Q.A light bulb and an open coil inductor are connected to an ac source through a key as shown in Fig. 7.9.

The source is $V=100\sqrt{2}\,\sin 200t$, so the peak voltage is $V_{max}=100\sqrt{2}\text{ V}$ and the angular frequency is $\omega=200\text{ rad/s}$.

Inductive reactance:

Impedance of the series $R$–$L$ circuit: …

We are dealing with an AC circuit where only a coil (inductor) is connected to a 220 V, 50 Hz supply. The problem asks for the inductance such that the maximum current is $3\sqrt{2}$ A. The key concept: in a purely inductive AC circuit, the voltage and current are related by inductive reactance, not resistance. The maximum (peak) values follow Ohm's law for reactance: $V_{\text{max}} = I_{\text{max}} X_L$.

Let’s work through it step by step.

1. Identify given quantities and convert to peak values The supply voltage is given as 220 V. In AC problems, unless stated otherwise, this is the RMS (root mean square) voltage. The relationship between RMS and peak (maximum) voltage is:

So:

The maximum current is directly given:

2. Recall the formula for inductive reactance For an inductor, the opposition to AC is called inductive reactance:

where $f = 50$ Hz is the frequency and $L$ is the inductance in henries.

3. Apply Ohm’s law for peak values in an inductive circuit For a pure inductor, the peak voltage and peak current are related exactly like Ohm’s law:

Substitute the expressions:

4. Simplify the equation Notice that $\sqrt{2}$ appears on both sides, so they cancel: $220 = 3 \times (2\pi \times 50 \times L)$ …

Concept & Intuition

A pure inductor opposes changes in current. When a sinusoidal voltage is applied, the current cannot rise instantly with the voltage — it must “wait” for the voltage to build up. This delay is exactly one-quarter of a cycle, or a phase lag of 90° (π/2 radians). So for a voltage $V(t) = V_0 \sin(\omega t)$, the current in a pure inductor is $i(t) = I_0 \sin(\omega t - \pi/2)$. The amplitude $I_0$ is given by Ohm’s law for AC: $I_0 = V_0 / X_L$, where $X_L = \omega L$ is the inductive reactance.

Step-by-step solution

1. Identify the given parameters

   Voltage: $V(t) = 200 \sin(250 t)$

   So amplitude $V_0 = 200$ V, angular frequency $\omega = 250$ rad/s.

   Inductance: $L = 0.02$ H.

2. Compute the inductive reactance

3. Find the current amplitude

4. Apply the phase relationship In a pure inductor, current lags voltage by $90^\circ$ (π/2 rad). Since voltage is a sine function, the current is: $$$$

The core idea here is Faraday’s law of electromagnetic induction: the induced emf in a coil equals the negative rate of change of magnetic flux through it. But we don’t have the emf directly — we have the current it drives through a known resistance. That’s where Ohm’s law steps in.

When a current flows because of an induced emf, the emf is simply $\mathcal{E} = I R$. Once we know the emf, Faraday’s law tells us the magnitude of the flux change over a given time interval. The negative sign (Lenz’s law) tells us direction, but the question asks only for the magnitude of the change.

Let’s go step by step.

1. Find the induced emf. The coil has resistance $R = 10~\Omega$ and carries an induced current $I = 2~\mathrm{A}$. By Ohm’s law, the induced emf is

2. Apply Faraday’s law. Faraday’s law states that the magnitude of the induced emf is

where $\Phi$ is the magnetic flux through the coil. For a small time interval $\Delta t$, if we assume the rate of change is constant,

3. Solve for the change in flux. Rearranging:

Here $\Delta t = 1~\mathrm{ms} = 1 \times 10^{-3}~\mathrm{s}$. So

$|\Delta \Phi| = 20 \times 10^{-3} = 2 \times 10^{-2}~\mathrm{Wb}.$ …

Concept and Intuition

When an AC source is connected to a purely inductive circuit (an ideal inductor with zero resistance), the relationship between voltage and current is not instantaneous — it’s governed by the inductor’s fundamental property: it opposes changes in current. This opposition is not a simple resistance; it’s a reactive effect that introduces a time shift, or phase difference, between voltage and current.

Think of it this way:

• For a resistor, voltage and current rise and fall together — they are in phase.
• For an inductor, the voltage depends on how fast the current is changing, not on the current itself.
  • When the current is at its peak, it’s momentarily not changing (slope = 0), so the voltage is zero.
  • When the current is crossing zero, it’s changing fastest, so the voltage is at its peak.

This swapping of peaks means the voltage reaches its maximum before the current does — hence voltage leads current by a quarter of a cycle (90°).

Step-by-Step Reasoning

1. Recall the defining equation for an ideal inductor The voltage across an inductor is proportional to the rate of change of current through it:

This is the key: voltage is not proportional to current itself, but to its derivative.

2. Assume a sinusoidal current Let the current be:

where $I_0$ is the peak current and $\omega$ is the angular frequency.

3. Compute the voltage Differentiate:

Using the identity $\cos(\omega t) = \sin(\omega t + 90^\circ)$, we get:

So the voltage is a sine wave with the same frequency, but shifted ahead by $90^\circ$ (or $\pi/2$ radians).

4. Interpret the phase relationship
   • Current: $i(t) = I_0 \sin(\omega t)$
   • Voltage: $v(t) = V_0 \sin(\omega t + 90^\circ)$ The voltage reaches its positive peak a quarter-cycle before the current does. In AC circuit terminology, we say voltage leads current by 90° (or current lags voltage by 90°). …

Concept & Intuition

In an AC circuit containing both resistance and inductance, the voltage and current are not in phase. The inductor causes the current to lag behind the voltage because it opposes changes in current. The resistor, however, keeps them in phase. The net phase difference depends on the ratio of inductive reactance ($X_L = \omega L$) to resistance ($R$). When $X_L = R$, the phase angle is exactly $45^\circ$ — a neat balance between the two effects.

Step-by-step solution

1. Find the angular frequency

   The source frequency is $f = \frac{50}{\pi}\,\mathrm{Hz}$.

   Angular frequency $\omega = 2\pi f = 2\pi \cdot \frac{50}{\pi} = 100\,\mathrm{rad/s}$.

2. Compute the inductive reactance

   $X_L = \omega L = 100 \times 1 = 100\,\Omega$.

3. Identify the resistance

   Given $R = 100\,\Omega$.

4. Phase difference formula for an LR circuit

   For a series RL circuit, the phase angle $\phi$ by which current lags voltage is:

   $\phi = \tan^{-1}\left(\frac{X_L}{R}\right)$ …

The key concept here is the difference between resistance and impedance.

A coil (inductor) has two properties:

• Resistance ($R$) — the same for DC and AC, due to the wire’s material.
• Inductance ($L$) — which only matters when the current changes, i.e., for AC.

For DC, the current is steady, so the inductor behaves like a plain resistor: opposition = $R = 20\ \Omega$.

For AC, the changing current induces a back emf that opposes the flow. This effect is called inductive reactance ($X_L = 2\pi f L$). The total opposition to AC is the impedance $Z = \sqrt{R^2 + X_L^2}$, which is always greater than $R$ alone (unless $X_L = 0$, which only happens at DC).

Thus, the coil offers more opposition to AC than to DC.

1. Identify the DC case

   For direct current, frequency $f = 0$, so inductive reactance $X_L = 2\pi f L = 0$.

   The only opposition is the ohmic resistance: $R = 20\ \Omega$.

2. Identify the AC case

   For alternating current, $f > 0$, so $X_L > 0$.

   The total opposition (impedance) is:

Since $X_L > 0$, we have $Z > R$.

3. Compare …

Maximum (peak) current $I_0=\sqrt2$ A, so rms current $I_{rms}=I_0/\sqrt2=1$ A.

Inductive reactance:

Since $X_L=2\pi f L$: …

For a pure inductor, if $i=I_0\sin\omega t$ then

$v=L\frac{di}{dt}=\omega L I_0\cos\omega t = \omega L I_0\sin\!\left(\omega t+\frac{\pi}{2}\right).$ …

[!TLDR]

Use $I_0=V_0/(2\pi f L)$ with peak voltage $V_0=220\sqrt2$ and peak current $0.9$ A to get $L\approx1.1$ H.

Concept

In a purely inductive AC circuit (CBSE Class-12 Alternating Current), the inductor limits current through its inductive reactance $X_L=\omega L=2\pi f L$. The maximum (peak) current relates to the peak voltage by $I_0=V_0/X_L$.

Solution

Given $V_{rms}=220$ V, $f=50$ Hz, maximum (peak) current $I_0=0.9$ A.

Peak voltage:

Inductive reactance needed:

Since $X_L=2\pi f L$, …

Step 1 — Read the circuit.

The figure shows a series LC circuit with:

• Inductor: $L = 0.5\ \text{mH}$
• Capacitor: $C = 20\ \mu\text{F}$

Step 2 — The concept behind resonance.

At resonance the inductive and capacitive reactances cancel exactly:

Since $\omega = 2\pi f$,

Step 3 — Convert to SI units (where most errors happen).

Step 4 — Compute $\sqrt{LC}$.

(The numbers are chosen so this comes out exact — a good sign we've converted correctly.)

Step 5 — Compute $f_r$. …