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5.5 Power in RL Series Circuit

Question What do we mean by an RL circuit? When we refer to a circuit with R and L, it means that it can either be a circuit with these elements alone or one containing all three receiver components: C, R, and L. The latter implies that the voltage drop across the coil is greater than that across the capacitor, i.e., the inductive effect is more important than the capacitive effect. Both types of power appear in an R-L circuit.

We know that current delays φ degrees, so that:

u = 2·U·sin ωt (5.22)

i = 2·I·sin(ωt - φ) (5.23)

We have that the instantaneous power is:

p(t) = u(t)·i(t) = 2·U·I·sin ωt·sin(ωt - φ) = U·I·cos φ - U·I·cos(2ωt - φ) (5.24)

Recalling the trigonometric functions:

2 sin A cos B = sin(A + B) + sin(A - B)

2 cos A cos B = cos(A + B) + cos(A - B)

2 sin A sin B = cos(A - B) - cos(A + B)

We can build the power triangle of the power wave (Fig. 5.7). The mean value of power indicates the value of the active power:

P = 1/T ∫₀ᵀ p(t)·dt = U·I·cos φ (5.25)

Integrating this equation, we have two separate integrals:

P = 1/T ∫₀ᵀ [U·I·cos φ - U·I·cos(2ωt - φ)]·dt = 1/T ∫₀ᵀ U·I·cos φ dt - 1/T ∫₀ᵀ U·I·cos(2ωt - φ)·dt = I₁ - I₂

I₁ = 1/T ∫₀ᵀ U·I·cos φ dt = 1/T·U·I·cos φ·∫₀ᵀ dt = 1/T·U·I·cos φ·(T - 0)

I₂ = 1/T ∫₀ᵀ U·I·cos(2ωt - φ)·dt = 1/T·U·I·(1/2ω)·sin(2ωt - φ)|₀ᵀ = 1/T·U·I·(1/2ω)·[sin(2ωt - φ)]₀ᵀ = 1/T·U·I·(1/2ω)·[sin(2ωt - φ) - sin(0 - φ)] = 1/T·U·I·(1/2ω)·[sin(2ωt - φ) - sin(-φ)] = 0

We know that:

sin 2ωt = 0 → sin(2ωt - φ) = sin(-φ)

P = I₁ - I₂ = U·I·cos φ - 0

According to these equations, we can plot the vector diagram of the R-L series circuit (Fig. 5.6).

Diagram of an electrical circuit and a phasor diagram. The circuit on the left includes a voltage source labeled "U," a resistor labeled "R," and an inductor labeled "L." Arrows indicate the direction of current flow. On the right, the phasor diagram shows vectors for voltage and current, with components labeled VR and VL, and an angle φ between them.

Fig. 5.6 Vector diagram of a R-L Circuit.

A right triangle diagram illustrating the relationship between real power, reactive power, and apparent power in an electrical system. The horizontal side is labeled "Real Power," the vertical side is labeled "Reactive Power," and the hypotenuse is labeled "Apparent Power." An angle φ (phi) is shown between the real power and apparent power, indicating the phase difference.

Fig. 5.7 Power-triangle in a R-L circuit.

Applying Kirchhoff second law to the circuit, we have that Fig. 5.6.

Ũ = ŨR + ŨL (5.26)

We get the power triangle by multiplying the voltage and current in the vector diagram (Fig.

练习题

In a pure inductive circuit, the average active power is zero because the inductor only stores and releases energy without dissipating it. However, in an RL series circuit, the active power is given by $P = U \cdot I \cdot cos φ$ . What is the fundamental reason that an RL circuit has non-zero active power?

A. The resistor in the RL circuit dissipates energy as heat, while a pure inductor only exchanges energy with the source

B. The phase angle

φ

in an RL circuit is always

90°

, making

cos φ = 1

C. The current in an RL circuit is always larger than in a pure inductive circuit

D. The voltage across the inductor in an RL circuit is always zero

In both a pure inductive circuit and an RL series circuit, the instantaneous power oscillates at twice the frequency of the voltage and current, and the oscillating component $cos (2 ω t - φ)$ has an average value of zero over one complete cycle.

Which of the following statements correctly describe reactive power in AC circuits?

A. In a pure inductive circuit, reactive power is given by

Q_{L} = U_{L} \cdot I = X_{L} \cdot I^{2}

B. In a pure capacitive circuit, reactive power is given by

Q_{C} = U_{C} \cdot I = X_{C} \cdot I^{2}

C. The reactive power of a capacitor opposes that of a coil, and their effects can cancel each other

D. Reactive power is measured in watts (W)

E. In an RL circuit, reactive power appears as the vertical component of the power triangle

In a purely capacitive circuit, current ___ voltage by $π /2$ radians, whereas in an RL series circuit, current lags voltage by an angle $φ$ .

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5.5 Power in RL Series Circuit