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In an ammeter, The deflecting torque is proportional to the current passing through it, and the instrument has full scale deflection of 80° for a current of 5 A. What deflection will occur for a current of 2.5 A when the instrument is spring-controlled?
1. 20°
2. 35° 
3. 45° 
4. 40° 

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Correct Answer - Option 4 : 40° 

Concept:

Deflecting torque of the PMMC meter is given as,

Td = NBAI = GI

Where,

Td = deflecting torque in N-m

B = flux density in air gap, Wb/m2

N = number of turns of the coils

A = effective area of the coil in m2

I= current passing through the meter, amperes

G = constant = NBA

The spring control provides a restoring (controlling) torque given as,

TC = Kθ (Nm)

K = spring constant (Nm/degree)

θ = angle of deflection in degree

For the final steady-state deflection condition

Td = Tc 

GI = Kθ 

Current I = (K/G) θ .....(1)

Calculation:

Given Deflecting torque is directly proportional to current, then the scale of the ammeter is linear and the meter is PMMC type.

Given data

I1 = 5 A, θ1 = 80° 

For I1 current let the deflection be θ1

I2 = 2.5 A , θ2 = ?

For I2 current let the deflection be θ2

From equation(1),

The current passing through the ammeter is directly proportional to the deflection of the pointer

I ∝ θ 

\(\frac{{{I_1}}}{{{I_2}}} = \frac{{{θ _1}}}{{{θ _2}}}\)

\(\frac{5}{{2.5}} = \frac{{80^\circ }}{{{θ _2}}}\)

θ2  = 40° 

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