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

LCR series circuit is connected to an ac source of voltage

V = V_{\circ} \sin ω t

. Using phasor diagrams, obtain an expression for the impedance and the phase angle between the voltage and the current.

Accepted Answer

Here, the ac voltage supply $V$ is applied across series LCR.

The supply voltage is $V = V_{\circ} \sin (ω t)$

From the phasor diagram:

The voltage across resistance $(V_{R})$ and the current through the series LCR $(i)$ are in phase.

But the voltage across inductor $(V_{L})$ leads the current by phase $\frac{π}{2}$ whereas the voltages across capacitor $(V_{C})$ lags behind the current by phase $\frac{π}{2}$ . If $V_{C} > V_{L}$ then the resultant will be their potential difference i.e. $V_{C} - V_{L}$ .

Here, $V_{R}$ and $V_{C} - V_{L}$ are mutually perpendicular.

So, the net voltage across series LCR: $V^{2} = V_{R}^{2} + {(V_{C} - V_{L})}^{2} \Rightarrow V = \sqrt{V_{R}^{2} + {(V_{C} - V_{L})}^{2}}$

Since, $V_{R} = R i$ , $V_{C} = X_{C} i = \frac{1}{ω C} i$ and $V_{L} = X_{L} i = ω L i$

Then, $V = \sqrt{{(R i)}^{2} + {(X_{c} i - X_{L} i)}^{2}}$

$\Rightarrow \frac{V}{i} = Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}} = \sqrt{R^{2} + {(\frac{1}{ω C} - ω L)}^{2}}$

Hence, the expression for the impedance of series LCR circuit is $\sqrt{R^{2} + {(\frac{1}{ω C} - ω L)}^{2}}$ .

The phase angle between the voltage and current will be: $\tan ϕ = \frac{X_{C} - X_{L}}{R} \Rightarrow ϕ = \tan^{- 1} (\frac{X_{C} - X_{L}}{R})$

Hence, the expression for phase angle between the voltage and the current is $\tan^{- 1} (\frac{X_{C} - X_{L}}{R})$ .

LCR series circuit is connected to an ac source of voltage $V = V_{\circ} \sin ω t$ . Using phasor diagrams, obtain an expression for the impedance and the phase angle between the voltage and the current.

Important Questions on Alternating Current

Learn and Prepare for any exam you want !

LCR series circuit is connected to an ac source of voltage V=V∘ sin ω t. Using phasor diagrams, obtain an expression for the impedance and the phase angle between the voltage and the current.

Important Questions on Alternating Current

Learn and Prepare for any exam you want !

LCR series circuit is connected to an ac source of voltage $V = V_{\circ} \sin ω t$ . Using phasor diagrams, obtain an expression for the impedance and the phase angle between the voltage and the current.