Question

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

Answer 1

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,

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

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

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

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

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

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

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

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

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

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

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

$\bold{RG=SR+SG}$

$\bold{RG-SG=SR}$

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

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