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 = 12 V / 2.4 ohms = 5 A.

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.

Total current I = V/R_eq. First, find R_eq = 1/(1/4 + 1/12) = 3Ω. Then, I = 24V / 3Ω = 8A.

First, find the equivalent resistance: 1/R_eq = 1/4 + 1/12 = 3/12 + 1/12 = 4/12, R_eq = 3Ω. Then, I = V/R_eq = 24V / 3Ω = 8A.

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.

The total resistance in parallel is given by 1/R_total = 1/R1 + 1/R2. Thus, 1/R_total = 1/6 + 1/3 = 1/2, so R_total = 2 ohms.

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.

First, find the equivalent resistance: 1/R_eq = 1/8 + 1/4 = 1/8 + 2/8 = 3/8, so R_eq = 8/3Ω. Then, I = V/R_eq = 16V / (8/3)Ω = 16 * 3/8 = 6A.

1/R_eq = 1/R1 + 1/R2 = 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.

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.

The capacitance (C) of a parallel plate capacitor is inversely proportional to the distance (d) between the plates. If d increases, C decreases.

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.

Total admittance Y = 1/R + j(ωC - 1/ωL). Calculate Y using given values.

At resonance, total current I = V/R. Calculate using V = 10V.

Quality factor Q = (1/R) * √(L/C) = (1/10) * √(0.2/50e-6) ≈ 3.

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

In a parallelogram, opposite angles are equal. Therefore, if one angle is 70 degrees, the opposite angle is also 70 degrees.

In a parallelogram, opposite angles are equal, so the opposite angle also measures 60 degrees.