a) When a current, is set up in a wire of radius r, the drift velocity is vd. If the same current is set up through a wire of radius 2r, what will be the drift velocity? 

(b) A generator has an e.m.f. of 440 Volt and internal resistance of 400 Ohm. Its terminals are connected to a load of 4000 Ohm. What is the voltage across the load?

                                                       OR

(a) In the circuit, Fig., what should be the value of r in ohm so that power developed in the resistor r will be maximum?

(b) In a Wheatstone bridge all the four arms have equal resistance R. If the resistance of galvanometer arm is also R, what is the equivalent resistance of combination?

$\left(\mathrm{a}\right)\mathrm{I}={\mathrm{nAev}}_{\mathrm{d}} \mathrm{or} {\mathrm{v}}_{\mathrm{d}} \propto 1/{\mathrm{\pi r}}^2=\frac{{\mathrm{v}}_{\mathrm{d}}}{4}$

(b) A battery of emf E and internal resistance r is connected to an external resistance R. The maximum power in the external circuit is 9W. The current flowing in the circuit under the condition is 3A. What is the value of E in volt?

$$\begin{gathered} \mathrm{Total} \mathrm{resistance} \mathrm{of} \mathrm{the} \mathrm{circuit} \\ =4000+400=4400\mathrm{\Omega } \\ \mathrm{Current} \mathrm{flowing} \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{440}{4400}=0.1 \mathrm{amp}. \\ \mathrm{Voltage} \mathrm{across} \mathrm{load}=\mathrm{Ri} \\ =4000\times 0.1=400 \mathrm{volt}. \end{gathered}$$

Maximum power consumed,

$$\begin{gathered} {\mathrm{P}}_{\max }={\left(\frac{\mathrm{E}}{\mathrm{r}+\mathrm{r}}\right)}^2\times \mathrm{r}=\frac{{\mathrm{E}}^2}{4\mathrm{r}}=9 ...\left(\mathrm{i}\right) \\ \mathrm{Current}, \mathrm{I}=\frac{\mathrm{E}}{\mathrm{r}+\mathrm{r}}=3, \mathrm{E}=6 \mathrm{r} ...\left(\mathrm{ii}\right) \\ \mathrm{Solving} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get} \mathrm{E}=6\mathrm{V}. \end{gathered}$$

OR

(a) The equivalent circuit with the distribution of current  

Using Kirchhoff's second law, in closed circuit ABEFA, we have, $6{\mathrm{I}}_1+3\left({\mathrm{I}}_1+{\mathrm{I}}_2\right)-\mathrm{V}=0$

or      $\mathrm{V}=9{\mathrm{I}}_1+3{\mathrm{I}}_2$

       Using Kirchhoff's second law, in closed circuit BCDEB, we have ${\mathrm{rI}}_3-6{\mathrm{I}}_1=0 \mathrm{or} {\mathrm{I}}_1=\frac{{\mathrm{I}}_2}{6}\mathrm{r}$

$\mathrm{From} \left(\mathrm{i}\right), 9\times \frac{{\mathrm{I}}_2}{6}\mathrm{r}+3{\mathrm{I}}_2=\frac{3}{2}{\mathrm{I}}_2\mathrm{r}+3{\mathrm{I}}_2=\frac{3{\mathrm{I}}_2}{2}(\mathrm{r}+2)$

or   ${\mathrm{I}}_2=\frac{2\mathrm{V}}{3(\mathrm{r}+2)}$ Power developed in resistor r,

$\mathrm{P}={\mathrm{I}}_2^2\mathrm{r}=\frac{4{\mathrm{V}}^2}{9{(\mathrm{r}+2)}^2}\times \mathrm{r}$

Power developed is maximum, when ${(\mathrm{r}+2)}^2$ is minimum.

or   ${(\mathrm{r}+2)}^2=0$

or   ${\mathrm{r}}^2+\mathrm{r}+4=0$

or   ${\mathrm{r}}^2-4\mathrm{r}+4+8\mathrm{r}=0$

or   ${(\mathrm{r}-2)}^2+8\mathrm{r}=0 \mathrm{or} \mathrm{r}=2\mathrm{\Omega }.$

(b) If all the four arms in a Wheatstone bridge have equal resistance R and the resistance of galvanometer arm is also R, then the equivalent resistance of combination is R.