Question

Model a torch battery of length $l$ to be made up of a thin cylindrical bar of radius $‘a’$ and a concentric thin cylindrical shell of radius $‘b’$ filled in between with an electrolyte of resistivity $\rho$ (see figure). If the battery is connected to a resistance of value $R$, the maximum joule heating in $R$ will take place for:

• A. $\begin{array}{l}R=\frac{\rho }{2\pi l}\left(\frac{b}{a}\right)\end{array}$
• B. $\begin{array}{l}R=\frac{2\rho }{\pi l}In\left(\frac{b}{a}\right)\end{array}$
• C. $\begin{array}{l}R=\frac{\rho }{\pi l}In\left(\frac{b}{a}\right)\end{array}$
• D. $\begin{array}{l}R=\frac{\rho }{2\pi l}In\left(\frac{b}{a}\right)\end{array}$

Answer 1

The correct option is D

$\begin{array}{l}R=\frac{\rho }{2\pi l}In\left(\frac{b}{a}\right)\end{array}$

Step 1: Given data:

Length of torch battery$=l$

Radius of thin cylindrical bar$=a$

Radius of thin concentric circle$=b$

Resistivity of electrolyte $=\rho$

Resistance$=R$

Step 2: Formula used:

Resistivity is defined as-

$\rho =R\frac{A}{L}$

Ohm's law-

$V=iR$

Power will be-

$H=i^{2}R$

Step 3: Calculation of internal resistance:

Consider internal resistance of battery is $r$

E.m.f of the battery is $\epsilon$

Current will be-

$i=\frac{\epsilon }{r+R}$

For maximum joule heating- $\frac{dH}{dr}=0$

$\begin{array}{rcl}H& =& i^{2}R\\ H& =& {\left(\frac{\epsilon }{r+R}\right)}^{2}R\\ H& =& \frac{{\epsilon }^{2}R}{{\left(r+R\right)}^2}\\ \frac{dH}{dr}& =& \frac{2{\epsilon }^{2}R\left(r+R\right)-{\left(r+R\right)}^2\times 0}{{\left(r+R\right)}^4}\\ \frac{2{\epsilon }^{2}R\left(r+R\right)-{\left(r+R\right)}^2\times 0}{{\left(r+R\right)}^4}& =& 0\\ 2{\epsilon }^{2}R\left(r+R\right)& =& 0\\ formaximumheatingr& =& R\end{array}$

Step 4: Calculation of resistance:

$\begin{array}{rcl}dR& =& \rho \frac{dr}{2\pi rl}\\ \int dR& =& \frac{\rho }{2\pi l}{\int }_a^b\frac{1}{r}dr\\ R& =& \frac{\rho }{2\pi l}{\left[\mathrm{Ln}\left(r\right)\right]}_a^b\\ R& =& \frac{\rho }{2\pi l}\mathrm{Ln}\left(\frac{b}{a}\right)\\ r& =& \frac{\rho }{2\pi l}\mathrm{Ln}\left(\frac{b}{a}\right)\end{array}$

Hence, option D is the correct answer.