Q.(a) A battery of emf E and internal resistance r is connected to an external resistance R. Show that power in the external circuit will be maximum when R = r. Determine the expression of maximum power.

Maximising P = I²R over R by calculus gives the classic result R = r, with P_max = E²/4r; resistance of a metal increases nearly linearly with temperature, characterised by its temperature coefficient α.

1. Maximum power transfer: Current in the circuit, $I=\dfrac{E}{R+r}$. Power delivered to the external resistance: $P=I^2R=\dfrac{E^2R}{(R+r)^2}$ To maximise P with respect to R, set $\dfrac{dP}{dR}=0$: $\dfrac{dP}{dR}=E^2\cdot\dfrac{(R+r)^2-R\cdot2(R+r)}{(R+r)^4}=E^2\cdot\dfrac{(R+r)-2R}{(R+r)^3}=E^2\cdot\dfrac{r-R}{(R+r)^3}$ Setting this to zero: $r-R=0\Rightarrow R=r$. (This is indeed a maximum, as P increases for R<r and decreases for R>r.) Maximum power: substituting $R=r$, $P_{max}=\dfrac{E^2\cdot r}{(2r)^2}=\dfrac{E^2}{4r}$
2. Temperature coefficient of resistance (α): It is defined as the fractional change in resistance per unit rise in temperature: $\alpha=\dfrac{R_2-R_1}{R_1(T_2-T_1)}$ …