Practice equations of phase relationship between current and voltage in series LCR circuit, resonance frequency & phasor diagram RLC and impedance of a series LCR circuit.
If current is zero at some instant, then the phasor diagrams will be as follows:
In resistor, the voltage and current are in same phase.
In inductor, the voltage leads the current by 90°.
In capacitor, the voltage lags behind the current by 90
On combining all these phasor diagrams, the following phasor diagram will be obtained :
The resultant phasor diagram of series LCR circuit will be as follows :
Case I : When V_{L} > V_{C}
Case II : When V_{C} > V_{L}
Peak Current, I_{0} :
Assume that, V_{L} > V_{C}
From right angled triangle ABC,
\(V_0^2=V_R^2+(V_L-V_C)^2\)
\(V_0=\sqrt {V_R^2+(V_L-V_C)^2}\)
\(V_0=\sqrt {(I_0R)^2+(I_0X_L-I_0X_C)^2}\)
\(V_0=I_0\sqrt {R^2+(X_L-X_C)^2}\)
\(I_0=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}\)
where,
\(\sqrt {R^2+(X_L-X_C)^2}\) is known as impedance. It is denoted by Z.
Thus, \(I_0=\dfrac {V_0}{Z}\)
The phase angle \(\phi\) between the current and voltage is given as follows:
From right angled triangle ABC,
\(tan\,\phi=\dfrac {V_L-V_C}{V_R}\)
\(\phi=tan^{-1}\left(\dfrac {V_L-V_C}{V_R}\right)\)
\(\phi=tan^{-1}\left(\dfrac {I_0X_L-I_0X_C}{I_0R}\right)\)
\(\phi=tan^{-1}\left(\dfrac {X_L-X_C}{R}\right)\)
When voltage source \( V = V_0 \,sin \,\omega t\)
\(I = I_0 \sin\, (\omega t –\phi)\)
\(=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}\;sin\left ( \omega t-tan^{-1}\dfrac {(X_L-X_C)}{R} \right)\)
A For purely resistive circuit, I = I0 sin \(\omega\)t
B For capacitive circuit, I = I0 sin ( \(\omega\)t + \(\pi/6\) )
C For inductive circuit, I = \(I_0\) sin ( \(\omega\)t + \(\pi/2\) )
D For inductive circuit, I = \(I_0\) sin ( \(\omega\)t - \(\pi/6\) )
If current is zero at some instant, then the phasor diagrams will be as follows:
In resistor, the voltage and current are in same phase.
In inductor, the voltage leads the current by 90°.
In capacitor, the voltage lags behind the current by 90°
On combining all these phasor diagrams, the following phasor diagram will be obtained:
The resultant phasor diagram of series LCR circuit will be as follows:
Case I : When V_{L} > V_{C}
Case II : When V_{C} > V_{L}
Peak Current, \(\text I_0\)
Assume that V_{L} > V_{C}
From right angled triangle ABC,
\(V_0^2=V_R^2+(V_L-V_C)^2\)
\(V_0=\sqrt {V_R^2+(V_L-V_C)^2}\)
\(V_0=\sqrt {(I_0R)^2+(I_0X_L-I_0X_C)^2}\)
\(V_0=I_0\sqrt {R^2+(X_L-X_C)^2}\)
\(I_0=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}\)
where, \(Z={\sqrt {R^2+(X_L-X_C)^2}}\) is known as impedance. It is denoted by \(Z\).
Thus, \(I_0=\dfrac{V_0}{Z}\)
The phase angle \(\phi\) between the current and voltage is given as follows:
From right angled triangle ABC,
\(tan\,\phi=\dfrac {V_L-V_C}{V_R}\)
\(\phi=tan^{-1}\left(\dfrac {V_L-V_C}{V_R}\right)\)
\(\phi=tan^{-1}\left(\dfrac {I_0X_L-I_0X_C}{I_0R}\right)\)
\(\phi=tan^{-1}\left(\dfrac {X_L-X_C}{R}\right)\)
When voltage source \(V = V_0 \,sin\, \omega t\)
(\(I = I_0\, sin\, (\omega t –\phi)\))
\(I=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}\;sin\left ( \omega t-tan^{-1}\dfrac {(X_L-X_C)}{R} \right)\)
\(V_L=I_0\,X_L\)
where,
_{\(I_0\) }=_{ }Peak Current
X_{L }= Inductive reactance
\(V_C=I_0\,X_C\)
where,
\(I_0\)_{ }=_{ }Peak Current
X_{C}_{ }= Capacitive reactance
\(V_R=I_0\,R\)
where,
\(I_0\)_{ }=_{ }Peak Current
R_{ }= Resistance of resistor
A VR =30 Volt, VL = 234.98 Volt, VC = 20 Volt
B VR =40 Volt, VL = 150.72 Volt, VC = 242.56 Volt
C VR =20 Volt, VL = 30 Volt, VC = 234.98 Volt
D VR =60 Volt, VL = 90 Volt, VC = 100 Volt
Inductive reactance, \(X_L=\omega\;L\)
Capacitive reactance, \(X_C=\dfrac {1}{\omega \;C}\)
Current in series LCR circuit is given by \(I=\dfrac {V}{\sqrt{ R^2+(X_L-X_C)^2}}\)
If frequency is varied, keeping everything else constant, then
\(X_L=X_C\)
\(\omega L=\dfrac {1}{\omega C}\)
\(R_1<R_2<R_3\)
Since, \(I_{max}=\dfrac {V_0}{R}\)
\(I_{max}\propto\dfrac {1}{R}\)
It can be concluded from the graph that at resonance frequency, maximum current is inversely proportional to resistance.
The current leads the voltage.
If current is zero at some instant, then the phasor diagrams will be as follows:
In resistor, the voltage and current are in same phase.
In inductor, the voltage leads the current by 90°
In capacitor, the voltage lags behind the current by \(90^\circ\)
On combining all these phasor diagrams, the following phasor diagram will be obtained:
The resultant phasor diagram of series LCR circuit will be as follows:
Case I : When V_{L} > V_{C}
Case II : When V_{C} > V_{L}
Peak Current, \(I_0\)
Assume that V_{L} > V_{C}
From right angled triangle ABC,
\(V_0^2=V_R^2+(V_L-V_C)^2\)
\(V_0=\sqrt {V_R^2+(V_L-V_C)^2}\)
\(V_0=\sqrt {(I_0R)^2+(I_0X_L-I_0X_C)^2}\)
\(V_0=I_0\sqrt {R^2+(X_L-X_C)^2}\)
\(I_0=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}\)
The phase angle \(\phi\) between the current and voltage is given as follows:
From right angled triangle ABC,
\(tan\,\phi=\dfrac {V_L-V_C}{V_R}\)
\(\phi=tan^{-1}\left(\dfrac {V_L-V_C}{V_R}\right)\)
\(\phi=tan^{-1}\left(\dfrac {I_0X_L-I_0X_C}{I_0R}\right)\)
\(\phi=tan^{-1}\left(\dfrac {X_L-X_C}{R}\right)\)
When voltage source \(V = V_0 \sin \omega t\)
\(I = I_0 \,sin\, (\omega t – \phi)\)
\(I=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}\;sin\left ( \omega t-tan^{-1}\dfrac {(X_L-X_C)}{R} \right)\)
\(I_0=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}\)
where,
\(Z=\sqrt {R^2+(X_L-X_C)^2}\)
The quantity Z is known as impedance of series LCR circuit.
where,
V_{0} \(\rightarrow\) Peak Voltage
Z \(\rightarrow\) Impedance
A \(420\,\Omega\)
B \(300\,\Omega\)
C \(313\,\Omega\)
D \(575.05\,\Omega\)
\(i_{rms}=\sqrt {\overline{i^2}}\)
\(I_{rms}=\dfrac {V_{rms}}{Z}\)
where,
Z = Impedance of the circuit
\(V_{rms}=\) rms voltage