Electrical Engineering All Engineering Career

A paper capacitor is usually available in the form of rolled foil. 

So, the correct option is:

rolled foil

A capacitor consists of two conductors separated by an insulator, which is called a dielectric. 

So, the correct option is:

two conductors separated by an insulator

The incorrect statement is:

"The stored energy in a capacitor increases with applied voltage."

The stored energy (\(W\)) in a capacitor is given by the formula:

\[ W = \dfrac{1}{2} CV^2 \]

Where:
- \(C\) is the capacitance,
- \(V\) is the voltage (applied).

As per the formula, the stored energy in a capacitor is proportional to the square of the applied voltage (\(V\)). So, an increase in applied voltage leads to an increase in stored energy, not a decrease.

Therefore, the correct answer is:

"The stored energy in a capacitor decreases with reduction in value of capacitance."

Inside a hollow metallic charged sphere, the electric field is zero. 

According to Gauss's law, the electric field inside a conductor is zero. This holds true for a hollow metallic sphere as well, regardless of whether it's charged or not. The charges on the surface redistribute themselves in such a way that the electric field inside cancels out, resulting in a net electric field of zero.

So, the correct option is:

zero

If an AC voltage is applied to a capacitive circuit, the alternating current can flow in the circuit because the varying voltage produces charging and discharging currents.

In an AC circuit with a capacitor, the voltage across the capacitor alternates, causing the capacitor to charge and discharge in each half-cycle of the AC waveform. As the voltage across the capacitor changes, current flows into or out of the capacitor, resulting in an alternating current in the circuit.

So, the correct option is:

Varying voltage produces the charging and discharging currents.

To obtain a resultant capacitance of \( \dfrac{2}{3}C \) using three capacitors each of capacity \( C \), we can use two capacitors in parallel and the third in series with this combination.

Let's denote the capacitance of each capacitor as \( C_1 \), \( C_2 \), and \( C_3 \). Then, for the parallel combination, the equivalent capacitance \( C_{\text{parallel}} \) is given by:

\[ C_{\text{parallel}} = C_1 + C_2 \]

Since all capacitors have the same capacitance, we have:

\[ C_{\text{parallel}} = C + C = 2C \]

Now, for the series combination, the equivalent capacitance \( C_{\text{series}} \) is given by:

\[ C_{\text{series}} = \dfrac{1}{{\dfrac{1}{{C_{\text{parallel}}}} + \dfrac{1}{{C_3}}}} \]

Substituting the values:

\[ C_{\text{series}} = \dfrac{1}{{\dfrac{1}{{2C}} + \dfrac{1}{C}}} = \dfrac{1}{{\dfrac{3}{2C}}} = \dfrac{2}{3}C \]

So, the resultant capacitance \( \dfrac{2}{3}C \) can be obtained by using two capacitors in parallel and the third in series with this combination. 

Therefore, the correct option is:

two in parallel and third in series with this combination

The relative permittivity (also known as dielectric constant) of free space, denoted by \( \varepsilon_0 \), is given by 1.

So, the correct option is:

1

The charge (\(Q\)) acquired by a capacitor can be calculated using the formula:

\[ Q = C \times V \]

Where:
- \(C\) is the capacitance (in farads),
- \(V\) is the potential difference (in volts).

Given:
- \(C = 80 \mu F = 80 \times 10^{-6} F\) (capacitance),
- \(V = 250\) V (potential difference).

Substituting the values into the formula:

\[ Q = 80 \times 10^{-6} \times 250 \]\[ Q = 20 \times 10^{-3} \]\[ Q = 0.02 \, \text{C} \]

So, the charge acquired by the capacitor is \(0.02 \, \text{C}\).

Therefore, the correct option is:

0.02 C

In a non-magnetized state, the permeability of a material is close to the permeability of free space (\( \mu_0 \)), which is approximately \( 4\pi \times 10^{-7} \) H/m (henries per meter). This value represents the permeability of vacuum or air, where magnetic fields propagate freely without any influence from magnetic materials.

So, the permeability in a non-magnetized state is approximately.

The magnetism of a magnet can be destroyed by all of the methods mentioned:

1. Heating: Heating a magnet beyond its Curie temperature can disrupt the alignment of its magnetic domains, thus demagnetizing it.

2. Hammering: Repeated mechanical shock or hammering can also disrupt the alignment of magnetic domains, causing demagnetization.

3. Inductive action of another magnet: Placing a magnet near another strong magnet can sometimes cause demagnetization, especially if the magnetic fields oppose each other in a certain way, effectively canceling each other out.

So, the correct answer is "by all above methods."