What is the phase difference between AC e.m.f and current in the following?
Pure resistor and pure inductor.

HintTo find the given values we have to treat all the voltmeters and ammeters as a set of resistances.
Then we have to obtain voltage differences and current across each section.

Complete step-by-step solution:
1. In the case of a pure resistor the e.m.f and current are both in the same phase. It is shown in the given diagram
   

Here by kirchoff’s loop law,
${V_m}\sin \omega t = iR$
$
  i = ({V_m}/R)\sin \omega t \\
  i = {i_m}\sin \omega t \\
 $
2. So the phase difference $\phi $ between both the e.m.f and current is ${0^o}$
In the case of pure inductor the e.m.f leads the current by ${90^o}$. It is shown in the given diagram

Here by the kirchoff’s law,
\[
  \int {di/dt} dt = {V_m}/L\int {\sin (\omega t)dt} \\
  i = {V_m}\cos (\omega L)/\omega L + cons\tan t \\
 \]

After the integration
$i = {i_m}\sin (\omega t - \pi /2)$

Hence the phase difference is ${90^o}$.

Note:- Here all the equations should be obtained step by step. The integration should be done independently and subsequently.