Class 12 Physics