I = V / R = 240V / 60Ω = 4A.

According to Ohm's law, if voltage is halved and resistance remains constant, the current also halves.

According to Boyle's Law, at constant temperature, pressure is inversely proportional to volume. Therefore, if the volume is doubled, the pressure is halved.

According to Boyle's Law, if the volume is halved, the pressure doubles at constant temperature.

According to Boyle's Law, if the volume is halved, the pressure doubles.

The angular separation (θ) is directly proportional to the wavelength (λ). Therefore, increasing λ increases the angular separation.

Fringe separation is directly proportional to the wavelength. If the wavelength increases, the fringe separation also increases.

The position of the first minimum in single-slit diffraction is given by a sin(θ) = λ. Increasing λ will cause θ to increase, moving the minimum away from the center.

At the midpoint, the electric fields due to +Q and -Q cancel each other out, resulting in a net electric field of 0.

C_eq = 1 / (1/C1 + 1/C2) = 1 / (1/4 + 1/4) = 2 µF.

In series, R_eq = R1 + R2 = 4Ω + 6Ω = 10Ω.

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.

1/R_eq = 1/4 + 1/8 = 2/8 + 1/8 = 3/8, so R_eq = 8/3Ω ≈ 2.67Ω.

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