AC Through Circuit Elements

AC Through a Resistor

An AC source of emf $E$ and frequency $f$ is connected to a pure resistor of resistance $R$ as shown in figure.

AC through a resistor of resistance R

The instantaneous emf across the resistor is given by \[E=E_0\sin\omega t\text{ __(1)}\] where, $\omega=2πf$, $f$ being the frequency of the source and $E_0$ is its maximum emf.

Let $I$ be the instantaneous current in the circuit at any time $t$, then, \[I=\frac{E}{R}=\frac{E_0}{R}\sin\omega t\] \[\therefore I=I_0\sin\omega t\text{ __(2)}\] where, $I_0=\frac{E_0}{R}$ is the peak value of ac.

From $\text{(1)}$ and $\text{(2)}$, we find that current and voltage are in same phase. Hence, in an ac circuit containing only resistor, the current and voltage are in same phase as shown in the figure given below.

Graph showing voltage and current in same phase in an ac circuit containing only resistor

The phasor diagram or vector diagram showing the relationship between alternating current and alternating emf in an ac circuit containing only resistor is shown below.

Phasor diagram between current and voltage of an ac circuit containing only resistor

AC Through an Inductor

An AC source of emf $E$ and frequency $f$ is connected to a pure inductor of inductance $L$ as shown in figure.

AC through an inductor of inductance L

The instantaneous emf across the inductor is given by \[E=E_0\sin\omega t\text{ __(a)}\] According to the Faraday’s laws of electromagnetic induction, an emf $(\varepsilon)$ is induced across the inductor given by \[\varepsilon=-L\frac{dI}{dt}\]

As there is no potential drop in the circuit, using Kirchhoff’s voltage law $(\sum V=\sum IR)$, the net emf in the circuit is given by \[E+\varepsilon=0\] \[E=L\frac{dI}{dt}\] \[E_0\sin\omega t\: dt=L\:dI\] Integrating both sides, we get \[\int E_0\sin\omega t\:dt=\int L\:dI\] \[-\frac{E_0\cos\omega t}{\omega}=LI+C\text{ __(b)}\] where $C$ is an arbitrary constant.

For $I=0$, $\omega t=\frac{π}{2}$, Hence, \[-\frac{E_0\cos\frac{π}{2}}{\omega}=L×0+C\] \[\therefore C=0\] Putting the value of $C$ in equation $\text{(b)}$, \[-\frac{E_0\cos\omega t}{\omega}=LI\] \[I=\frac{E_0}{\omega L}\sin\left(\omega t-\frac{π}{2}\right)\] We have, inductive reactance $(X_L)$$=\omega L$. \[\therefore I=\frac{E_0}{X_L}\sin\left(\omega t-\frac{π}{2}\right)\] \[\therefore I=I_0\sin\left(\omega t-\frac{π}{2}\right)\] where, $I_0=\frac{E_0}{X_L}$ is the peak value of ac.

Comparing $\text{(a)}$ and $\text{(c)}$, we find that current $I$ lags behind voltage $E$ by a phase angle of $\frac{π}{2}$. Conversely, the voltage $E$ leads the current $I$ by a phase angle of $\frac{π}{2}$.

Graph of an ac circuit containing only inductor: voltage leads current by a phase angle of π/2

The phasor diagram or vector diagram of an ac circuit containing only an inductor is shown below.

Phasor diagram of an ac circuit containing only inductor: voltage leads current by a phase angle of π/2

AC Through a Capacitor

An AC source of emf $E$ and frequency $f$ is connected to a pure capacitor of capacitance $C$ as shown in figure.

AC through a capacitor of capacitance C

The instantaneous emf across the resistor is given by \[E=E_0\sin\omega t\text{ __(i)}\] Let $q$ be the charge on the capacitor at any instant. Then, we can write \[q=CE\] \[q=CE_0\sin\omega t\] The instantaneous current $(I)$ is given by \[I=\frac{dq}{dt}\] \[=\frac{d}{dt}(CE_0\sin\omega t)\] \[=\omega CE_0\cos\omega t\] \[=\omega CE_0\sin\left(\omega t+\frac{π}{2}\right)\]

We have, capacitive reactance $(X_C)$$=\frac{1}{\omega C}$. \[\therefore I=\frac{1}{X_C}E_0\sin\left(\omega t+\frac{π}{2}\right)\] \[\therefore I=I_0\sin\left(\omega t+\frac{π}{2}\right)\text{ __(ii)}\] where, $I_0=\frac{E_0}{X_C}$ is the peak value of ac.

Comparing $\text{(i)}$ and $\text{(ii)}$, we find that current $I$ leads voltage $E$ by a phase angle of $\frac{π}{2}$. Conversely, the voltage $E$ lags behind the current $I$ by a phase angle of $\frac{π}{2}$.

Graph of an ac circuit containing only capacitor: voltage lags behind current by a phase angle of π/2

The phasor diagram or vector diagram of an ac circuit containing only a capacitor is shown below.

Phasor diagram of an ac circuit containing only capacitor: voltage lags behind current by a phase angle of π/2

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