A galvanometer coil has resistance 20 Ω and the metre shows full scale deflection for a current of 2 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

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  1. A 6.66 mΩ shunt resistor is to be connected in series with the galvanometer
  2. A 10.55 mΩ shunt resistor is to be connected in series with the galvanometer
  3. A 10.55 mΩ shunt resistor is to be connected in parallel with the galvanometer
  4. A 6.66 mΩ shunt resistor is to be connected in parallel with the galvanometer

Answer (Detailed Solution Below)

Option 4 : A 6.66 mΩ shunt resistor is to be connected in parallel with the galvanometer
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Detailed Solution

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CONCEPT:

  • galvanometer can be converted into an ammeter by connecting a shunt resistance in parallel to it.
  • The shunt resistance should have very low resistance. 
  • So, the ammeter (the parallel combination of galvanometer and shunt resistance) will have low resistance.
  • To convert a galvanometer into an ammeter of current rating ‘I’, a small resistance ‘S’ (shunt resistance) is connected in parallel across the galvanometer.

Capture 15

\({V_g} = \left( {I - {I_g}} \right)S = {I_g}{R_g}\)

Where Vg is the voltage across the galvanometer, I is the current in the circuit, Ig is the current in the galvanometer, Rg is the resistance of the galvanometer and S is the resistance of the shunt.

CALCULATION:

Given:

Galvanometer coil has resistance Rg = 20 Ω

Full-scale deflection current Ig = 2 mA

Ammeter range I = 6 A

S = IgRg/(I - Ig)

⇒ S = 0.002 × 20/(6 - 0.002)

⇒ S = 6.66 mΩ 

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