Question

A circuit connected to an ac source of emf $e = {e_0}\sin \left( {100t} \right)$ with t in seconds, gives a phase difference of $\dfrac{\pi }{4}$ between the emf e and current i. Which of the following circuits will exhibit this?
A. RC circuit with $R = 1k\Omega {\text{ and C = 1}}\mu {\text{F}}$
B. RL circuit with $R = 1k\Omega {\text{ and L = 1mH}}$
C. RL circuit with $R = 1k\Omega {\text{ and L = 10mH}}$
D. RC circuit with $R = 1k\Omega {\text{ and C = 10}}\mu {\text{F}}$

Answer 1

Hint: In case of DC circuit if there is resistor we can measure using an ammeter but if there are inductors and capacitors in the AC circuit we can’t measure the obstruction for the flow of current hence there comes a term inductive reactance and capacitive reactance and they depend on source angular frequency too.
Formula used:
$\eqalign{
  & {X_C} = \dfrac{1}{{\omega C}} \cr
  & {X_L} = \omega L \cr} $
$\tan \phi = \dfrac{{{X_L}}}{R}$
$\tan \phi = \dfrac{{{X_C}}}{R}$

Complete answer:
In case alternating currents there will be angular velocity of a wave and we can find frequency of an alternating wave from that. That frequency determines the capacitive reactance and inductive reactance and total impedance of the AC circuit. If a capacitor is present in the AC circuit then we include the capacitive reactance and if not then only resistance and inductor reactance determine the impedance.
Normally AC voltages will be in the form $e = {e_0}\sin (\omega t)$ where ${e_0}$ is the peak voltage while $\omega $ is the angular frequency.
In AC circuits there will be some phase difference between the current and potential. Here that phase difference is given as ${45^0}$.
So in the given circuit we have the inductor and the resistor or inductor and capacitor, hence we have the inductive reactance (${X_L}$) and the resistance(R) and capacitive reactance(${X_C}$) which are related to the phase difference as
$\tan \phi = \dfrac{{{X_L}}}{R}$
$\tan \phi = \dfrac{{{X_C}}}{R}$
$\eqalign{
  & {X_C} = \dfrac{1}{{\omega C}} \cr
  & {X_L} = \omega L \cr} $
Where L is the inductance and C is capacitance.
Resistance is given as 1 kilo ohm i.e 1000 ohms and angular frequency is given as 100 rad/sec
Consider there is only inductive reactance (${X_L}$) and the resistance(R) in the circuit then
$\eqalign{
  & \tan \phi = \dfrac{{{X_L}}}{R} \cr
  & \Rightarrow \tan \left( {\dfrac{\pi }{4}} \right) = \dfrac{{\omega L}}{{1000}} \cr
  & \Rightarrow 1 = \dfrac{{\omega L}}{{1000}} \cr
  & \Rightarrow 1000 = 100L \cr
  & \Rightarrow L = 10H \cr} $
Consider there is only capacitive reactance (${X_C}$) and the resistance(R) in the circuit then
$\eqalign{
  & \tan \phi = \dfrac{{{X_C}}}{R} \cr
  & \Rightarrow \tan \left( {\dfrac{\pi }{4}} \right) = \dfrac{{\dfrac{1}{{\omega C}}}}{{1000}} \cr
  & \Rightarrow 1 = \dfrac{{\dfrac{1}{{\omega C}}}}{{1000}} \cr
  & \Rightarrow 1000 = \dfrac{1}{{\omega C}} \cr
  & \Rightarrow C = \dfrac{1}{{100 \times 1000}} \cr
  & \Rightarrow C = {10^{ - 5}}F = 10 \times {10^{ - 6}}F = 10\mu F \cr} $
Hence only 10 microfarad will satisfy the given conditions.

Hence option D will be the answer.

Note:
In order to remember whether current leads voltage or voltage leads current one should remember the terms IPL and CCL. IPL means in the inductive circuit potential leads the current and CCL means in capacitive circuit current leads the potential. While current and voltage across the resistor will be in phase always.