Name: AC Voltage Applied to a Resistor
Uploaded: 2019-07-19
Duration: 7 min 45 s
Description: A purely resistive circuit has very little inductance at its typical frequency. This video explains various aspects of it and also shows the plotting of curr...

Question

Arrive at the expression for the impedance of a series LCR circuit using phasor diagram method and hence write the expression for the current through the circuit.

Accepted Answer

Consider a series LCR circuit connected to an AC source.

Applied AC voltage is $V = V_{m} \sin (ω t)$

Let the current through the series circuit be $I = I_{m} \sin (ω t + ϕ)$

Where $ϕ$ is the phase difference between voltage and current.

Let $\vec{I}$ be the phasor representing current and $\vec{V_{R}}, \vec{V_{L}}, \vec{V_{C}}$ and $\vec{V}$ be the phasors representing voltage across $R, L, C$ and source respectively.

Let's assume that $\vec{V_{C}} > \vec{V_{L}}$

$V_{R} = I R$

$V_{L} = I X_{L}$

$V_{C} = I X_{C}$

According to Kirchoff's loop rule $V_{L} + V_{C} + V_{R} = V$

The phasor relation is represented as $\vec{V_{L}} + \vec{V_{C}} + \vec{V_{R}} = \vec{V}$

Applying Pythagorean theorem to the right angle triangle,

$V^{2} = {(V_{R})}^{2} + {(V_{C} - V_{L})}^{2}$

$V^{2} = {(I)}^{2} [{(R)}^{2} + {(X_{C} - X_{L})}^{2}]$

$V = I {[{(R)}^{2} + {(X_{C} - X_{L})}^{2}]}^{\frac{1}{2}}$

$I = \frac{V}{\sqrt{{(R)}^{2} + {(X_{C} - X_{L})}^{2}}}$

$I = \frac{V}{Z}$

Where, $Z$ is called impedance of the circuit.

$Z = \sqrt{{(R)}^{2} + {(X_{C} - X_{L})}^{2}}$

Arrive at the expression for the impedance of a series LCR circuit using phasor diagram method and hence write the expression for the current through the circuit.

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