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NEET PHYSICSMedium

A galvanometer having a resistance of 8 Ω8\ \Omega is shunted by a wire of resistance 2 Ω2\ \Omega. If the total current is 1 A1\ \text{A}, the part of it passing through the shunt will be:

A

0.25 A

B

0.8 A

C

0.2 A

D

0.5 A

Step-by-Step Solution

  1. Circuit Configuration: A galvanometer (resistance GG) converted into an ammeter has a low resistance shunt (SS) connected in parallel with it .
  2. Current Division: The total current II splits between the galvanometer (IgI_g) and the shunt (IsI_s). Since they are in parallel, the potential difference across them is equal: IgG=IsSI_g G = I_s S
  3. Conservation of Charge: The total current is the sum of the individual currents: I=Ig+Is  ⟹  Ig=I−IsI = I_g + I_s \implies I_g = I - I_s.
  4. Substitution: Substitute IgI_g into the voltage equation: (I−Is)G=IsS(I - I_s) G = I_s S IG−IsG=IsSIG - I_s G = I_s S IG=Is(S+G)IG = I_s (S + G) Is=I(GS+G)I_s = I \left( \frac{G}{S + G} \right)
  5. Calculation: Given G=8 ΩG = 8\ \Omega, S=2 ΩS = 2\ \Omega, and I=1 AI = 1\ \text{A}: Is=1×(82+8)=810=0.8 AI_s = 1 \times \left( \frac{8}{2 + 8} \right) = \frac{8}{10} = 0.8\ \text{A}.
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