Overview

$l_1=\frac{25}{5+R}$

$l_2=\frac{5}{R+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\begin{array}{l}?l_1=l_2\Rightarrow 4R=4\\ \Rightarrow R=1\Omega \end{array}$

$\frac{R_1{R}_2}{R_1+R_2}=3$

$\Rightarrow \frac{\left({\rho }_1\frac{\mathcal{l}}{A}\right)\left({\rho }_2\frac{\mathcal{l}}{A}\right)}{\left(\frac{{\rho }_1\mathcal{l}}{A}\right)+\left(\frac{{\rho }_2\mathcal{l}}{A}\right)}=3$

$=\frac{3*\left[\frac{\pi }{4}*{\left(2*10^{-3}\right)}^2\right]\left[12+51\right]*10^{-6}*10^{-2}}{12*10^{-6}*10^{-2}*51*10^{-6}*10^{-2}}$

$=97$

$R_{eq}=0.6\Omega +\frac{\frac{4*12}{4+12}*\frac{6*8}{6+8}}{\frac{4*12}{4+12}+\frac{6*8}{6+8}}\Omega$

$=0.6\Omega +\frac{3*\frac{48}{14}}{3+\frac{48}{14}}\Omega$

$=0.6\Omega +\frac{\left(144/14\right)\Omega }{\left(\frac{90}{14}\right)}$

$\begin{array}{l}=0.6\Omega +1.6\Omega \\ =2.2\Omega \end{array}$

p = $\frac{v^2}{R_{eq}}=\frac{{\left(2.2V\right)}^2}{2.2\Omega }=2.2W$

In galvanometer

$\Rightarrow \left(l-l_g\right)S=l_{g}G$

$\frac{I_g}{l}=\frac{S}{S+G}$

$\Rightarrow \frac{1}{3}=\frac{S}{S+G}\Rightarrow S+G=3S\Rightarrow G=2S$

R_w = 2R_y

$\Rightarrow \rho \left(\frac{2x}{\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=\frac{2\rho \left(1-x\right)}{A}$

$\frac{4\rho x}{A}=\frac{2\rho \left(1-x\right)}{A}$

4x = 2 – 2x 6x = 2

x = $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$6$}\right.=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ $\therefore \frac{L_x}{L_y}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{1}{3}*\frac{3}{2}$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Based on theoretical data.

J (current density) = 4 * 10⁶ Am^-2

Area between radial distance  $\frac{R}{2}toR$

A =  $\pi \left[R^2-\frac{R^2}{4}\right]=\frac{3R^2}{4}\pi$

I = AJ

=  $\frac{3R^2}{4}\pi *4*10^6$

= 3R² * 10⁶ πAmp

= 3 (4 * 10^-3)² * 10⁶ πAmp

= 3 * 16 * 10^-6 * 10⁶π Amp

= 48πA

R shunt Resistance

Given

CD = 3m

CF = 2m

Current through potentiometer wire

I = $\frac{v}{R_0}=\frac{VA}{\rho L}$ $\frac{}{{}_{}}\frac{}{}$ (1)

A Area of potentiometer wire

$$  $\rho \to$ Resistivity of the wire

L Length of the potentiometer wire

Case 1 When 8 $\Omega$ $$ shunt is bal aced at 3m length

$V_{CD}=I{R}_{CD}=\frac{V}{\rho L}A*\frac{\rho CF}{A}$

$V_{CD}=\frac{V\mathcal{l}}{L}=\frac{V*3}{L}$

VCD = VAB = E – I1r

= $-\frac{E}{R+r}r=\frac{V}{L}*3$ $\frac{}{}\frac{}{}$

$$ $\Rightarrow E\left[1-\frac{r}{8+r}\right]=\frac{V}{L}*3$  (2)

Case 2 When 4 $$ $\Omega$ shunt is balanced at 2m length

$V_{CF}=I{R}_{CF}=\frac{V}{\rho L}A*\frac{\rho CF}{A}=\frac{V*2}{L}$

$V_{CF}=V_{AB}=E-I_{2}r$

= $E-\frac{E}{R+r}r$ $\frac{}{}$

= $E\left[1-\frac{r}{4+r}\right]$ $\frac{}{}$

$\frac{}{}\frac{}{}$  $\Rightarrow E\left(1-\frac{r}{4+r}\right)=\frac{V*2}{L}$ (3)

$\left(2\right)÷\left(3\right)$

$\frac{1-\frac{r}{8+r}}{1-\frac{r}{4+r}}=\frac{3}{2}$

$\Rightarrow \left(\frac{8}{8+r}\right)*\left(\frac{4+8}{4}\right)=\frac{3}{2}$

$\Rightarrow 4\left(4+r\right)=3\left(8+r\right)$

$\Rightarrow 16+4r=24+3r$

r = 24 – 16

r = 8 $\Omega$

$\Rightarrow I_{g}G=\left(l-l_g\right)8\Rightarrow l_g*72=\left(l-l_g\right)8\Rightarrow 9l_g=l-l_g$

$10l_g=l{l}_g=\frac{1}{10}l$

% of I which passes through galvanometer

$$ $l_g=10\mathrm{\%}$  of I

$H=l^{2}Rt$

$\mathrm{\%}errorinH=\frac{\Delta H}{H}*100=2\left(\frac{\Delta l}{l}\right)*100+\left(\frac{\Delta R}{R}\right)*100+\left(\frac{\Delta t}{t}\right)*100=2*2+1+3=8\mathrm{\%}$