Current Electricity

According to the question, the circuit diagram is as shown below

Power dissipated in circuit,

$ \begin{equation*} P=\frac{E^2}{R_{\mathrm{eff}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

The given circuit is a balanced Wheatstone bridge.

So, its equivalent resistance is given by

$ \begin{aligned} R_{\mathrm{eff}} & =(4+4)\|(4+4)=8\| 8 \\ & =\frac{8 \times 8}{8+8}=4 \Omega \end{aligned} $

From Eq. (i), $P=\frac{(12)^2}{4}=36 \mathrm{~W}$

The equivalent resistance between point $A$ and $B$ is calculated as

According to given situation, circuit diagram of meter bridge is given as

At balance condition of meter bridge,

$ \begin{aligned} & \frac{S}{R} =\frac{25}{(100-25)} \\ \Rightarrow & \frac{S}{1} =\frac{25}{75} \\ \Rightarrow & S =\frac{1}{3} \Omega=0.33 \Omega \end{aligned} $

Given, length of wire, $P=l$

Length of wire, $Q=\frac{l}{2}$

Diameter of wire, $P=d$

Diameter of wire, $Q=\frac{d}{2}$

Resistance $R=\rho \cdot \frac{l}{A}=\rho \cdot \frac{l}{\pi d^2 / 4}$

$ \begin{aligned} & \begin{aligned} \therefore \frac{\text { Resistance of } Q}{\text { Resistance of } P} & =\left(\frac{\rho \cdot \frac{l}{2}}{\frac{\pi}{4}\left(\frac{d}{2}\right)^2}\right) / \frac{\rho l}{\frac{\pi}{4} d^2} \\ =\frac{\frac{1}{2}}{\frac{1}{4}}= & \frac{4}{2}=\frac{2}{1} \\ \Rightarrow \text { Resistance of } Q & =\text { Resistance of } P \times 2 \\ & =10 \times 2=20 \Omega \end{aligned} \end{aligned} $

The given circuit diagram is

Equivalent resistance of given circuit

$ \begin{aligned} & =(100 \Omega| | 100 \Omega)+100 \Omega+100 \Omega \\ & =\frac{100 \times 100}{100+100}+200=50+200=250 \Omega \end{aligned} $

Equivalent emf of circuit

$ =20-10=10 \mathrm{~V} $

Current in circuit,

$ I=\frac{E}{R_{\mathrm{eq}}}=\frac{10}{250}=0.04 \mathrm{~A} $

Current density is related to electric field as

$ \mathbf{J}=\frac{\mathbf{E}}{\rho} $

where, $\rho=$ resistivity.

Here, $I=5 \mathrm{~A}, A=1 \mathrm{~mm}^2=1 \times 10^{-6} \mathrm{~m}^2$ and $\rho=3 \times 10^{-8} \Omega-\mathrm{m}$

So, electric field inside the conductor is

$ \begin{aligned} \mathbf{E} & =\mathbf{J} \cdot \rho=\frac{I}{A} \cdot \rho \\ & =\frac{5 \times 3 \times 10^{-8}}{1 \times 10^{-6}}=0.15 \mathrm{~V} / \mathrm{m} \end{aligned} $

For balance condition of Wheatstone bridge,

$ \frac{R_2}{R_4}=\frac{R_1}{R_3} \quad \text { or } \quad \frac{R_2}{R_1}=\frac{R_4}{R_3} $

We are given with following ammeter

Now, $\quad V_{A B}=I_g G=\left(I-I_g\right) S$

$ \begin{aligned} \left(100 \times 10^{-6}\right) 100 & =\left(I-100 \times 10^{-6}\right) 0.1 \\ I & -100 \times 10^{-6}=10^{-1} \\ I & =0.1+100 \times 10^{-6} \\ & =0.1+0.0001=0.1001 \mathrm{~A} \\ & =100.1 \mathrm{~mA} \end{aligned} $

So, circuit current is 100.1 mA .