Question

An ac circuit consists of a series combination of circuit elements X and Y. The current is ahead of the voltage in phase by \[\left( {\dfrac{\pi }{4}} \right)\]. If element X is a pure resistor of 100Ω
(a) Name the circuit element Y.
(b) Calculate the rms value of current, if rms value of voltage is 141V.
(c) What will happen if the ac source is replaced by a dc source?

Answer 1

Hint:
For a circuit having
i) resistor, current and voltage will be in the same phase.
ii) inductor, voltage is ahead of the current.
iii) capacitor, current is ahead of voltage.

Formula used:
i) The phase difference between the current and voltage is given by
\[\phi = {\tan ^{ - 1}}\left( {\dfrac{{{X_C} - {X_L}}}{R}} \right)\]
Where, Φ= phase difference
$X_C$= Capacitive reactance
$X_L$= Inductive reactance
R= resistance
ii) Impedance Z of a C-R circuit is given by:
\[Z = \sqrt {{X_C}^2 + {R^2}} \]
iii) R.M.S value of current is given by
\[{I_V} = \dfrac{{{E_V}}}{Z}\]

Step by step solution:
i) As the current is ahead of voltage then the circuit has a capacitor element with the resistor.

ii) Phase difference between current and voltage is given \[\left( {\dfrac{\pi }{4}} \right)\].
Hence, putting this in the formula and taking $X_L$=0 we get,
\[
  \phi = {\tan ^{ - 1}}\left( {\dfrac{{{X_C} - {X_L}}}{R}} \right) \\
  {X_L} = 0 \\
  \tan \phi = \dfrac{{{X_C}}}{R} \\
  \tan \left( {\dfrac{\pi }{4}} \right) = \dfrac{{{X_C}}}{R} \\
  \dfrac{{{X_C}}}{R} = 1 \\
  {X_C} = R = 100\Omega \\
\]

iii) Now putting the value in the formula of impedance you can calculate impedance.
\[
  Z = \sqrt {{X_C}^2 + {R^2}} \\
  Z = \sqrt {{{100}^2} + {{100}^2}} \\
  Z = 100\sqrt 2 \\
  Z = 100 \times 1.41\Omega \\
  Z = 141\Omega \\
\]

iv) The rms value of voltage is given 141V. Now calculate the rms value of current using formula
\[
  {I_V} = \dfrac{{{E_V}}}{Z} = \dfrac{{141}}{{141}} \\
  {I_V} = 1A \\
\]

v) For dc source ω=0. Putting the value in the formula of capacitive reactance you get:
\[
  \omega = 0 \\
  {X_C} = \dfrac{1}{{{\omega _C}}} \\
  {X_C} = \dfrac{1}{0} = \infty \\
  i = \dfrac{e}{Z} = \dfrac{{141}}{\infty } \\
  i = 0A \\
\]

vi) Hence the current will be zero if we apply a dc source.

Note: Students must read simultaneously the circuits with different circuit elements to understand the phase difference in each case and the impedance changes in each case.