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 \(V = V_0 \,sin\, \omega t\)
(\(I = I_0\, sin\, (\omega t –\phi)\))
\(I=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}\;sin\left ( \omega t-tan^{-1}\dfrac {(X_L-X_C)}{R} \right)\)
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
\(V_L=I_0\,X_L\)
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.