In a capacitor circuit,the there is 90 degree phase shift between current and voltage in capacitor(when supplied sinusoidalvoltage and it varies from 0 to 90 degrees in a RC circuit.Can somebody tell me why exactly there is a phase difference between 0 to 90 degrees in a RC circuit while a purely capacitive circuit has a phase difference of 90 degrees
[Physics] Why is there a phase difference in RC circuit
capacitanceelectric-circuitselectric-currentelectrical-resistancevoltage
Related Solutions
No, a phase difference of 180º between applied voltage and current is not possible in a standard LCR circuit.
However, for reference, it is possible to construct electronic circuits that mimic negative resistance or negative impedance. In the case of negative resistance, $R = \dfrac{-V}{I}$. Here is a wikipedia article about it.
This will, for a voltage sine wave produce a 180º phase shift in the current. These types of circuit can use inductors and capacitors and without going through all the possibilities there is probably a way of adding L, C and Rs around an op-amp to achieve it too.
Well the condition will be different for each component. Resonance doesn't matter. The current must obviously be identical in all three series elements. The Voltage across the resistance will be in phase with that current, and the Voltage across the capacitance must lag the current by 90 degrees, while the Voltage across the inductance must lead the current by 90 degrees. This makes the inductance and capacitance Voltages 180 degrees out of phase. At resonance they will be equal and opposite, so they cancel leaving just the whole source Voltage across the resistance.
Best Answer
This phenomenon can be explained by taking a look at the electrical impedance, which is caused by the R and C components. The electrical impedance $Z$ is composed as \begin{align} Z = R + i X \end{align} where $X$ is the reactance. This is for capacitors $X_C = -\frac{1}{\omega C}$ and for inductors $X_L = \omega L$. We can make a sketch of this in a complex plane, which looks like
Now you can see, that it depends on the ratio of $R$ and $X$ what the phase difference $\phi$ between the real electrical current ($R$ - direction) and the voltage ($Z$ - direction). The actual dependence being \begin{align} \varphi= \arctan{\left(\frac{X}{R}\right)} \, . \end{align} (in the electrical engineers notation with $j$ as the imaginary unit) From this you can also see that if you have nearly no electrical resistance $R$ you end up with a $90°$ phase difference.
I hope this answers your questions.