Use the slider above to adjust the parameters of the ac power supply with emf $V_d = V_p \sin (\omega_d t + \phi_d)$. The subscript $d$ stands for "driving voltage".
$V_p$: amplitude, or the peak voltage.
$\omega_d$: angular frequency.
$\phi_d$: phase (usually set to zero in this course).
Notice the current and the voltage are in phase, rising and falling together.
Observe how the amplitude of the current does not change with the frequency.
Purely Capacitive AC Circuit
Purely Capacitive AC Circuit
$I = \omega_d C V_{peak} \sin(\omega_d t + \frac{\pi}{2})= \omega_d C V_{peak} \cos(\omega_d t )$, where $\omega_d = 2\pi f$.
Notice the current leads the voltage by $\frac{\pi}{2}$, i.e. $\phi = +\frac{\pi}{2}$ in $I= I_{peak} \sin(\omega_d t + \phi)$.
Observe how the amplitude of the current changes with the frequency and capacitance.
Purely Inductive AC Circuit
Purely Inductive AC Circuit
$I = \frac{V_{peak}}{\omega_d L} \sin(\omega_d t - \frac{\pi}{2}) + I_0= -\frac{V_{peak}}{\omega_d L} \cos(\omega_d t ) + I_0$, where $\omega_d = 2\pi f$.
Notice the current lags the voltage by $\frac{\pi}{2}$, i.e. $\phi = -\frac{\pi}{2}$ in $I= I_{peak} \sin(\omega_d t + \phi)$.
The $\pm$ signs on the inductor indicates the direction of the induced emf. Observe that the induced emf always opposes the emf of the ac power supply.
Observe how the amplitude of the current changes with the frequency and inductance.
The constant current $I_0$ is usually not included in most textbook descriptions, but is in theory present because this circuit has no resistance (by assumption) so any constant current (i.e. direct current) can flow in the background forever.
You can generate a non-zero $I_0$ (which shifts the current curve vertically) if you change any of the setting in the middle of the simulation. To remove the effect of $I_0$ simply press the reset button.
RCL AC Circuit
RCL AC Circuit
Observe how the amplitude of the current changes with different settings.
Observe the phase difference $\phi$ between the voltage and the current. See that $V$ and $I$ are not always "in phase" except at resonance.
The voltage curves that are plotted are $V_{total} = V_{ac generator}$, $V_R = IR$, $V_C= \frac{q}{C}$, and $V_L = L \frac{dI}{dt}$. Note the absence of the negtaive sign in $V_L$, we choose this sign convention here so it is consistent with $V_C$, which is positive when $q>0$ and opposing the power supply. With this convention, we have $V_{total} = V_R + V_C + V_L$.
Theory:
$L\frac{dI}{dt} + R I + \frac{q}{C} = V_d(t)$, where $V_d(t) = V_{d, peak} \sin(\omega_d t)$ is the driving voltage from the a.c. power supply.
Resonance when $\omega_d = \frac{1}{\sqrt{LC}}$.
$I(t) = \frac{V_{d, peak}}{Z} \sin (\omega_d t + \phi)$.
Adjust the frequency to match with the resonant frequency. Compare with the prediction $I_{peak} = V_{peak}/R$ and $\phi = 0$ (i.e. no phase difference between $V$ and $I$, they are "in phase").
Observe how $V_C$ and $V_L$ cancel each other at resonance. At other frequencies they always have the opposite signs.
Choose a large frequency, observe that the current lags the voltage by $\frac{\pi}{2}$, just like a purely inductive circuit in an earlier simulation.
Choose a low frequency, observe that the current leads the voltage by $\frac{\pi}{2}$, just like a purely capacitive circuit in an earlier simulation.
Show and hide different voltage curves to see which component dominates at different frequencies.
See how many properties of this circuit you can quantitatively or quanlitatively verify.
LC Circuit
LC Circuit
Observe how the period and frequency of the current and voltages change with different settings.
Observe the phase difference $\phi$ between the voltage and the current. See that $V$ and $I$ are not "in phase".
We choose our sign convention here so that $V_C + V_L = 0$.
$V_C$ is reset to $2V$ whenever the reset button is pressed.
Theory:
$L\frac{dI}{dt} + \frac{q}{C} = 0$.
Resonant frequency is $\omega_0 = \frac{1}{\sqrt{LC}}$.