Solution:

Measuring instruments do not measure instantaneous or maximum values of current and voltage, but the "effective values" of current and voltage.

Power in AC Circuits

1. Power in an AC Circuit with a Resistor

$$P = R I_m^2 \sin^2(\omega t), \quad [P] = \text{W}$$

2. Power in an AC Circuit with Impedance

• Active (real) power

  $$P = U I \cos \varphi, \quad [P] = \text{W}$$

• Apparent power

  $$S = U I, \quad [S] = \text{VA}$$

• Reactive power

  $$Q = U I \sin \varphi, \quad [Q] = \text{VAr}$$

where $U, I$ are RMS values.

3. Relationship Between AC Powers

$$S^2 = P^2 + Q^2$$

4. Power Factor

$$\cos \varphi = \frac{P}{U I}, \quad 0 < \varphi < \frac{\pi}{2}$$

5. Efficiency of an Electric Motor

$$\eta = \frac{P_2}{P_1} \cdot 100\%$$

where:

• $P_1 =$ input active power (electric power)

• $P_2 =$ output power (mechanical power of motor)

 This is the standard AC power triangle:

• $P$ = real power (W)

• $Q$ = reactive power (VAr)

• $S$ = apparent power (VA)

Solution:

Analysis:

R = 20Ώ,  U_ef = 24V  f = 50s^-1,  I_ef = ?

• The equation for the instantaneous value of current in the circuit is i = 1.7.sin(100π.t) [A]
• The effective value of the current in the circuit is I_ef = 1.2A.

Solution:

The instantaneous voltage equals the effective voltage at t₁ =T/8 and t₂ = 3T/8.

Solution:

Analysis:

U_ef = 156V,  f = 50Hz = 50s^–1,  U_m = ?,  t = ?

The instantaneous value of the AC voltage reaches the effective value after t = 2.5ms.

Solution:

Solution:

Analysis:

• If a coil is connected in the circuit and the frequency increases, the active power decreases (f is in the denominator).
• If a capacitor is connected in the circuit and the frequency increases, the active power also increases (f is in the numerator).

Solution:

Analysis:

P = 5000W,  U = 220V,  cosφ = 0.85,  η = 0.8

The current flowing through the single-phase motor is I = 33.4A

Solution:

Analysis:

U = 48V,  I = 6A,  P₁ = 200W,  φ₁ = ?, cos φ₂ = 0.98

The active power increases 1.41 times (by 41%). 

Solution:

Analysis:

U = 3000V,  f = 50Hz,  R = 40Ώ,  L = 95·10^-3H,  I = ?, φ = ?,  P = ?

The circuit current is I = 60A, the phase shift φ = 36°53‘, and the active power is 1.44·10⁵W.

Solution:

Analysis:

t = 5 min = 300s,  I = 10A,  U = 220V,  W = 0.125 kWh (1kWh = 3.6·10⁶J) W = 0.125·3.6·10⁶J = 0.45·10⁶J

The motor power factor is cos φ = 0.68.