Physics, asked by rsnail2009, 10 months ago

Derivation of G=RS÷(R-S) where G is resistance of galvanometer

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Answered by bestwriters
27

Derivation of G = RS ÷ (R - S):

The resistance of the galvanometer is used with half deflection method.

The below given is the simple circuit to find the resistance of the galvanometer.

From the diagram:

  • \bold{K_{1}} and \bold{K_{2}} are keys
  • G is the resistance of the galvanometer
  • E is EMF of the cell
  • I is current
  • H.R.B is high resistance box
  • L.R.B is low resistance box

Condition 1:

\bold{K_{1}} is closed and \bold{K_{2}} is opened

Current flowing through galvanometer is

\bold{i=\frac{E}{R+G}\rightarrow \left ( 1 \right )}

Where,

R is resistance through H.R.B

Current flowing through moving coil galvanometer is

\bold{i=k\theta}

Where,

\bold{k=\frac{c}{nab}\rightarrow \left ( 2 \right )}

Condition 2:

\bold{K_{1}} is opened and \bold{K_{2}} is closed

Current flowing through galvanometer is

\bold{I=\frac{E}{R+\frac{GS}{G+S}}\rightarrow \left ( 3 \right )}

Now, the deflection of the galvanometer is maintained half.

\bold{\Rightarrow I_{1}=\frac{\theta}{2}\rightarrow \left ( 4 \right )}

From equation (2) and (4),

\bold{I_{1}=\frac{i}{2}\rightarrow \left ( 5 \right )}

Thus,

\bold{I_{2}=I-I_{1}=I-\frac{i}{2}\rightarrow \left ( 6 \right )}

By parallel combination of resistors

\bold{I_{1}G=I_{2}S}

On substituting equation (5) and (6) in above equation, we get,

\bold{\frac{i}{2}\times G=\left ( I-\frac{i}{2} \right )\times S}

On substituting equation (1) and (3) in above equation, we get,

\bold{\left ( \frac{E}{2\left ( R+G \right )} \right )G=\left ( \frac{E}{R+\frac{GS}{G+S}}-\frac{E}{2\left ( R+G \right )} \right )S}

\bold{\left ( \frac{E}{2\left ( R+G \right )} \right )G=E\left ( \frac{1}{R+\frac{GS}{G+S}}-\frac{1}{2\left ( R+G \right )} \right )S}

\bold{\left ( \frac{G}{2\left ( R+G \right )} \right )=\frac{S}{R+\frac{GS}{G+S}}-\frac{S}{2\left ( R+G \right )}}

\bold{\frac{G}{2\left ( R+G \right )}+\frac{S}{2\left ( R+G \right )}=\frac{S}{R+\frac{GS}{G+S}}}

\bold{\frac{G}{2\left ( R+G \right )}+\frac{2}{2\left ( R+G \right )}=\frac{S\left ( G+S \right )}{R\left ( G+S \right )+GS}}

\bold{\frac{G+S}{2\left ( R+G \right )}=\frac{S\left ( G+S \right )}{R\left ( G+S \right )+GS}}

\bold{R(G+S)+GS=2S(R+G)}

\bold{RG+RS+GS=2S(R+G)}

\bold{RG+S(R+G)=2S(R+G)}

\bold{RG=2S(R+G)-S(R+G)}

\bold{RG=S(R+G)}

\bold{RG=SR+SG}

\bold{RG-SG=SR}

\bold{G(R-S)=SR}

\bold{\therefore G=\frac{SR}{R-S}}

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