In a capacitive circuit the voltage lags current by 90 degrees. In an inductive circuit voltage leads current by 90 degrees. In a resistive circuit voltage and current are in phase. If we have a circuit with an AC voltage source, a capacitor, an inductor and a resistor, at what point does the phase change happen? Is it just at the particular component? If so is there 0 phase difference from one component to the next?

  • \$\begingroup\$ It depends on the configuration \$\endgroup\$
    – BeB00
    Apr 22, 2018 at 1:10
  • \$\begingroup\$ Lets say all components are in series. \$\endgroup\$ Apr 22, 2018 at 1:44
  • \$\begingroup\$ tinyurl.com/ya9gogf9 \$\endgroup\$
    – BeB00
    Apr 22, 2018 at 1:49
  • \$\begingroup\$ That shows the phase difference at each component. How about between the components? \$\endgroup\$ Apr 22, 2018 at 3:18
  • \$\begingroup\$ There is no voltage drop across an ideal conductor. \$\endgroup\$
    – Chu
    Apr 22, 2018 at 8:20

2 Answers 2


The voltage/current phase relationship is true for each component individually.

In a series RLC circuit, they all share current. This means the total observed terminal voltage is the sum of all the individual voltages. The inductor voltage is always opposed to the capacitor voltage, they are always 180 degrees apart from each other.

At low frequencies, where the capacitor voltage is high and so dominates the total, and the voltage will lag the current. At high frequencies, where the inductor voltage is high, the current will lag the voltage.

At one specific intermediate frequency where the magnitude of the capacitor voltage and the inductor voltage are equal, they will cancel each other out. Voltage and current will now be in phase, voltage will be a minimum, and the circuit is said to be in resonance.

You can go through a similar line of reasoning for parallel connected components. These share the same voltage, and so the currents through the individual components add up to make the total. At resonance the total current is now a minimum.


Lets say all components are in series

The series impedance is: -

$$R+j\omega L+ \dfrac{1}{j\omega C}$$

At low frequencies, the capacitor impedance dominates so the impedance tends to be: -

$$\dfrac{1}{j\omega C}$$

The "+j" in the denominator is the same as "-j" in the numerator and mathematically "-j" (in an impedance equation) implies a 90 degrees phase shift of current relative to voltage.

At high frequencies, the impedance of the inductor dominates and so its "+j" in the numerator implies the current lags voltage by 90 degrees.

At resonance: -

$$j\omega L + \dfrac{1}{j\omega C} = 0$$ i.e. the two impedances cancel and this leads to: -

$$j^2\omega^2 LC = -1$$

And, because \$j^2\$ = -1 we have the resonance frequency: -

$$\omega = \frac{1}{\sqrt{LC}}$$

At this frequency the impedance is purely dictated by the resistor and the phase shift is zero.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.