Derive an expression for impedance of an a.c. circuit consisting of an inductor and resistor.
Answer
595.8k+ views
Hint Total opposition that a circuit or a part of a circuit presents to electric current. Impedance includes both resistance and reactance we can find the formula by using both the reactance of inductor as well as resistor.
Complete step by step solution
Impedance :The measure of the opposition of an electric current to the energy flow when the voltage is applied.For example the impedance is a line of resistance within an electrical current.
Consider a circuit containing a resistor of resistance R and an inductor of inductance L connected in series.
As the applied voltage is given by
$V = {V_0}\sin \omega t$
Let ${V_R}$ be the voltage across resistor
${V_L}$ be the voltage across inductor
As we know Voltage ${V_R}$ and currently I are in the same phase.
Whereas ${V_L}$ leads current by $\dfrac{\pi }{2}$
Which means ${V_R}$ and ${V_L}$ are mutually perpendicular.
The applied voltage is obtained by the resultant of ${V_R}$ and ${V_L}$
So,
$V = \sqrt {{V_R}^2 + {V_L}^2} $
And ${V_R} = Ri$,
${V_L} = {X_L}i = \omega Li$
Where L is impedance,
i is current,
and $\omega $ is frequency
Here ${X_L}$ is called inductive reactance
So putting values of ${V_R}$ and ${V_L}$
We get
$V = \sqrt {{{(Ri)}^2} + {{({X_L}i)}^2}} $
So impedance
$Z = \dfrac{V}{i} = \sqrt {{R^2} + {X_L}^2} $
Hence
$Z = \sqrt {{R^2} + {{(\omega L)}^2}} $
Note Impedance would be different if the capacitor is also in the circuit. In that case the reactance of the capacitor is also considered along with the inductor and resistance. Remember the impedance can never be greater than R.
Complete step by step solution
Impedance :The measure of the opposition of an electric current to the energy flow when the voltage is applied.For example the impedance is a line of resistance within an electrical current.
Consider a circuit containing a resistor of resistance R and an inductor of inductance L connected in series.
As the applied voltage is given by
$V = {V_0}\sin \omega t$
Let ${V_R}$ be the voltage across resistor
${V_L}$ be the voltage across inductor
As we know Voltage ${V_R}$ and currently I are in the same phase.
Whereas ${V_L}$ leads current by $\dfrac{\pi }{2}$
Which means ${V_R}$ and ${V_L}$ are mutually perpendicular.
The applied voltage is obtained by the resultant of ${V_R}$ and ${V_L}$
So,
$V = \sqrt {{V_R}^2 + {V_L}^2} $
And ${V_R} = Ri$,
${V_L} = {X_L}i = \omega Li$
Where L is impedance,
i is current,
and $\omega $ is frequency
Here ${X_L}$ is called inductive reactance
So putting values of ${V_R}$ and ${V_L}$
We get
$V = \sqrt {{{(Ri)}^2} + {{({X_L}i)}^2}} $
So impedance
$Z = \dfrac{V}{i} = \sqrt {{R^2} + {X_L}^2} $
Hence
$Z = \sqrt {{R^2} + {{(\omega L)}^2}} $
Note Impedance would be different if the capacitor is also in the circuit. In that case the reactance of the capacitor is also considered along with the inductor and resistance. Remember the impedance can never be greater than R.
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