Understanding Voltage and Current Phase Difference

Question

I am having trouble understanding the concept of voltage and current having a phase difference in AC circuits. I have recently begun exploring AC circuits and have just started delving into the technical details. While AS Further Mathematics has helped me keep up with the mathematics, I'm struggling to grasp the implications of current leading or lagging to voltage.

Initially, we learn that voltage drives current (through a resistor), but I recently encountered a perspective suggesting the relationship is arbitrary - that voltage could alternatively be viewed as a consequence of current. However, the fundamental relationship maintains a direct correspondence between the two, which is straightforward to understand when it comes to a load.

In AC circuits, Ohm's Law appears to hold until reactive components are introduced. My confusion arises when considering how there can be a positive potential difference while current flows in the opposite direction. How does this phase difference affect a load's behavior, and what are the implications for power?

Furthermore, how would these phase-shifted signals interact with more complex components like op-amps and transistors?

Ohm's Law holds with reactive components, too. You just need to move from domain \$\mathbb{R}\$ to domain \$\mathbb{C}\$. For non-linear (often \$\ln\$ or \$\exp\$ or \$x^n\$ in nature), it's done by linearizing around an operating point, adjusting time by \$\text{d}t\$, moving along that instantaneous linear slope to a new operating point, then repeating over and over. Closed solutions to non-linear equations are hard. A few are amenable to the product-log (LambertW) function. But that's in just a few cases. Also, graph theory applies. — periblepsis
– periblepsis, Commented May 10 at 10:23
It's important to be aware that Ohm's law isn't a general law like most things called "laws" in physics. It applies specifically and only to ohmic (defined somewhat circularly as "things that follow Ohm's law", but it's a category that includes most metals and bulk semiconductors) resistive elements, and can be extended to apply to linear reactive elements as well. It ceases to apply when nonlinear elements, such as semiconductor junctions or saturating magnetics, get involved (though there are ways to use it as an approximation in limited cases). — Hearth
– Hearth, Commented May 10 at 14:27
What you learned is true about the current in the resistor (voltage drives current through a resistor). But in the AC circuit, we have two additional components. The capacitor and the inductor. For example, when a capacitor is fully charged, no current is flowing through the capacitor (V_cap = max, I_cap = 0A). On the after hand, if we connect an empty capacitor (V_cap = 0V) to a voltage source, the maximum current will flow (I_cap = max ). As you can see, we have a phase shift. electronics.stackexchange.com/questions/287394/… — G36
– G36, Commented May 10 at 16:02

Justme · Accepted Answer · 2025-05-10 10:46:46Z

If you think of a square wave voltage driving a resistance, you find out current and voltage have no phase offset as they are directly proportional based on Ohm's law like you say.

Now think of the same square wave charging and discharging a capacitor. There is no current when square wave is high or low, but during the edge of the transition, current must flow to move charge in order to change the capacitor voltage. So you see the curret is based on the derivative of the voltage.

When the driving signal is a sine wave at some frequency, you can have an RC filter where the driving sine signal is charging the capacitor through a resistance. In a fixed frequency case you can simply convert the capacitance to some impedance at this frequency.

As you may see, the current through the resistor depends on the voltage difference over the resistor, where on one side is the driving signal voltage and on the other side is the capacitor voltage.

And as you see it always taks time to charge the capacitor through the resistor so the capacitor voltage lags in respect to the driving signal. So safe to say, compared to the driving sine wave, RC filter output has same frequency, but lower amplitude, and it lags in phase, because after the peak of the driving sine wave, the voltage is still higher than the capacitor voltage, and the capacitor does not start to discharge until driving sine wave voltage goes below the capacitor voltage. So the driving current also is not in phase with the driving voltage.

So Ohm's law also works for any complex impedance, resistors just have purely real impedance and capacitors have purely imaginary impedance.

Power lost in the resistor is obviously voltage over the resistor times the current through the resistor. The capacitor does not dissipate power, it just charges up and stores energy until dicharged. However the current used to charge and discharge the capacitor still flows in the wires and for example electric companies need to have big enough wires to allow useless reactive load currents to flow in addition to the useful loads that actually use the energy.

To op-amps and transistors it will just see the instantaneous output of the RC filter for example. They do not know or care it has gone through some circuit that shifts phase.

I think the questioner is more mathy than electronic in view. Probably wants the math. For example, the complex domain version of Ohm's Law? That said, I really like your answer, too. So I don't care, +1 anyway. — periblepsis
– periblepsis, Commented May 10 at 11:00
What if you took a purely resistive load (lets say its a heater), and the current going through was leading the voltage. there will be a specific instant where the voltage is positive but the current is negative. So the power comes out to be negative. how does the heater behave at that instant where power is negative? technically since there is a current flowing through, it should generate heat. However, the negative sign begs to differ. — VALKYRIEiSr
– VALKYRIEiSr, Commented May 11 at 7:00
@VALKYRIEiSr Such a scenario cannot happen, because in a purely resistive element, current and voltage are always in phase as there is no reactive elements that could shift the phase. — Justme
– Justme, Commented May 11 at 8:06
@Justme But if this was an RC circuit, where R would be the load. Althought I understand you wouldnt be driving something like a Heater through a RC circuit, this is a purely hypothetical scenario, because I dont understand how the load would respond in such a case. — VALKYRIEiSr
– VALKYRIEiSr, Commented May 11 at 8:51
@VALKYRIEiSr Exactly - there is an RC circuit, but all resistors simply obey Ohm's law, current is defined only by resistance and voltage over the resistor. But the capacitor defines the voltage over resistor in an RC series circuit, because the capacitor voltage changes based on the charge it stores, and how the voltage changes depends on the current, and the current in turn is defined by the resistance, but anyway the phase shift in the resistor is caused by capacitor and the phase of resistor voltage is shifted compared to source voltage, but current and voltage in resistor has no shift. — Justme
– Justme, Commented May 11 at 10:03

Hearth · Accepted Answer · 2025-05-10 15:07:43Z

Ohm's law can be extended to apply to linear reactive components by bringing complex numbers into it. In engineering, this is usually done via the Laplace transform (or sometimes phasor analysis, which is easier to use when dealing with a single frequency; this answer will use the Laplace transform however):

If you take the Laplace transform of your circuit,

resistors remain as they are, an impedance of \$Z_R=R\$,
capacitors become an impedance of \$Z_C=\frac{1}{sC}\$,
and inductors become an impedance of \$Z_L=sL\$.

All of these follow Ohm's law in exactly the same \$V=IZ\$ form, but the inclusion of the Laplace variable \$s\$ results in a phase shift. If you're familiar with the Laplace domain, you probably know that \$s\$ is the differentiation operator and \$\frac{1}{s}\$ is the integration operator; both of these operators shift a sinusoid's phase by 90° in opposite directions.

This only applies to linear inductors and capacitors; nonlinear inductors include any non-air-core inductors, while nonlinear capacitors include any class-II ceramic capacitors. These can both be approximated as linear for small signals, but become increasingly nonlinear as signal amplitude increases.

As for op amps and transistors, these are more complicated. Op amps, provided they aren't driven outside their normal operating range, can be simplified down to ideal gain blocks with feedback defining their gain, but real op amps have nonidealities that limit this analysis.

Transistors are even worse—they are very nonlinear and can't generally be approximated as simple gain blocks. For small-signal, medium-frequency analysis, there are certain approximations that can be used (look into the hybrid-pi model, for instance), but a complete analysis necessarily includes nonlinear elements, and gets very messy. Simple full models include the Ebers-Moll model for BJTs and the Shichman-Hodges model for FETs, but the very small transistors used in modern devices tend to be poorly modelled by these, requiring very complex models like BSIM instead.

Note that Ohm's law is not as general a law as most things called "laws" in physics. It only applies to ohmic materials (which are, perhaps somewhat circularly, defined as materials for which Ohm's law applies), a category that includes most conductors and bulk semiconductors, but which does not include semiconductor junctions, metal-semiconductor Schottky junctions (it's possible for a metal-semiconductor junction to be ohmic, however, depending on conditions beyond the scope of this answer), or insulators undergoing any form of breakdown (including electric arcs). The key parameter of resistance also is not constant for a given material; it varies with temperature in all cases (though the degree differs between materials) and in the special case of bulk semiconductors it varies significantly with illumination as well. In some special cases, it can also vary with magnetic field as well.

I know this is a little off topic: The reason I brought up op amps is because I have been trying to wrap my head around the effects of out-of-phase voltage & current in transimpedance amplifiers. I have been trying to intuitively understand this. The current from the source creates an out-of-phase voltage across the feedback resistor Rf which the non-inverting input identifies and the output is adjust accordingly. The output would just be identical to the input(amplified and inverted however the voltage-current phase difference remains the same). Then why are there issues such as oscillations. — VALKYRIEiSr
– VALKYRIEiSr, Commented May 11 at 6:48
@VALKYRIEiSr Oscillations are due to feedback. If you want more information, I suggest asking a separate question--as you said, a little off topic for here. There's no rule against asking more questions! — Hearth
– Hearth, Commented May 11 at 13:22

Circuit fantasist · Accepted Answer · 2025-05-13 12:53:43Z

I am having trouble understanding the concept of voltage and current having a phase difference in AC circuits.

Since you want to understand the idea of phase shift, I offer you an intuitive approach to explanation illustrated with multiple CircuitLab simulations.

Current-supplied circuit

... voltage could alternatively be viewed as a consequence of current.

Some devices, for example resistors, can be controlled by both voltage and current.

Resistor

If you drive a resistor with an AC current source...

schematic

^{simulate this circuit – Schematic created using CircuitLab}

... the voltage across it instantaneously follows the current (there is no phase difference) because there is no accumulation of charge.

Capacitor

Other devices, like capacitor, are more conveniently (without conflict) controlled by current.

schematic

^{simulate this circuit}

As you can see from the graphs below, the voltage waveform is shifted exactly 90° to the right (or, as it is commonly said, the current leads the voltage by 90°). This can be explained as follows:

Since the current source is ideal, the moment the current changes its direction does not depend on the voltage across the capacitor, and consequently, on the frequency of the input signal.

Voltage-supplied circuit

Current sources are most easily made using a voltage source Vin and a resistor R1 in series (see Schematic 2.1.1 below). However, these sources are imperfect because the maximum voltage across the load R2 they can create (the so-called compliance voltage) is equal to their internal voltage.

AC circuit operation

Resistor: So, if you drive a resistor with such a resistor-type AC current source...

schematic

^{simulate this circuit}

... the voltage across resistor R2...

... and the current through it replicate Vin without any phase shift.

Capacitor: But if we replace the resistor with a capacitor, an interesting effect occurs – now the current is determined not only by Vin but by the difference Vin - Vc.

schematic