Efficiency (η) = 1 - (T_cold / T_hot) = 1 - (300 K / 500 K) = 0.4 or 40%.

Efficiency (η) = 1 - (T_cold / T_hot) = 1 - (300 K / 500 K) = 0.4 or 40%.

Total resistance R = 8Ω + 8Ω = 16Ω. Current I = V/R = 24V / 16Ω = 1.5A.

First, find the equivalent resistance: 1/R_eq = 1/4 + 1/8 = 3/8, so R_eq = 8/3Ω. Then, I = V/R_eq = 24V / (8/3)Ω = 9A.

First, find the equivalent resistance: 1/R_eq = 1/5 + 1/10 = 3/10, R_eq = 10/3Ω. Then, I = V/R_eq = 20V / (10/3)Ω = 6A.

Using Ohm's law, I = V/R = 9V / 3Ω = 3A.

Power is given by P = V²/R = (9V)² / 3Ω = 81V² / 3Ω = 27W.

Power P = V²/R = (9V)² / 3Ω = 81 / 3 = 27W.

Power is given by P = I²R. If I is doubled, P becomes (2I)²R = 4I²R, which means power quadruples.

According to Ohm's law, I = V/R. If R is halved, I will double.

According to Ohm's law, I = V/R. If V is doubled, I also doubles.

In a damped harmonic oscillator, the amplitude decreases over time due to energy loss.

The central maximum has the maximum intensity because all light waves arrive in phase at this point, leading to constructive interference.

Fringe width (β) is given by β = λD/d. If λ is halved, β will also be halved.

Frequency (f) = (1/2π)√(k/m). If mass (m) is doubled, frequency decreases.

First, find the equivalent resistance: 1/R_eq = 1/6 + 1/3 = 1/2, so R_eq = 2Ω. Then, I = V/R_eq = 12V / 2Ω = 6A.

Using the formula 1/R_eq = 1/R1 + 1/R2, we get 1/R_eq = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, thus R_eq = 2Ω.

Y = 1/R + j(ωC - 1/ωL); calculate Y for given values.

Quality factor Q = (1/R) * √(L/C) = (1/50) * √(0.2/50e-6) = 2.

In a purely capacitive circuit, the current leads the voltage by 90°.

In a purely capacitive circuit, the current leads the voltage by 90 degrees.

f0 = 1 / (2π√(LC)) = 1 / (2π√(0.1 × 100 × 10^-6)) = 159.15 Hz.

Resonant frequency f₀ = 1 / (2π√(LC)) = 1 / (2π√(0.1 * 100e-6)) = 159.15 Hz.

If inductive reactance (XL) is greater than capacitive reactance (XC), the circuit is inductive.

Resonant frequency f = 1/(2π√(LC)) = 1/(2π√(0.1 * 100e-6)) = 159.15 Hz.

Resonant frequency f0 = 1/(2π√(LC)) = 1/(2π√(0.1*100e-6)) ≈ 159Hz.

For resonance in a series RLC circuit, the condition XL = XC must be met.

When light reflects off a denser medium, a phase change of π occurs, leading to destructive interference for certain path differences.

If destructive interference occurs for blue light, the color seen will be the complementary color, which is red.

At the center of the thin film interference pattern, the color that appears is typically green due to the constructive interference of wavelengths in that range.