Derive an expression for the impedance of a series LCR, circuit, when an AC voltage is applied to it.

Explanation

Let I be the instantaneous current in the circuit. Since the components are connected in series the current through each component is same.
Let `vecI , vecV _(R), vecV _(L) and vecV_(c )` be the phasors representing current and voltage across resistor, inductor and capacitor respectively as shown in phasor diagram .
`vecV _(R)` is in phase with `vecI and V _(Rm) = i _(m) R`
`vecV_() ` leads `vecI ` by `pi /2 and V _(Lm) = i _(m) X _(L)`
`vecV _(C)` lags behind `vecI` by `pi/2 ` rad and `V _(CM) = i_(m) X _(C)`
Let `X _(C) gt X _(L)` then th4e phasor OD represents the resultant of `vecV _(L) and vecV _(C) i.e., (vecV _(C) - vecV _(L)).` The diagonal of the parallelogram OE represents the resultant of `vecV _(R), vecV _(L) and vecV _(C).`
Applying Pythagorean Theorem, we get
`OE ^(2) = OA ^(2) + AB ^(2)`
`v _(m ) ^(2) = v _(Rm ) ^(2)+ ( v _(Cm ) - v _(Lm)) ^(2)`
`= (i _(m) R ) ^(2) + (i_(m) X _(C) - i _( m ) x _(L)) ^(2)`
`= i _(m) ^(2) [ R ^(2) + (X _(C) - X _(L))^(2) ]`
`i _(m) ^(2) = ( v _(m) ^(2))/( [ R ^(2) + (X _(c) - X _(L ))^(2) ])`
` i _(m )= ( v _(m))/( sqrt ( R ^(2) + (X _(c ) - X _(L)) ^(2)))`
According to Ohm.s law, the term `(sqrt ( R ^(2) + ( X _(c) - X _(L ))^(2)))` plays the role of resistance and it is called impedance. It is denoted by Z.
` Z = sqrt( R ^(2) + ( X _(C) - X _(L)) ^(2))`
`Z = ( v _(m))/( i _(m))`