Current Electricity

Given, that x is greater than 2 $\Omega$, when the bridge is balanced then,

$\frac{R}{l}=\frac{x}{100-l}$ ..... (i)

(from metre bridge balanced bridge formula)

$\Rightarrow 100R-Rl=lx$

$\Rightarrow 200-2l=lx~~(\because R=2\Omega)$

$l=\frac{200}{x+2}$ ..... (ii)

Where, $l$ = length of segment of wire from one end where null point is obtained.

Now, these resistances are interchanged, the Jockey shifts 20 cm. So,

$\frac{x}{l+20}=\frac{2}{80-l}$ ...... (iii)

From equation (iii) we have,

$80x=l(x+2)+40$

$\Rightarrow 80x=\left(\frac{200}{x+2}\right)(x+2)+40$

$\Rightarrow x=\frac{240}{80}=3\Omega$

$\therefore x=3\Omega$

Given, current carrying circular coil does not have an electric field and a gravitational field.

But it has a magnetic field and induced electric field if the magnetic field due to current in the coil is changing with time.

Resistance of $1^{\text {st }}$ half part is $\mathrm{R}_1=\frac{\rho l_1}{\mathrm{~A}_1}=\frac{\rho(1 / 2)}{\pi(4 r)^2}$

Resistance of $2^{\text {nd }}$ part is, $\mathrm{R}_2=\frac{\rho l_2}{\mathrm{~A}_2}=\frac{\rho(1 / 2)}{(2 r)^2 \pi}$

$Thus, \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\frac{\rho(1 / 2)}{\pi 16 r^2}}{\frac{\rho(1 / 2)}{4 \pi r^2}}=\frac{1}{4} $

(A) Power loss : $P_1=I^2 R_1$ and $P_2=I^2 R_2$,

So $\frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{1}{4} \Rightarrow \mathrm{P}_1 .4=\mathrm{P}_2$

(B) Voltage drop : $\mathrm{V}_1=\mathrm{IR}_1$ and $\mathrm{V}_2=\mathrm{IR}_2$

So $\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{1}{4}$

(C) Current density : $\mathrm{J}_1=\frac{\mathrm{I}}{\mathrm{A}_1}$ and $\mathrm{J}_2=\frac{\mathrm{I}}{\mathrm{A}_2}$

So $\frac{\mathrm{J}_1}{\mathrm{~J}_2}=\frac{\mathrm{A}_2}{\mathrm{~A}_1}=\frac{\pi r^2}{\pi(2 r)^2}=\frac{1}{4}$

(D) Electric field: $\frac{E_1}{E_2}=\frac{J_1}{J_2}=\frac{1}{4} \quad(J=\sigma E)$

At null points, the Wheatstone bridge will be balanced.

Therefore,

$\frac{\mathrm{X}}{r_{1}}=\frac{\mathrm{R}}{r_{2}} \text { (or) } \mathrm{X}=\mathrm{R} \frac{r_{1}}{r_{2}}$

Where $\mathrm{R}$ is a constant and $r_{1}$ and $r_{2}$ are variable. The maximum fraction error is

${{\Delta x} \over x} = {{\Delta {r_1}} \over {{r_1}}} + {{\Delta {r_2}} \over {{r_2}}}$

$\begin{array}{rr} & \mathrm{R}=\mathrm{R}_{1}, \mathrm{R}=\mathrm{R}_{2}, \mathrm{R}=\mathrm{R}_{3} \\ \text { Here, } & \Delta r_{1}=\Delta r_{2}=y \text { (say), }\end{array}$

$y$ can be considered as least count of the scale,

Then, $ \frac{\Delta x}{x}=y\left[\frac{r_{2}+r_{1}}{r_{1} r_{2}}\right]$

For $\frac{\Delta x}{x}$ to be minimum, $r_{1} \times r_{2}$ should be maximum [Since $r_{1}+r_{2}=\mathrm{c}$ (constant $)$ ]

Let,

$\begin{aligned} \mathrm{E} & =r_{1} \times r_{2}=r_{1} \times\left(c-r_{1}\right) \\ & =r_{1} c-r_{1}^{2} \\ \therefore \frac{d \mathrm{E}}{d r_{1}} & =c-2 r_{1}=0 \text { or } \frac{c}{2}=r_{1}\end{aligned}$

(or) $r_{2}=c / 2 \text { (or) } r_{1}=r_{2}$