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Q.(a) Show that the steady deflection φ produced by a moving coil galvanometer is proportional to the current flowing through it.

(2)
(b) Write down the expression for current sensitivity and voltage sensitivity.
(2)
(c) What happens to current sensitivity when number of turns of the galvanometer coil is increased? (1)
Kerala DhseKerala DHSE Plus Two Board 2026Subjective· 5mImportance★★★★★
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In a moving-coil galvanometer, the deflecting magnetic torque is balanced by a restoring spring torque proportional to the deflection, which forces ϕ∝I\phi \propto I; from this, current and voltage sensitivity follow directly, and both grow with more turns N (current sensitivity strictly proportionally).

  1. Deflection proportional to current: A rectangular coil of N turns, area A, carrying current I, is suspended in a radial magnetic field B (arranged via curved pole pieces so that B is always in the plane of the coil, i.e. always perpendicular to the coil's normal). This makes the magnetic torque on the coil constant in magnitude for any deflection angle: τdeflecting=NIAB\tau_{deflecting} = NIAB (unlike the sin⁡θ\sin\theta-dependence of a torque in a uniform field, the radial field design keeps this torque the same at every angle.) As the coil rotates by angle ϕ\phi, a spring (or suspension fibre) provides a restoring torque proportional to the twist: τrestoring=kϕ\tau_{restoring} = k\phi where k is the torsional constant of the spring. At equilibrium (steady deflection), the two torques balance: NIAB=kϕ⟹ϕ=NABk INIAB = k\phi \quad\Longrightarrow\quad \boxed{\phi = \frac{NAB}{k}\,I} Since N, A, B, k are all fixed constants of the instrument, this shows ϕ∝I\phi \propto I — the deflection is directly proportional to the current, which is why the galvanometer scale can be linear.
  2. Sensitivities: Current sensitivity: Is=ϕI=NABk\text{Current sensitivity: } I_s = \frac{\phi}{I} = \frac{NAB}{k} Voltage sensitivity: Vs=ϕV=ϕIR=NABkR\text{Voltage sensitivity: } V_s = \frac{\phi}{V} = \frac{\phi}{IR} = \frac{NAB}{kR} …

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