LR and CR Circuits

AC Through Inductor and Resistor in Series (LR Circuit)

Let an inductor of inductance $L$ and a resistor of resistance $R$ be connected in series to an ac source of emf $E$ and frequency $f$ as shown in the figure given below. This type of circuit is known as LR circuit.

AC Through Inductor and Resistor in Series (LR Circuit)

The potential difference across inductor \[V_L=IX_L\] In the inductor, voltage $V_L$ leads current $I$ by a phase angle of $\frac{π}{2}$. [From: AC Through an Inductor (AC Through Circuit Elements)]

The potential difference across resistor \[V_R=IR\] In the resistor, voltage $V_R$ and current $I$ are in same phase. [From: AC Through a Resistor (AC Through Circuit Elements)]

Phasor diagram of an ac circuit containing inductor and resistor in series (LR Circuit)

Since $V_R$ and $I$ are in same phase, $V_R$ is represented by $OA$ in the direction of $I$. $V_L$ is represented by $OB$ perpendicular to the direction of $I$ because $V_L$ leads $I$ by a phase angle of $\frac{π}{2}$. As the resultant of $V_L$ and $V_R$ is equal to $E$, by using Pythagoras theorem in $\Delta OAC$, \[OC=\sqrt{OA^2+OB^2}\] \[\therefore E=\sqrt{V_R^2+V_L^2}\]

\[E=\sqrt{I^2R^2+I^2X_L^2}\] \[E=I\sqrt{R^2+X_L^2}\] Hence, the current in the circuit is given by \[I=\frac{E}{\sqrt{R^2+X_L^2}}\] Also, if $Z$ be the impedance of the circuit, then \[Z=\frac{E}{I}=\sqrt{R^2+X_L^2}\] If the voltage $E$ leads the current $I$ by an angle $\theta$, then \[\tan\theta=\frac{AC}{OA}=\frac{V_L}{V_R}=\frac{X_L}{R}\] \[\therefore\theta=\tan^{-1}\left(\frac{X_L}{R}\right)\] The phase relation between current and voltage in LR circuit is graphically shown below.

Phase relation between voltage and current in LR circuit

AC Through Capacitor and Resistor in Series (CR Circuit)

Let a capacitor of capacitance $C$ and a resistor of resistance $R$ be connected in series to an ac source of emf $E$ and frequency $f$ as shown in the figure given below. This type of circuit is known as CR circuit.

AC Through Capacitor and Resistor in Series (CR Circuit)

The potential difference across capacitor \[V_C=IX_C\] In the capacitor, voltage $V_C$ lags behind current $I$ by a phase angle of $\frac{π}{2}$. [From: AC Through a Capacitor (AC Through Circuit Elements)]

The potential difference across resistor \[V_R=IR\] In the resistor, voltage $V_R$ and current $I$ are in same phase. [From: AC Through a Resistor (AC Through Circuit Elements)]

Phasor diagram of an ac circuit containing capacitor and resistor in series (CR circuit)

Since $V_R$ and $I$ are in same phase, $V_R$ is represented by $OA$ in the direction of $I$. $V_C$ is represented by $OB$ perpendicular to the direction of $I$ because $V_C$ lags behind $I$ by a phase angle of $\frac{π}{2}$. As the resultant of $V_C$ and $V_R$ is equal to $E$, by using Pythagoras theorem in $\Delta OAC$, \[OC=\sqrt{OA^2+OB^2}\] \[\therefore E=\sqrt{V_R^2+V_C^2}\]

\[E=\sqrt{I^2R^2+I^2X_C^2}\] \[E=I\sqrt{R^2+X_C^2}\] Hence, the current in the circuit is given by \[I=\frac{E}{\sqrt{R^2+X_C^2}}\] Also, if $Z$ be the impedance of the circuit, then \[Z=\frac{E}{I}=\sqrt{R^2+X_C^2}\] If the voltage $E$ lags behind the current $I$ by an angle $\theta$, then \[\tan\theta=\frac{AC}{OA}=\frac{V_C}{V_R}=\frac{X_C}{R}\] \[\therefore\theta=\tan^{-1}\left(\frac{X_C}{R}\right)\] The phase relation between current and voltage in CR circuit is graphically shown below.

Phase relation between voltage and current in CR circuit

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