The energy released in a nuclear reaction is referred to as nuclear energy, which is a result of changes in the binding energy of the nucleus.

The mass defect is the difference in mass between the reactants and products in a nuclear reaction, which is related to the binding energy.

When a p-n junction diode is forward biased, the depletion region narrows, allowing current to flow easily through the junction.

When a p-n junction diode is reverse-biased, the depletion region widens, preventing current flow.

An electric field is formed at the p-n junction due to the diffusion of charge carriers.

The depletion region is the area around the p-n junction where charge carriers are depleted, creating an electric field.

The depletion region is the area around the p-n junction where charge carriers have recombined, leaving behind immobile ions.

The equivalent resistance for resistors in parallel is given by 1/R_eq = 1/R1 + 1/R2 + 1/R3. Thus, 1/R_eq = 1/2 + 1/3 + 1/6 = 1. Therefore, R_eq = 1Ω.

Using the formula for resistors in parallel, 1/R_eq = 1/R1 + 1/R2 + 1/R3, we find R_eq = 1 / (1/6 + 1/12 + 1/18) = 3 ohms.

1/R_eq = 1/R1 + 1/R2 + 1/R3 = 1/2 + 1/3 + 1/6 = 1. Therefore, R_eq = 3Ω.

The formula for equivalent resistance in parallel is 1/R_eq = 1/R1 + 1/R2 + 1/R3. Thus, 1/R_eq = 1/6 + 1/3 + 1/2 = 1/6 + 2/6 + 3/6 = 6/6 = 1, so R_eq = 1 ohm.

The equivalent resistance in parallel is given by 1/R_total = 1/R1 + 1/R2 + 1/R3. Thus, 1/R_total = 1/2 + 1/3 + 1/6 = 1Ω.

For parallel resistors, 1/R_eq = 1/R1 + 1/R2. Thus, 1/R_eq = 1/4 + 1/6 = 5/12, so R_eq = 12/5 = 2.4 ohms.

First, find the equivalent resistance: 1/R_eq = 1/4 + 1/6 => R_eq = 2.4 ohms. Then, I = V/R_eq = 12V / 2.4Ω = 5 A.

First, find the equivalent resistance: 1/R_eq = 1/4 + 1/6 => R_eq = 2.4 ohms. Then, I = V/R_eq = 12 V / 2.4 ohms = 5 A.

First, find the equivalent resistance: 1/R_eq = 1/4 + 1/6 => R_eq = 2.4Ω. Then, I = V/R_eq = 12V / 2.4Ω = 5A.

Using the formula for resistors in parallel, 1/R_eq = 1/R1 + 1/R2, we have 1/R_eq = 1/6 + 1/3 = 1/6 + 2/6 = 3/6. Therefore, R_eq = 6/3 = 2 ohms.

1/R_eq = 1/R1 + 1/R2 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, thus R_eq = 2Ω.

In a parallel circuit, if one resistor fails, the total current decreases because the total resistance increases.

In a parallel circuit, removing a resistor decreases the total resistance.

In a parallel combination, the total capacitance is simply the sum of the individual capacitances: C_eq = C1 + C2 + C3.

Capacitance is directly proportional to the area of the plates. Doubling the area will double the capacitance.

The electric potential V is directly proportional to the distance d between the plates, so if d is doubled, V also doubles.

Capacitance C = ε₀ * A / d. If d is halved, C doubles.

In a parallel RLC circuit, increasing frequency generally increases the total current due to lower reactance.

Increasing the intensity of light increases the number of photons hitting the surface, which in turn increases the number of emitted electrons.

If the frequency of light is just above the threshold frequency, the emitted electrons will have zero kinetic energy, as all the energy is used to overcome the work function.

Increasing the stopping potential indicates that the emitted electrons have higher kinetic energy, as more energy is required to stop them.

The stopping potential does not affect the energy of the emitted electrons; it only affects the number of electrons reaching the detector.

Increasing the stopping potential increases the maximum kinetic energy of the emitted electrons, as the stopping potential is directly related to the kinetic energy of the electrons.