A bulb and a capacitor are connected in series to an a.c. source. A dielectric slab is now introduced between the plates of the capacitor. The intensity of the bulb will be:
Increases
Analyzing Bulb Intensity in an AC Circuit with a Dielectric
This question asks what happens to the intensity of a bulb connected in series with a capacitor to an AC source when a dielectric slab is introduced into the capacitor. The intensity of the bulb is directly related to the power dissipated by it, which in turn depends on the current flowing through the circuit.
Let's break down the components and their behavior in an AC circuit:
- The bulb acts as a resistor (R).
- The capacitor has capacitance (C).
- The circuit is connected to an AC voltage source.
In a series AC circuit containing resistance and capacitance, the flow of current is limited by the total opposition to the current, which is called impedance (Z). The impedance of an R-C series circuit is given by the formula:
$\qquad Z = \sqrt{R^2 + X_C^2}$
where R is the resistance of the bulb and $X_C$ is the capacitive reactance of the capacitor.
The capacitive reactance $X_C$ is the opposition offered by the capacitor to the AC current and is given by:
$\qquad X_C = \frac{1}{\omega C}$
where $\omega$ is the angular frequency of the AC source and C is the capacitance.
Effect of Introducing a Dielectric Slab
Now, consider what happens when a dielectric slab is introduced between the plates of the capacitor. A dielectric material increases the capacitance of the capacitor. The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon A}{d}$, where $\epsilon$ is the permittivity of the medium between the plates. When a dielectric with relative permittivity $\epsilon_r$ is introduced, the permittivity of the medium becomes $\epsilon = \epsilon_r \epsilon_0$, where $\epsilon_0$ is the permittivity of free space. Thus, the new capacitance $C'$ becomes:
$\qquad C' = \epsilon_r C$
Since for any dielectric material, the relative permittivity $\epsilon_r$ is greater than 1 ($\epsilon_r > 1$), introducing a dielectric slab increases the capacitance of the capacitor ($C' > C$).
How Increased Capacitance Affects the Circuit
Let's trace the effects of this increase in capacitance on the circuit:
- Capacitive Reactance ($X_C$): As capacitance C increases, the capacitive reactance $X_C = \frac{1}{\omega C}$ decreases.
- Impedance (Z): The total impedance of the circuit is $Z = \sqrt{R^2 + X_C^2}$. Since R (resistance of the bulb) remains constant and $X_C$ decreases, the total impedance Z of the circuit decreases.
- Current (I): The current flowing through the series circuit is given by Ohm's law for AC circuits: $I = \frac{V}{Z}$, where V is the voltage of the AC source (assumed constant). Since the impedance Z decreases and V is constant, the current I flowing through the circuit increases.
- Bulb Intensity: The intensity of the bulb is proportional to the power dissipated by it, which is given by $P = I^2 R$. Since the current I increases and the resistance R of the bulb is constant, the power dissipated by the bulb increases. Therefore, the intensity of the bulb increases.
| Action | Effect on Capacitance (C) | Effect on Capacitive Reactance (X<sub>C</sub>) | Effect on Impedance (Z) | Effect on Current (I) | Effect on Bulb Intensity |
|---|---|---|---|---|---|
| Introduce Dielectric | Increases ($\epsilon_r > 1$) | Decreases ($X_C \propto 1/C$) | Decreases ($Z = \sqrt{R^2 + X_C^2}$) | Increases ($I = V/Z$) | Increases ($P = I^2R$) |
In summary, introducing a dielectric slab into the capacitor increases its capacitance, which reduces the capacitive reactance. This reduction in reactance lowers the total impedance of the series circuit. A lower impedance allows a larger current to flow from the AC source. The increased current through the bulb results in greater power dissipation and thus increased intensity.
Revision Table: AC Circuit Concepts
| Concept | Description | Formula |
|---|---|---|
| Resistance (R) | Opposition to current flow in a resistor; independent of frequency. | Unit: Ohms ($\Omega$) |
| Capacitance (C) | Ability of a capacitor to store charge. | $C = \frac{Q}{V}$ (for DC), $C = \frac{\epsilon A}{d}$ (parallel plate) |
| Capacitive Reactance (X<sub>C</sub>) | Opposition to AC current flow offered by a capacitor; dependent on frequency and capacitance. | $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$ |
| Impedance (Z) | Total opposition to AC current flow in a circuit containing R, L, C components. | For R-C series: $Z = \sqrt{R^2 + X_C^2}$ |
| Current (I) in AC | Ratio of voltage to impedance in an AC circuit. | $I_{rms} = \frac{V_{rms}}{Z}$ |
| Power Dissipation in Resistor | Energy lost per unit time in a resistor. | $P = I^2R$ (in AC, using RMS current) |
| Dielectric Constant (ε<sub>r</sub>) | Factor by which capacitance increases when a dielectric fills the space between plates, compared to vacuum. | $\epsilon_r = \frac{C_{dielectric}}{C_{vacuum}}$ |
Additional Information: Dielectrics and Capacitors
A dielectric material is an electrical insulator that can be polarized by an applied electric field. When placed between the plates of a capacitor, the electric field polarizes the dielectric, causing the molecules within it to align in a way that opposes the field from the charges on the plates. This reduces the net electric field between the plates. Since the voltage across the capacitor is proportional to the electric field ($V = Ed$), a reduced field means a lower voltage for the same amount of charge stored ($Q$). From the definition of capacitance ($C = Q/V$), a lower voltage for the same charge means a higher capacitance.
The relative permittivity ($\epsilon_r$), also known as the dielectric constant (though permittivity is more accurate), quantifies how much a material increases the capacitance compared to a vacuum. Different materials have different dielectric constants; for example, vacuum has $\epsilon_r = 1$, air is slightly greater than 1, paper is around 3-4, and ceramics can be much higher.
Introducing a dielectric into a capacitor while it is connected to a voltage source (like our AC source) causes more charge to flow onto the plates to maintain the voltage, as the capacitor can now store more charge at that voltage. This increased charge capacity leads to a higher current flow in the AC circuit, as observed in our problem.
Important Questions from Electrostatic Potential and Capacitance
The waves used by artificial satellites for communication purposes are:
The shape of a wavefront when light emerges out of a convex lens after a parallel beam of light is incident on it:
A dielectric material placed in uniform electric field, which of the following option is NOT CORRECT:
Eight identical spherical drops, each having a potential of 9V, are combined together to form a single large drop. The potential of this large drop will be:
A uniformly charged conducting sphere of radius 1.3 m has a surface charge density of 70 μC m-2. What is the total electric flux leaving the surface of the sphere?
