when a 12 ohm resistor is connected in series with a moving coil galvanometer than it subtraction reduces form 50 division to 10 division find the resistance of the galvanometer

Dear student,
1311657‚Äč

 

Here let the value of current from one division be i A. Thus the current through ammeter I=50i and through Galvanometer Ig=10i  S=IgGIIgG=(IIg)SIg=50i10i10i×12Ω  =4×12Ω  =48Ω

Regards

  • 0

Answer

We know that for a galvanometer the deflection is proportional to the current passing through it.



Thus initially , [V=RG​∗50]
now , when a 12 ohm resistor is connected the equation will become,

[V=(RG​+12)∗10]


[(RG​+12)∗10=50∗RG​]


[RG​=3] ohm
 
  • 1

We know that for a galvanometer the deflection is proportional to the current passing through it.



Thus initially , [V=RG​∗50]
now , when a 12 ohm resistor is connected the equation will become,

[V=(RG​+12)∗10]


[(RG​+12)∗10=50∗RG​]


[RG​=3] ohm
 
  • 1
What are you looking for?