Correct option is (c)

To understand the impact of placing a dielectric between the plates of the capacitor on the glow of the bulb, we need to consider the properties and behavior of capacitors in an AC circuit.

When a capacitor is connected in series with a bulb in an AC circuit, the impedance of the capacitor plays a significant role in determining the current through the circuit. The impedance $Z_C$ of a capacitor in an AC circuit is given by:

$Z_C=\frac{1}{\omega C}$

where:

• $Z_C$ is the capacitive reactance (impedance of the capacitor).
• $\omega$ is the angular frequency of the AC supply ( $\omega =2\pi f$ , where $f$ is the frequency).
• $C$ is the capacitance of the capacitor.

When we place a dielectric between the plates of the capacitor, the capacitance $C$ increases. The capacitance with a dielectric can be described as:

$C^{\prime }=\kappa C$

where:

• $C^{\prime }$ is the new capacitance with the dielectric present.
• $\kappa$ is the dielectric constant ( $\kappa >1$ ).

Since $C^{\prime }>C$, the new capacitive reactance $Z_C^{\prime }$ can be given as:

$Z_C^{\prime }=\frac{1}{\omega C^{\prime }}$

Because $C^{\prime }=\kappa C$, we have:

$Z_C^{\prime }=\frac{1}{\omega \kappa C}=\frac{1}{\kappa }\left(\frac{1}{\omega C}\right)=\frac{Z_C}{\kappa }$

Since $\kappa >1$, $Z_C^{\prime }<Z_C$. This means the impedance of the capacitor decreases when a dielectric is placed between its plates. In a series circuit, the overall impedance decreases when the impedance of one component decreases, leading to an increase in the current through the circuit.

Thus, with an increase in current, the bulb will glow brighter. Therefore, the correct option is:

Option C: increases

Correct option is (b)

To find the inductance of the coil, we need to analyze the given circuit and use the information provided. The circuit consists of a resistor and an inductor in series, connected to an AC supply. Here are the given values:

1. Resistance, $R=90\mathrm{\Omega }$

2. Supply voltage, $V_{\text{total}}=120\mathrm{V}$

3. Frequency, $f=60\mathrm{Hz}$

4. Voltage across the resistor, $V_R=36\mathrm{V}$

First, we calculate the current through the resistor (which is the same as the current through the inductor, since they are in series) using Ohm's law:

$I=\frac{V_R}{R}=\frac{36}{90}=0.4\mathrm{A}$

Next, we find the total impedance $Z$ of the series combination from the total supply voltage:

$V_{\text{total}}=I\cdot Z$

$Z=\frac{V_{\text{total}}}{I}=\frac{120}{0.4}=300\mathrm{\Omega }$

We know that the total impedance in a series circuit consisting of a resistor and an inductor is given by:

$Z=\sqrt{R^2+(X_L{)}^2}$

where $X_L$ is the inductive reactance. Rearrange this to solve for $X_L$:

$X_L=\sqrt{Z^2-R^2}=\sqrt{(300{)}^2-(90{)}^2}=\sqrt{90000-8100}=\sqrt{81900}\approx 286\mathrm{\Omega }$

Now, we use the inductive reactance formula to find the inductance $L$:

$X_L=2\pi fL$

$L=\frac{X_L}{2\pi f}=\frac{286}{2\pi \cdot 60}\approx \frac{286}{376.99}\approx 0.76\mathrm{H}$

Thus, the inductance of the coil is:

Option B: 0.76 H

Correct option is (a)

$\begin{array}{rl}& V_L=I(\omega L)=31.4\\ & \begin{array}{rl}\Rightarrow I& =\frac{31.4}{2\times 3.14\times 50\times 10\times {10}^{-3}}\\ & =10\mathrm{A}\end{array}\end{array}$

Correct option is (c)

To understand which answer is correct, we must consider the relationship between voltage and current in various types of circuits, namely circuits with pure inductors, pure capacitors, pure resistors, and a combination of inductors and capacitors (LC circuits).

In a purely resistive circuit, the voltage and current are in phase, meaning when the voltage is maximum, the current is also maximum. Therefore, option C (pure resistor) cannot result in the instantaneous current being zero when the instantaneous voltage is maximum.

In a purely inductive circuit, the current lags the voltage by ${90}^{\circ }$, or in other words, when the voltage is at its maximum, the current is zero. This is because the inductor opposes changes in current, leading to this phase difference.

In a purely capacitive circuit, the current leads the voltage by ${90}^{\circ }$. This means when the voltage is at its maximum value, the current through the capacitor is zero because the current reaches its maximum or minimum before the voltage does.

In an LC circuit (inductor-capacitor), under certain conditions like at its resonant frequency, the circuit behaves as if it is purely resistive in nature where voltage and current are in phase. However, it's more nuanced because the question likely implies a condition not at resonance but rather at a general characteristic of LC circuits. In LC circuits, aside from the resonance condition, the presence of both inductive and capacitive components can lead to scenarios where the inductive and capacitive reactances cancel each other, particularly in the case where oscillations are involved, and indeed, there can be moments when the voltage is maximum and the current is minimum (zero in the ideal case) due to the energy being alternately stored in the magnetic field of the inductor and the electric field of the capacitor.

Thus, reflecting on the scenarios:

• Pure inductor - Fits the description: instantaneous current is zero when the instantaneous voltage is maximum.
• Pure capacitor - Also fits the description for the reason described, taking into account ideal conditions.
• Pure resistor - Does not fit as voltage and current are in phase.
• Combination of an inductor and capacitor (LC circuit) - Can fit the description under certain non-resonance conditions where the behavior of energy exchange between L and C at a specific instant can result in the instant current being zero when the voltage is maximum, but this is more complex and dependent on specific conditions unlike the straightforward lag or lead in purely inductive or capacitive circuits.

Therefore, the correct answer is Option C: A, B, and D only.

Correct option is (c)

$\begin{array}{rl}& E=25\sin (1000t)\\ \\ & \cos \theta =\frac{1}{\sqrt{2}}\end{array}$

LR circuit

$\begin{array}{rl}& \text{ Initially }\frac{R}{{\omega }_{1}L}=\frac{1}{\tan \theta }=\frac{1}{\tan {45}^{\circ }}=1\\ \\ & X_L={\omega }_{1}L\\ \\ & {\omega }_2=2{\omega }_1,\text{ given }\\ \\ & \tan {\theta }^{\prime }=\frac{{\omega }_{2}L}{R}=\frac{2{\omega }_{1}L}{R}\\ \\ & \tan {\theta }^{\prime }=2\\ \\ & \cos {\theta }^{\prime }=\frac{1}{\sqrt{5}}\end{array}$