Q.A variable frequency a.c. source is connected to a capacitor. How will the displacement current change with decrease in frequency?

Concept & Intuition

In an AC circuit, the total RMS current $I_{\text{rms}}$ can be split into two perpendicular components:

• Active current $I_a$: in phase with the voltage, responsible for real power.
• Wattless (reactive) current $I_r$: 90° out of phase with the voltage, responsible for reactive power.

These form a right triangle:

The power factor is $\cos\phi = \frac{I_a}{I_{\text{rms}}}$. So if we know the total current and the reactive component, we can find the active component and then the power factor.

Step-by-step solution

1. Identify given quantities

   RMS current: $I_{\text{rms}} = 2\ \text{A}$

   Wattless (reactive) current: $I_r = \sqrt{3}\ \text{A}$

2. Find the active current

   Using the Pythagorean relation:

   $$$$

Understanding Impedance in LCR Circuits

In a series LCR circuit, the total impedance determines the phase relationship between current and voltage. The impedance is:

The term $\left(\omega L - \frac{1}{\omega C}\right)$ is the net reactance:

• $\omega L$ is the inductive reactance $X_L$
• $\frac{1}{\omega C}$ is the capacitive reactance $X_C$

The condition $\frac{1}{\omega C} < \omega L$ means $X_C < X_L$, so the net reactance is positive and inductive.

Analyzing Each Option

Let me work through what this condition implies:

1. Phase angle and its sign

The phase angle $\phi$ between voltage and current is given by:

Since $\omega L > \frac{1}{\omega C}$, the numerator is positive. Assuming $R > 0$ (as in any real circuit), we have $\tan \phi > 0$, which means $\phi$ is positive, not negative.

2. The circuit's character

When $X_L > X_C$, the inductive reactance dominates. The circuit behaves as if it has a net inductance, making it predominantly inductive. This confirms option (C) is correct.

3. Current-voltage relationship

In an inductive circuit (positive phase angle), the voltage leads the current, or equivalently, the current lags the voltage. This confirms option (D) is correct.

4. Power factor

The power factor is:

Since we have a series circuit with resistance $R$ (which must be present for any real circuit), the power factor is: …

Concept & Intuition

A light bulb emits electromagnetic radiation, but not all the electrical power becomes light — only the efficiency fraction does. That radiated power spreads uniformly over a sphere of radius $r$. The intensity (power per area) at distance $r$ is linked to the peak electric field $E_0$ by the formula for a plane wave in vacuum:

We work backwards: from the bulb’s useful power → intensity → peak field.

Step-by-step solution

1. Find the actual radiated power The bulb’s total power is 60 W, but only 16% is converted to electromagnetic radiation.

2. Intensity at distance $r = 2$ m The radiation spreads uniformly over a sphere of radius 2 m. The surface area is $4\pi r^2$.

3. Relate intensity to peak electric field For an electromagnetic wave in vacuum, the time‑averaged intensity is

We know $c = 3 \times 10^8$ m/s and $\epsilon_0$ from the given constant:

It’s cleaner to use the relation in terms of the given constant. Recall that

Actually, we can solve directly:

Substitute $I = 0.6/\pi$:

$E_0^2 = \frac{2 \times (0.6/\pi)}{\epsilon_0 c} = \frac{1.2}{\pi} \cdot \frac{1}{\epsilon_0 c}.$ …

Concept and Intuition

Amplitude Modulation (AM) is a technique where the amplitude of a high-frequency carrier wave is varied in accordance with the instantaneous amplitude of a low-frequency message signal. This process allows the message signal to be transmitted over long distances.

Mathematically, if a carrier signal is given by $c(t) = A_c \sin(\omega_c t)$ and a message signal by $m(t) = A_m \sin(\omega_m t)$, then the amplitude-modulated wave $s(t)$ can be represented as:

Here, $A_c$ is the amplitude of the carrier wave and $A_m$ is the amplitude of the message signal. $\omega_c$ and $\omega_m$ are the angular frequencies of the carrier and message signals, respectively.

The modulation index, denoted by $\mu$, is a crucial parameter in AM. It is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier signal:

Substituting $A_m = \mu A_c$ into the equation for $s(t)$:

Using the trigonometric identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$, we can rewrite the second term:

So, the AM wave can be expressed as:

This equation shows that an AM wave consists of three frequency components:

1. The carrier wave, with frequency $\omega_c$ and amplitude $A_c$.
2. A lower sideband (LSB) wave, with frequency $(\omega_c - \omega_m)$ and amplitude $\frac{\mu A_c}{2}$.
3. An upper sideband (USB) wave, with frequency $(\omega_c + \omega_m)$ and amplitude $\frac{\mu A_c}{2}$.

The term "amplitude of the sidebands" typically refers to the amplitude of each individual sideband component, which is $\frac{\mu A_c}{2}$. However, in some contexts, especially when options are provided, it might refer to the amplitude of the modulating signal ($A_m = \mu A_c$) or the sum of the amplitudes of the two sideband components ($2 \times \frac{\mu A_c}{2} = \mu A_c$). We will explore this ambiguity in the steps below.

Step-by-step Solution

1. Identify the given values:

   • Amplitude of the carrier signal, $A_c = 30$ V.
   • Modulation index, $\mu = 0.4$.

2. Calculate the amplitude of the modulating signal ($A_m$):

   The modulation index is defined as $\mu = \frac{A_m}{A_c}$.

   Therefore, the amplitude of the modulating signal is $A_m = \mu A_c$.

3. Determine the amplitude of each sideband: From the expanded form of the AM wave, the amplitude of the lower sideband (LSB) and the upper sideband (USB) is $\frac{\mu A_c}{2}$. …

Concept & Intuition

In an LCR series circuit connected to an AC source, the average power dissipated per cycle is not simply $V_{\text{rms}}^2 / Z$ — that would give the apparent power. The real power is dissipated only in the resistor, because the inductor and capacitor store and return energy each cycle without net loss. So the key is to find the rms current through the circuit and then use $P = I_{\text{rms}}^2 R$. The impedance $Z$ tells us how the total voltage relates to current, but only $R$ matters for power.

Step-by-step solution

1. Identify the source voltage and its rms value The source is $v(t) = 150 \sin(80\pi t)$ volts. The peak voltage is $V_0 = 150~\text{V}$. The rms voltage is

2. Find the rms current using the impedance For any series AC circuit, the rms current is

where $Z$ is the total impedance. Here $Z = 75~\Omega$, so

3. Compute the average power Only the resistor dissipates power. The average power is

With $R = 25~\Omega$ and $I_{\text{rms}} = \sqrt{2}~\text{A}$,

4. Check the other options …

The concept first: the potentiometer is a voltage ruler

A potentiometer works because the driver cell sets up a uniform potential gradient $k$ (volts per cm) along the wire. When the galvanometer shows null deflection at length $\ell$, the unknown p.d. exactly equals the p.d. across that stretch of wire:

Because $k$ is the same for every measurement on the same setup, length is a faithful stand-in for voltage. So any relation among the voltages translates, term for term, into the same relation among the lengths.

Step-by-step

Step 1 — Translate each measurement.

Step 2 — Use the fact that potentials add along the path $A \to B \to C$.

Step 3 — Solve for $V_{BC}$.

$V_{BC} = V_{AC} - V_{AB} = k(8) - k(64) = -56k$ …

Concept and Intuition

When the battery is disconnected before inserting the dielectric, the capacitor plates are isolated, so the charge $Q$ on them cannot change. The dielectric reduces the electric field inside the capacitor (by a factor equal to the dielectric constant $K$), which lowers the voltage between the plates and therefore reduces the stored electrostatic energy. Energy is conserved overall: the decrease in stored energy must appear as heat (or other forms of energy) — that’s the “energy dissipated.” The problem asks for that amount.

A common pitfall is to treat this as if the battery remained connected (where voltage stays constant and charge changes). Here, because the battery is removed, the charge is constant, not the voltage. That changes the calculation completely.

Step-by-step solution

1. Find the initial charge and energy The capacitor is fully charged by the battery. Capacitance $C_0 = 8\,\mu\mathrm{F}$, battery voltage $V_0 = 20\,\mathrm{V}$. Charge stored:

Initial stored energy:

2. After inserting the dielectric (battery disconnected) The dielectric constant $K = 8$. With the battery disconnected, charge $Q$ remains $160\,\mu\mathrm{C}$. The new capacitance becomes:

The new voltage across the plates:

The new stored energy:

3. Energy dissipated The energy lost by the capacitor is the difference between initial and final stored energy: $$$$

The key idea is that the bulb’s brightness depends on the power it dissipates, which is proportional to the square of the current through it. In an RC series circuit, the total impedance is $Z = \sqrt{R^2 + X_C^2}$, where $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$. As frequency $f$ increases, $X_C$ decreases, so $Z$ decreases. With the source voltage fixed, the current $I = V/Z$ increases, and the bulb glows more intensely.

1. Identify the circuit behavior The bulb acts as a resistor $R$. The capacitor has reactance $X_C = \frac{1}{\omega C}$. They are in series, so the total impedance is

The source voltage $V$ is constant.

2. Effect of increasing frequency

   Angular frequency $\omega = 2\pi f$. When $f$ increases, $\omega$ increases, so $X_C = 1/(\omega C)$ decreases. This makes the denominator $Z$ smaller.

3. Current through the bulb

   By Ohm’s law for AC circuits,

Since $Z$ decreases, $I$ increases.

4. Power dissipated in the bulb …

The key idea is that in an LC circuit (no resistance), the current oscillates sinusoidally. The rate of change of current, $\frac{di}{dt}$, is directly related to the voltage across the inductor by Faraday's law: $V_L = L \frac{di}{dt}$. Since the inductor and capacitor are in series, the voltage across the inductor equals the voltage across the capacitor at every instant (with opposite sign). So the maximum $\frac{di}{dt}$ occurs when the capacitor voltage is maximum — which is the initial voltage $50\ \mathrm{V}$, because there's no resistance to dissipate energy.

Let's work through it step by step.

1. Identify the circuit and the relevant physics. After the battery is removed and the coil is connected, we have a pure LC circuit. The capacitor is initially charged to $V_0 = 50\ \mathrm{V}$ with capacitance $C = 2\ \mu\mathrm{F} = 2 \times 10^{-6}\ \mathrm{F}$. The inductor has $L = 10\ \mathrm{mH} = 10^{-2}\ \mathrm{H}$. There is no resistor, so the total energy oscillates between capacitor and inductor without loss.

2. Relate $\frac{di}{dt}$ to the voltage. For an inductor, $V_L = L \frac{di}{dt}$. In the series LC loop, Kirchhoff's voltage law gives $V_C + V_L = 0$, so $V_L = -V_C$. Therefore,

The magnitude of the rate of change of current is $| \frac{di}{dt} | = \frac{|V_C|}{L}$.

3. Find when $|V_C|$ is maximum. In an undamped LC oscillation, the capacitor voltage varies sinusoidally between $+V_0$ and $-V_0$. The maximum magnitude is $V_0 = 50\ \mathrm{V}$, which occurs at $t=0$ (and again every half-period). So the maximum $|V_C|$ is simply the initial voltage. …

The interval $1\text{ ms}$ is only a fraction of the AC period, so the average power is not the full-cycle value; the energy must be found by direct integration of the instantaneous power.

1. Instantaneous power.

With $\varepsilon(t)=10\sin(250\pi t)$ (volts, $t$ in seconds) and $R=100~\Omega$,

2. Energy integral.

3. Evaluate.

Since $\sin(0.5\pi)=1$, …