How do you analyse a series RL circuit: impedance, phase angle and power factor?

In one line

In a series RL circuit the impedance is \( Z = \sqrt{R^2 + X_L^2} \), the current lags the voltage by \( \phi = \tan^{-1}(X_L/R) \), and the power factor is \( \cos\phi = R/Z \) lagging.

Explained simply

Most real loads are like this: a fan or motor coil has both resistance (wire heats up) and inductance (the coil's magnetic field). The resistor's voltage is in step with the current, while the inductor's voltage is 90° ahead. Because they are 90° apart, they add like the sides of a right triangle, not like ordinary numbers.

Technically, in series the same current \( I \) flows through both. \( V_R = IR \) is in phase with \( I \); \( V_L = IX_L \) leads \( I \) by 90°. The supply voltage is their phasor sum.

How it works

1. Take current \( I \) as the reference phasor (horizontal).
2. Draw \( V_R \) along \( I \) and \( V_L \) vertically up (leading by 90°).
3. Supply \( V = \sqrt{V_R^2 + V_L^2} \); dividing each side by \( I \) gives the impedance triangle \( Z, R, X_L \).
4. In complex form \( Z = R + jX_L \); current \( \bar I = \bar V/\bar Z \).
5. Power: \( P = VI\cos\phi = I^2R \) (only the resistor consumes real power).

Key formula

\( Z = \sqrt{R^2 + X_L^2}, \quad \phi = \tan^{-1}\dfrac{X_L}{R}, \quad \cos\phi = \dfrac{R}{Z}, \quad P = VI\cos\phi \)

\( Z \) = impedance (\( \Omega \)), \( R \) = resistance (\( \Omega \)), \( X_L = 2\pi fL \) = inductive reactance (\( \Omega \)), \( \phi \) = phase angle by which current lags, \( P \) = real power (W), \( V, I \) = RMS values.

Quick example

\( R = 30\,\Omega \), \( X_L = 40\,\Omega \), 230 V supply. \( Z = 50\,\Omega \), \( I = 4.6 \) A, \( \cos\phi = 30/50 = 0.6 \) lagging (\( \phi = 53.13^\circ \)). \( P = 230 \times 4.6 \times 0.6 = 634.8 \) W, which equals \( I^2R = 4.6^2 \times 30 \). \( V_R = 138 \) V and \( V_L = 184 \) V add to more than 230 V arithmetically, but \( \sqrt{138^2+184^2} = 230 \) V.

Common mistakes

• Adding \( V_R \) and \( V_L \) arithmetically.
• Forgetting to write 'lagging' with the power factor.
• Using \( P = VI \) instead of \( VI\cos\phi \).

How MU asks it

Very common 7–8 mark numerical: "A coil of resistance R and inductance L is connected to 230 V, 50 Hz. Find impedance, current, power factor and power", often with a phasor diagram. For full marks: compute \( X_L \) first, show the impedance triangle, give the power factor with 'lagging', draw the phasor diagram with \( I \) as reference, and quote units for every answer.