Phasors & AC Voltage Applied to an Inductor

Phasors: Rotating Vectors for Oscillating Quantities

For a resistor, v and i are in phase — but for inductors and capacitors they are not. To track phase relationships, we use phasors.

A phasor is a vector rotating counter-clockwise about the origin with angular speed $\omega$. The vertical projection of the voltage phasor $\vec{V}$ (length $v_m$) and current phasor $\vec{I}$ (length $i_m$) at time t give the instantaneous values $v(t)$ and $i(t)$. The angle between the phasors is the phase difference between v and i — frozen as both rotate together.

Key Point (NCERT footnote): voltage and current are not actually vectors — they are scalars. It just happens that harmonically varying scalars combine mathematically like the projections of rotating vectors, so phasors give us a familiar (vector-addition) rule for adding oscillating quantities. A favourite 'true/false' exam line!

For the resistor: $\vec{V}$ and $\vec{I}$ point the same way at every instant — phase difference zero.

AC Across a Pure Inductor

Connect $v = v_m\sin\omega t$ to an inductor of self-inductance L (winding resistance negligible). Kirchhoff's loop rule with the back emf (Chapter 6):

$$v - L\frac{di}{dt} = 0 \quad\Rightarrow\quad \frac{di}{dt} = \frac{v_m}{L}\sin\omega t$$

Integrating (and dropping the constant — the current oscillates symmetrically about zero, so no steady component exists):

$$i = -\frac{v_m}{\omega L}\cos\omega t = i_m\sin\left(\omega t - \frac{\pi}{2}\right)$$

$$\boxed{i_m = \frac{v_m}{X_L}, \qquad X_L = \omega L}$$

$X_L$ is the inductive reactance — it limits the current exactly as resistance does, has the dimension of resistance, and is measured in ohms. It is directly proportional to both L and the frequency.

The headline: in a pure inductor, the current LAGS the voltage by $\pi/2$ — one quarter cycle ($T/4$). The current phasor trails the voltage phasor by 90 degrees.

[NEET Important] Memory hook for both reactive elements: 'L lags, C leads' (for the current relative to voltage).

An Inductor Consumes No (Average) Power

The instantaneous power delivered to the inductor:

$$p_L = iv = i_m\sin\left(\omega t - \frac{\pi}{2}\right) \times v_m\sin\omega t = -\frac{i_m v_m}{2}\sin 2\omega t$$

The average of $\sin 2\omega t$ over a complete cycle is zero, so

$$\boxed{\bar{P}_L = 0}$$

Physically: for a quarter cycle the source feeds energy into the inductor's magnetic field; in the next quarter the field returns it. The energy sloshes back and forth, never dissipating — the inductor limits current without consuming power (unlike a resistor).

[JEE Tip] $X_L = \omega L$ has two instructive limits: for DC ($\omega = 0$), $X_L = 0$ — an ideal inductor is a plain wire to steady current; at high frequency it chokes the current off. Frequency-dependence questions on $X_L$ (and $X_C$, next section) are near-guaranteed in JEE Main and NEET.

Solved Examples

Example 1: The 25 mH inductor (NCERT Example 7.2)

A pure inductor of 25.0 mH is connected to a 220 V, 50 Hz source. Find the inductive reactance and the rms current.

Solution:

1. Reactance: $X_L = 2\pi\nu L = 2 \times 3.14 \times 50 \times 25 \times 10^{-3} = 7.85\ \Omega$.
2. Current: $I = \frac{V}{X_L} = \frac{220}{7.85} = 28$ A.
3. A mere 7.85 ohm of reactance — yet no heat is generated in this ideal inductor.

Example 2: Reactance scales with frequency [NEET Numerical]

Find the reactance of a 0.1 H inductor at 50 Hz and at 500 Hz.

Solution:

1. At 50 Hz: $X_L = 2\pi \times 50 \times 0.1 = 31.4\ \Omega$.
2. At 500 Hz: ten times the frequency, ten times the reactance: $X_L = 314\ \Omega$.
3. $X_L \propto \nu$ — a straight line through the origin on an $X_L$ vs $\nu$ graph.

Example 3: Writing i(t) completely [JEE Numerical]

A voltage $v = 283\sin(314\,t)$ V is applied to a pure 25 mH inductor. Write the full expression for the current.

Solution:

1. $X_L = \omega L = 314 \times 0.025 = 7.85\ \Omega$.
2. $i_m = v_m/X_L = 283/7.85 = 36$ A.
3. Current lags by $\pi/2$: $i = 36\sin\left(314\,t - \dfrac{\pi}{2}\right)$ A.

Example 4: Lag in milliseconds [NEET Numerical]

On the 50 Hz mains, by how much time does the current in a pure inductor reach its peak after the voltage peaks?

Solution:

1. The lag is a quarter cycle: $\Delta t = T/4$.
2. $T = 1/50 = 20$ ms.
3. Answer: $\Delta t = 5$ ms — the current peaks 5 ms after the voltage, every cycle.

Example 5: Energy sloshing, quantified [JEE Numerical]

A 50 mH inductor carries an AC current of peak value 2.0 A. Find the maximum energy stored in it, and the average power it consumes.

Solution:

1. Max stored energy: $U = \frac{1}{2}Li_m^2 = \frac{1}{2} \times 0.05 \times 4 = 0.1$ J — held momentarily at each current peak.
2. Average power: zero — the same 0.1 J is returned to the source every quarter cycle.
3. Energy circulates; nothing burns.

Example 6: The DC limit [JEE Numerical]

An ideal 0.2 H inductor is connected to (a) a 12 V DC battery, (b) a 12 V (rms), 50 Hz source. Compare the opposition it offers.

Solution:

1. (a) DC: $\omega = 0 \Rightarrow X_L = 0$ — an ideal inductor offers no opposition to steady current (real coils are limited only by their winding resistance).
2. (b) AC: $X_L = 2\pi \times 50 \times 0.2 = 62.8\ \Omega$, so $I = 12/62.8 \approx 0.19$ A.
3. Same element, drastically different behaviour — reactance is a frequency phenomenon.

Example 7: Why does the current lag?

Explain physically why the current in an inductor lags the applied voltage.

Solution:

1. An inductor's back emf $-L\,di/dt$ opposes changes in current (Chapter 6 — electrical inertia).
2. When the applied voltage is at its peak, it is still 'pushing the current up' against this inertia; the current reaches its own peak only a quarter cycle later.
3. Mathematically the integration of $\sin\omega t$ gives $-\cos\omega t$ — automatically a 90-degree lag.

Example 8: Phasors are not vectors

Voltage and current are represented by rotating vectors. Are they vector quantities? Justify.

Solution:

1. No — voltage and current are scalars.
2. Harmonically varying scalars happen to add with the same mathematics as projections of rotating vectors; phasors simply exploit that coincidence to give us an easy addition rule.
3. NCERT flags this explicitly — quote it when asked, and earn the easiest mark of the paper.

Example 9: The iron rod and the bulb (NCERT Example 7.5)

A light bulb in series with an open-coil inductor glows on AC. An iron rod is inserted into the coil. What happens to the glow, and why?

Solution:

1. The rod's iron is magnetised by the coil's field, greatly increasing the magnetic field inside — so the inductance L increases.
2. $X_L = \omega L$ rises, so a larger share of the source voltage drops across the inductor, leaving less for the bulb.
3. The glow decreases. (Pull the rod out and it brightens again — induction made visible.)

Example 10: Where did the energy go?

Over one full cycle, the source connected to a pure inductor does zero net work — yet current flowed throughout. Reconcile.

Solution:

1. During the quarter cycles when $|i|$ grows, the source feeds energy into the field ($p > 0$).
2. During the quarter cycles when $|i|$ falls, the collapsing field drives the source backwards, returning that energy ($p < 0$).
3. The ledger balances exactly each half cycle: $\bar P = -\frac{i_mv_m}{2}\langle\sin 2\omega t\rangle = 0$. Current without consumption.