
In series LCR circuit voltage leads the current when (Given that\[{\omega _0}\]=resonant angular frequency)
A. \[\omega < {\omega _0}\]
B. \[\omega = {\omega _0}\]
C. \[\omega > {\omega _0}\]
D. None of these
Answer
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Hint: There are two possible arrangements in which we can arrange the components of a circuit- series and parallel. In a series circuit, all the components are connected one after the other in a continuous arrangement.
Formula used:
The inductive reactance is,
\[{X_L} = \omega L\]
Here, $\omega$ is the angular frequency and $L$ is the inductance.
The capacitive reactance is,
\[{X_C} = \dfrac{1}{{\omega C}}\]
Here, $\omega$ is the angular frequency and $C$ is the capacitance.
The impedance of the LCR circuit is,
\[Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} \]
Here, $X_L$ is the inductive reactance, $X_C$ is the capacitive reactance and $R$ is the resistance.
Complete step by step solution:
An LCR circuit is also known as inductance-capacitance-resistance circuit as it contains an inductor, capacitor and a resistor connected in series to each other. When the circuit is at high angular frequency, inductive reactance becomes greater than the capacitive reactance. Inductive reactance is represented as \[{X_L}\] and capacitive reactance is written as \[{X_C}\].
In a series LCR circuit, the inductive reactance becomes,
\[{X_L} = \omega L\]
The capacitive reactance is given by,
\[{X_C} = \dfrac{1}{{\omega C}}\]
As a result, the impedance of the LCR circuit which is given by the formula written below will also become very large.
\[Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} \]
Substituting the values given in the above formula, we get
\[Z = \sqrt {{R^2} + {{(\omega L - \dfrac{1}{{\omega C}})}^2}} \]
When the resonance is high, the voltage will lead the current. If \[{\omega _0}\] is the resonant angular frequency in a series LCR circuit and if the voltage leads the current, then \[\omega > {\omega _0}\].
Hence, Option C is the correct answer.
Note:It is important to remember that the resistance or the opposition offered by the inductor to the flow of electric current in an AC circuit is known as inductive reactance which has low value if the frequency is low and vice versa. On the other hand, the opposition to the flow of current in an AC circuit is known as capacitive reactance which behaves as a resistor. In this case since the voltage is leading the current, therefore, it will be an inductive circuit.
Formula used:
The inductive reactance is,
\[{X_L} = \omega L\]
Here, $\omega$ is the angular frequency and $L$ is the inductance.
The capacitive reactance is,
\[{X_C} = \dfrac{1}{{\omega C}}\]
Here, $\omega$ is the angular frequency and $C$ is the capacitance.
The impedance of the LCR circuit is,
\[Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} \]
Here, $X_L$ is the inductive reactance, $X_C$ is the capacitive reactance and $R$ is the resistance.
Complete step by step solution:
An LCR circuit is also known as inductance-capacitance-resistance circuit as it contains an inductor, capacitor and a resistor connected in series to each other. When the circuit is at high angular frequency, inductive reactance becomes greater than the capacitive reactance. Inductive reactance is represented as \[{X_L}\] and capacitive reactance is written as \[{X_C}\].
In a series LCR circuit, the inductive reactance becomes,
\[{X_L} = \omega L\]
The capacitive reactance is given by,
\[{X_C} = \dfrac{1}{{\omega C}}\]
As a result, the impedance of the LCR circuit which is given by the formula written below will also become very large.
\[Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} \]
Substituting the values given in the above formula, we get
\[Z = \sqrt {{R^2} + {{(\omega L - \dfrac{1}{{\omega C}})}^2}} \]
When the resonance is high, the voltage will lead the current. If \[{\omega _0}\] is the resonant angular frequency in a series LCR circuit and if the voltage leads the current, then \[\omega > {\omega _0}\].
Hence, Option C is the correct answer.
Note:It is important to remember that the resistance or the opposition offered by the inductor to the flow of electric current in an AC circuit is known as inductive reactance which has low value if the frequency is low and vice versa. On the other hand, the opposition to the flow of current in an AC circuit is known as capacitive reactance which behaves as a resistor. In this case since the voltage is leading the current, therefore, it will be an inductive circuit.
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