Effect on phase angle due to resistance

Question

When an ac voltage source is applied to a resistor, current and voltage are in phase in the circuit.

When an ac voltage is applied to a pure inductor, voltage leads the current in the circuit by 90 degree.

But

When we apply ac voltage to a circuit where both resistor and inductor are in series, voltage leads the current not by 90 degree but by another angle.

My question is that when resistor is alone in circuit it doesnt effect the phase angle, then how can it causes change in phase angle of the overall current when it is with an inductor?

Answer 1

Consider this diagram where \$|R| = |X_L|\$: -

The following has to be true: -

• The voltage across the resistor (red arrow) and the voltage across the inductor (blue arrow) have to spatially add-up to equal the supply voltage (black arrow). That addition is shown by the dotted lines.

• The red and blue voltage arrows have to be 90 degrees apart because both inductor (blue) and resistor (red) share the same current.

• The current through both components (orange arrow) has to be 90 degrees lagging the inductor voltage (blue) AND simultaneously in phase with resistor voltage (red).

When |R| = |\$X_L\$|, there can be no other diagram that complies with the above requirements.

If L dominates or R dominates you get these scenarios: -

If you take these two scenarios to the extremes, you should see that: -

• Current lags supply voltage by 90 degrees (no resistance)
• Current becomes in phase with supply voltage (no inductance)

If we set the combined series impedance magnitude of R and L, \$\sqrt{R^2 +X_L^2}\$ to be a constant, we will see that the trajectories for \$V_R\$ and \$V_L\$ follow circular paths: -

For the scenarios above, the current will have a constant amplitude and be in phase with \$V_R\$.

In case you didn't realize, these are called phasor diagrams and can be re-arranged like so (for example): -

This should allow you to see why the impedance magnitude of a series inductor and resistor (same current) is \$\sqrt{R^2+X_L^2}\$. This is because the angle inside a semi-circle is always 90°.

Answer 2

There isn't a great answer to this question. But I will point out that the voltage across an inductor and the current through the inductor are ALWAYS 90 degrees out of phase. That is the rule that the inductor enforces.

Likewise, the voltage across a resistor and the current through the resistor are always in phase. The resistor enforces this condition by its nature, just like the inductor.

If you lump them together in series this is still true. What happens is, the supply ends up supplying voltage and current which are not in phase when measured at the supply. But the angle between voltage and current at the resistor will still be 0 degrees, and 90 degrees measured at the inductor.

Each element in the circuit has its rule that it tries to enforce. Fortunately, the supply is flexible about phase angle. It only enforces voltage, and the phase angle of the current at the supply is free to be whatever it needs to be to make the rest of the circuit happy.

Answer 3

You have to use superposition. So you get a resulting direction (angle). Imagine the following metaphors:

resistor: go straight, inductance: go right, capacity: go left,

Resistor + inductance: go straight + go right = diagonal, Resistor + inductance + capacity: go straight + go right + go left = go straight (for inductance = capacity)

Answer 4

It is better to explain that by calculating it. Let's consider the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

The transfer function is given by:

$$\frac{V_{out}}{V_{in}}=\frac{Ls}{R+Ls}$$

$$\frac{I_{in}}{V_{in}}=\frac{1}{R+Ls}$$

The voltage phase is given by:

$$\angle \frac{V_{out}}{V_{in}}=\arctan(\frac{\omega}{0})-\arctan(\frac{L\omega}{R})$$

The current phase is given by:

$$\angle \frac{I_{in}}{V_{in}}=\arctan(\frac{0}{1})-\arctan(\frac{L\omega}{R})$$

Taking the difference:

$$\angle \frac{V_{out}}{V_{in}} - \angle \frac{I_{in}}{V_{in}}=\arctan(\frac{\omega}{0})-\arctan(\frac{L\omega}{R}) - \arctan(\frac{0}{1})+\arctan(\frac{L\omega}{R})$$

$$\angle \frac{V_{out}}{V_{in}} - \angle \frac{I_{in}}{V_{in}}=\arctan(\frac{\omega}{0}) - \arctan(\frac{0}{1}) = 90°C$$

As you can see, the resistor does not play a role in the phase angle difference.