Q 1 :    

A galvanometer has a coil of resistance 200Ω with a full scale deflection at 20 μA. The value of resistance to be added to use it as an ammeter of range (0-20) mA is             [2024]

  • 0.40Ω

     

  • 0.20Ω

     

  • 0.50Ω

     

  • 0.10Ω

     

(B)     G=200 Ω and ig=20 μA

          i=ig(GS+1)

          20×10-3=20×10-6(200S+1)

          200S=999S0.2Ω

 



Q 2 :    

A galvanometer of resistance 100Ω when connected in series with 400Ω measures a voltage of up to 10 V. The value of resistance required to convert the galvanometer into an ammeter to read up to 10 A is x×10-2Ω. The value of x is           [2024]

  • 800

     

  • 2

     

  • 20

     

  • 200

     

(3)

For voltmeter, 

[IMAGE 190]

ig=10400+100=20×10-3A

for ammeter

[IMAGE 191]

For ammeter

Let shunt resistance =S

igR=(i-ig)S

20×10-3×100=10SS=20×10-2Ω



Q 3 :    

Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across A and B, their readings are V1?, V2 and V3?. Choose the correct option.           [2024]

[IMAGE 192]

  • V1=V2

     

  • V1V3-V2

     

  • V1+V2>V3

     

  • V1+V2=V3

     

(4)

From KVL, V1+V2-V3=0V1+V2=V3

 



Q 4 :    

A galvanometer having coil resistance 10Ω shows a full scale deflection for a current of 3 mA. For it to measure a current of 8A, the value of the shunt should be:    [2024]

  • 3×10-3Ω

     

  • 4.85×10-3Ω

     

  • 3.75×10-3Ω

     

  • 2.75×10-3Ω

     

(3)

Given G=10Ω, Ig=3 mA and I=8A

In case of conversion of galvanometer into ammeter.

[IMAGE 193]

We have IgG=(I-Ig)S

S=IgGI-Ig=(3×10-3)×108-0.003=3.75×10-3Ω



Q 5 :    

The deflection in moving coil galvanometer falls from 25 divisions to 5 divisions when a shunt of 24Ω is applied. The resistance of galvanometer coil will be ____    [2024]

  • 12Ω

     

  • 96Ω

     

  • 48Ω

     

  • 100Ω

     

(2)

Let x= current/division.

Without shunt,

[IMAGE 194]

After applying shunt,

[IMAGE 195]

Now 5x×G=20x×24

G=4×24=96Ω