Current Electricity

According to given figure and by using Kirchhoff's current law,

Input current at P = Output current from P

$\therefore 4+5=I+3+5$

$\Rightarrow 9=8+I$

$\Rightarrow I=9-8$

= 1A from P to Q

Let length of each wire be $l$.

$\therefore$ Total length $=10 l$

At, lst balance point, $l_1: l_2=6 l: 4 l$

and as in case of potentiometer, $\frac{\mathrm{emf}_1}{\mathrm{emf}_2}=\frac{l_1}{l_2}$

i.e. $\varepsilon \propto l$

If $l$ increases, emf also increases and emf is only increased by increasing resistance $(R)$ in main circuit.

Given, length of germanium, $l=0.925 \mathrm{~cm} =0.925 \times 10^{-2} \mathrm{~m} $

Area of cross-section,

$A=1 \mathrm{~mm}^2$

$\begin{aligned} & =\left(10^{-3} \mathrm{~m}\right)^2 \\ & =10^{-6} \mathrm{~m}^2 \\ n_i & =2.5 \times 10^{19} \mathrm{~m}^{-3} \\ \mu_n & =0.19 \mathrm{~m}^2 / \mathrm{V}-\mathrm{s} \\ \mu_e & =0.39 \mathrm{~m}^2 / \mathrm{V}-\mathrm{s} \end{aligned}$

$\text { Conductivity, } \begin{aligned} \sigma & =n_i e\left(\mu_e+\mu_n\right) \\ & =2.5 \times 10^{19} \times 1.6 \times 10^{-19}(0.39+0.19) \\ & =2.32 \mathrm{~mho~m}^{-1} \end{aligned}$

Resistivity, $\rho=\frac{1}{\sigma}=0.431 \Omega-\mathrm{m}$

$\text { and resistance, } \begin{aligned} R & =\frac{\rho l}{A} \\ & =\frac{0.431 \times 0.925 \times 10^{-2}}{10^{-6}} \\ & =0.3986 \times 10^4=3.9 \times 10^3 \\ & =4 \times 10^3 \Omega=4 \mathrm{k} \Omega \end{aligned}$

Given, emf of cell $=1.8 \mathrm{~V}$

Current, $I=17 \mathrm{~A}$

Resistance of ammeter, $R=0.06 \Omega$

Let internal resistance be $r$.

$\begin{aligned} & \text { Since, } \mathrm{emf}=I(R+r) \\ & \therefore \quad 1.8=17(0.06+r) \\ & \Rightarrow \quad \frac{1.8}{17}=0.06+r \\ & \Rightarrow \quad r=\frac{1.8}{17}-0.06=0.046 \Omega \\ & \end{aligned}$