RL Series Circuit

Electrical Circuit Analysis > Series and Parallel ac Circuits > AC Series Configuration

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The behavior of an RL series circuit can be described using various electrical properties such as resistance, inductance and impedance. Understanding these properties is essential for analyzing and designing circuits in electrical engineering.

In an RL series circuit, the resistor and the inductor are connected in series, i.e. the current flowing through the circuit passes through both components one after the other. The induction coil in the circuit generates a magnetic field that stores energy in the form of an electromagnetic field. The resistor, on the other hand, resists the current flow and dissipates the energy stored in the circuit. Lets explore the RL series circuit in the following example.

Phasor notations:

Refer to [Fig. 1].

$$e = 141.1 \sin wt \Rightarrow E = 100V \angle 0^\circ$$

Fig. 1: RL series circuit configuration.

Fig. 2: Applying phasor notation to the RL series circuit configuration of Fig. 1.

where

$$ Z_T =Z_1 + Z_2 \\ =3 \angle 0^\circ + 4 \angle 90^\circ\\ =3Ω + j4\\ Z_T = 5 \angle 53.13^\circ $$

Impedance diagram

See [Fig. 3].

Fig. 3: Impedance diagram for the RL series circuit configuration of Fig. 1.

Current:

$$I = {V \over Z_T} = {100 V \angle 0^\circ \over 5 \angle 53.13^\circ} = 20A \angle -53.13^\circ$$

Ohms laws:

$$ \begin{split} V_R &= I Z_R = (20A \angle -53.13^\circ)(3Ω \angle 0^\circ)\\ &=(60V \angle -53.13^\circ)\\ V_L &= I Z_L = (20A \angle -53.13^\circ)(4Ω \angle 90^\circ)\\ &=(80V \angle 36.87^\circ)\\ \end{split}$$

Kirchhoff's voltage law:

$$\begin{split} \sum V &= E - V_R -V_L = 0\\ E &= V_R +V_L\\ &= 60V \angle -53.13^\circ + 80V \angle 36.87^\circ\\ &= (36 -j48)+ (64 + j48)\\ &= 100V + j0 = 100V \angle 0^\circ \\ \end{split}$$

as applied.

Phasor diagram:

Note that for the phasor diagram of [Fig. 4], I is in phase with the voltage across the resistor and lags the voltage across the inductor by $90^\circ$.

Fig. 4: Phasor diagram for the RL series circuit configuration of Fig. 1.

Power:

The total power in watts delivered to the circuit is

$$ \begin{split} P_T &= EI \cos \theta_T\\ &= (100 V)(20V) \cos 53.13^\circ\\ &= (2000V) (0.6)\\ \end{split}$$

where E and I are effective values and $\theta_T$ is the phase angle between E and I, or

$$ \begin{split} P_T &= I^2 R \\ &= (20 V)^2(3 Ω)=(400) (3) \\ &= 1200W\\ \end{split}$$

where I is the effective value, or, finally

$$ \begin{split} P_T &= P_R + P_L\\ &= V_RI \cos \theta + V_LI cos \theta\\ &= (60)(20) \cos 0^\circ + (80)(20) \cos 90^\circ \\ &= 1200 W +0 = 1200W\\ \end{split}$$

Power factor:

The power factor $F_p$ of the circuit is $\cos 53.13^\circ =0.6$ lagging, where $53.13^\circ$ is the phase angle between E and I.
If we write the basic power equation $P = EI \cos \theta$ as follows:

$$\cos \theta = {P \over EI}$$

where E and I are the input quantities and P is the power delivered to the network, and then perform the following substitutions from the basic series ac circuit:

$$ \begin{split} \cos \theta &= {P \over EI} = {I^2 R \over EI} \\ &={I R \over E} = {R \over E/I} = {R \over Z_T}\\ \end{split}$$

we find

$$\bbox[10px,border:1px solid grey]{F_p = \cos \theta = {R \over Z_T}}$$

Further supporting the fact that the impedance angle $\theta$ is also the phase angle between the input voltage and current for a series ac circuit. To determine the power factor, it is necessary only to form the ratio of the total resistance to the magnitude of the input impedance. For the case at hand,

$$F_p = \cos \theta = {R \over Z_T} = {3Ω \over 5Ω} =0.6 \text{lagging}$$

as found above.

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