Question

What is the phase relation between current and voltage in an ideal inductor connected to an ac source? Draw a phasor diagram for the circuit.

Hint:

Remember the acronym ELI: in an EMF source connected to an L (inductor), the voltage (E) leads the current (I). The angle of separation is always exactly \(90^{\circ}\) or \(\pi/2\) radians for ideal components.

Answer 1

Concept: When an alternating voltage source \(v = V_0 \sin(\omega t)\) is connected across a purely ideal inductor of inductance \(L\), a back electromotive force (EMF) is induced within its windings according to Faraday's and Lenz's laws: \[ e = -L \frac{di}{dt} \] By applying Kirchhoff's loop rule to this single-element circuit: \[ v + e = 0 \quad \Rightarrow \quad V_0 \sin(\omega t) = L \frac{di}{dt} \]

Step 1: Deriving the phase relation mathematically.
To find the instantaneous current \(i\), we rearrange the loop differential equation: \[ di = \frac{V_0}{L} \sin(\omega t) dt \] Integrating both sides with respect to time: \[ i = \int \frac{V_0}{L} \sin(\omega t) dt = -\frac{V_0}{\omega L} \cos(\omega t) \] Using the trigonometric identities to map this into a standard positive sine wave representation: \[ -\cos(\omega t) = \sin\left(\omega t - \frac{\pi}{2}\right) \] Substituting this back gives: \[ i = I_0 \sin\left(\omega t - \frac{\pi}{2}\right) \] Where the peak current value is defined as \(I_0 = \frac{V_0}{\omega L} = \frac{V_0}{X_L}\).

Step 2: Conclusion of phase difference and sketching phasor parameters.
Comparing our alternating voltage equation and alternating current equation: \[ v = V_0 \sin(\omega t) \] \[ i = I_0 \sin\left(\omega t - \frac{\pi}{2}\right) \] It is clear that the alternating current lags behind the alternating voltage by a phase angle of \(\frac{\pi}{2}\) radians (\(90^{\circ}\)). Alternatively, we state that the voltage leads the current by exactly \(\frac{\pi}{2}\) radians. The phasor representation diagram reflects this geometry clearly. The voltage phasor \(\vec{V}_0\) is drawn at an angle \(\omega t\), while the current phasor \(\vec{I}_0\) is positioned exactly \(90^{\circ}\) behind it in a clockwise direction.