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Derive an expression for the impedance of a series LCR, circuit, when an AC voltage is applied to it. |
Answer» Solution : 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))` |
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