Circuit containing Inductor and Resistor in Series (L-R Series Circuit )

Mathematical Analysis of L-R Series Circuit :
Alternating Current Circuit Containing Inductor and Resistor in series
Let us consider, a circuit containing inductor $L$ resistor $R$ and these are connected in series. If an alternating voltage source is applied across it then the resultant voltage of the L-R circuit

$V=\sqrt{ V_{L} ^{2} + V^{2}_{R}} \qquad(1)$

We know that:

$V_{R} = iR$
$V_{L} = iX_{L}$

So from equation $(1)$

$V=\sqrt{\left( iX_{L} \right)^{2} + \left(iR\right)^{2}} $

$V=i\sqrt{\left( X_{L} \right)^{2} + R^{2}} $

$\frac{V}{i}=\sqrt{\left( X_{L} \right)^{2} + R^{2}} $

$Z=\sqrt{\left( X_{L} \right)^{2} + R^{2}} \qquad(2)$

$Z \rightarrow$ Impedance of L-R circuit.
$X_{L} \rightarrow$ Inductive Reactance which has value $\omega L$

So from equation $(2)$, we get

$Z=\sqrt{\left( \omega L \right)^{2} + R^{2}} \qquad(3)$

The phase of resultant voltage:
Phasor Diagram For L-R Circuit
If the phase of resultant voltage from from current is $\phi$ then

$tan \phi = \frac{X_{L} }{R} \qquad(4)$

$tan \phi = \frac{\omega L }{R} $

$\phi = tan^{1} \left(\frac{\omega L }{R}\right) $

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