Class 12 Physics, Chapter 7 Alternating current Class 12 Physics, Chapter 7 Alternating current | Page 12

In figure 12( b) we have assumed that V L is greater than V C which makes i lags behind V. If V C > V L then, i lead V
In this phasors diagram OA represent V R, AD represent V C and AC represent V L. So in this case as we have assumed that V L > V C, there resultant will be( V L-V C) represented by vector AD
Vector OB represent resultant of vectors V R and( V L-V C) and this vector OB is the resultant of all the three, which is equal to applied voltage V, thus
is called impedance of the circuit From phasors diagram 12( b), current i lag behind resultant voltage V by an phase angle given by,
From equation( 20) three cases arises( i) When ωL > 1 / ωC then tanφ is positive i. e. φ is positive and voltage leads the current i( ii) When ωL < 1 / ωC, then tanφ is negative i. e. φ is negative and voltage lags behind the current i( iii) When ωL = 1 / ωC, then tanφ is zero i. e. φ is zero and voltage and current are in phase Again considering case( iii) where ωL = 1 / ωC, we have
� which is the minimum value Z can have.
This is the case where X L = X C, the circuit is said to be in electric resonance where the impedance is purely resistive and