Equal Peak Voltages in a Circuit: Is It Always True?

[etotheipi]

One correction to post 34=the electrical current is always the same in each component of the series circuit. It is the voltages that are out of phase with the current for the inductor and capacitor. For the resistor, the voltage drop across the resistor is in phase with the current. The idea is good, but it might need a minor edit or two.

What I wrote is correct! For that first part I was considering the same voltage source applied to each component in turn (not all of the components in series) and showing the different resulting current responses in each case.

Although, I will edit to make this more clear

 

[Frigus]

It can be good to have some introduction to ac circuits and their operation to at least get a basic understanding of how the electrical circuits work that we use on an everyday basis.

I have studied about RMS voltage and current of ac circuit and compared it with DC circuit of equivalent voltage which generates same power.
Is this what is being talked in this post?or is it so something else?

@Hemant, I feel that you might have been slightly left behind and that perhaps some of the discussion here and some of my earlier comments as well were not helpful for building your understanding. I'd like to try and summarise, just in case you're still a bit confused.

First consider an oscillating voltage $v = v_0 \cos{(\omega t + \phi_0)}$ applied to either a resistor, capacitor or inductor. Through the component a current will flow, oscillating each time at the same angular frequency but with perhaps an additional phase $\varphi$, that is $i = i_0 \cos{(\omega t + \phi_0 + \varphi)}$. Let's consider the AC response for resistors, capacitors and inductors in turn:$$\begin{align*}

\mathrm{R}: &\, i = \frac{v}{R} = \frac{v_0}{R} \cos{(\omega t + \phi_0)} \\ \\

\mathrm{C}: &\, i = \frac{dq}{dt} = C \frac{dv}{dt} = -\omega C v_0 \sin{(\omega t + \phi_0)} = \omega C v_0 \cos \left(\omega t + \phi_0 + \frac{\pi}{2} \right) \\ \\

\mathrm{I}: &\, i = i(0) + \int_0^{t} \frac{di}{d\xi} d\xi = i(0) + \int_0^t \frac{v_0}{L} \cos{(\omega \xi + \phi_0)} d\xi = \frac{v_0}{\omega L} \sin{(\omega t + \phi_0)} = \frac{v_0}{\omega L} \cos \left( \omega t + \phi_0 - \frac{\pi}{2} \right)
\end{align*}$$You may notice that the resistor response is in phase with the voltage, whilst the capacitor and inductor responses are $\pi / 2$ ahead and behind the voltage respectively.

It is convenient to introduce the complex voltage and complex current, defined by $V = V_0 e^{i \omega t}$ and $I = I_0 e^{i \omega t}$ respectively, where $V_0 = v_0 e^{i \phi_0}$ and $I_0 = i_0 e^{i (\phi_0 + \varphi)}$ are complex amplitudes encoding the magnitude and initial phase. The $\mathrm{Re}$ function maps from the complex voltage/current and standard voltage/current, that is, $v = \mathrm{Re}(V)$ and $i = \mathrm{Re}(I)$. One helpful property to remember is that the $\mathrm{Re}$ function commutes with time-differentiation; given some $z: \mathbb{R} \rightarrow \mathbb{C}$ with $t \mapsto z(t)$, you can write $z(t) = a(t) + ib(t)$ and thus$$\mathrm{Re} \left( \frac{\mathrm{d}z}{\mathrm{d}t} \right) = \mathrm{Re} \left( \frac{\mathrm{d}a}{\mathrm{d}t} + i \frac{\mathrm{d}b}{\mathrm{d}t} \right) = \frac{\mathrm{d}a}{\mathrm{d}t} = \frac{\mathrm{d}\left( \mathrm{Re}(z) \right)}{\mathrm{d}t} $$What this means is that you can do all the differentiations in complex numbers, and convert everything back only at the very end.

Now, consider a "black box" consisting of some network of resistors, inductors and capacitors. It may be assigned a (complex) impedance $Z$, which you may think of loosely as a complex analogue of resistance, and is defined as the ratio $Z := V / I$. For example, for the resistor you simply have $I = V/R$ and thus$$Z_R = \frac{V}{I} = R$$then for the capacitor, you have $I = C \frac{dV}{dt} = i\omega C V_0 e^{i \omega t} = i\omega C V$ and thus$$Z_C = \frac{V}{I} = \frac{1}{i\omega C }$$and for the inductor, you have $V = L \frac{dI}{dt} = i\omega L I_0 e^{i \omega t} = i\omega L I$ and thus$$Z_I = \frac{V}{I} = i \omega L$$How do you combine impedances? Let's just do one example, e.g. say $N$ impedances $\{Z_1, \dots, Z_N \}$ in parallel. The complex voltage $V$ across each of these is the same, and the sum of the complex currents through each must equal the total current into the parallel network, hence$$\sum_{k=1}^{N} I_k = V\sum_{k=1}^{N} \frac{1}{Z_k} = \frac{V}{Z_{\mathrm{eff}}}$$which upon division by $V$ yields the familiar formula.

You might wonder, do all of Kirchoff's laws, and the rules you know for analysing circuits hold also on the complex amplitudes $V_0$ and $I_0$? Yes! For example, consider the current law applied to a junction with $N$ inputs, $$(\forall t \in D) \quad \mathrm{Re}\left( \sum_{k=1}^N (I_0)_k e^{i \omega t} \right) \implies \sum_{k=1}^{N} (I_0)_k = 0$$because the only way for the sum to be zero for all values of $t$ is if the sum of the complex current amplitudes themselves vanish. The same analysis holds for the voltage law.

Why is this formalism useful? Well, let's consider your example of a circuit containing a resistor, capacitor and inductor in series, connected across an oscillating voltage source. You already know from the "impedances in series" principle that the resulting impedance is$$Z = Z_R + Z_L + Z_C = R + i \left(\omega L - \frac{1}{\omega C} \right)$$From this it's not hard to write down the current,$$I = \frac{V}{Z} = \frac{V}{R + i \left(\omega L - \frac{1}{\omega C} \right)} = \frac{V e^{- i \phi_Z}}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}}$$where the phase shift $\phi_Z$ is given by$$\phi_Z = \arctan \left( \frac{\omega L - \frac{1}{\omega C}}{R} \right)$$With this we can now work out some interesting quantities. For instance, what is the rate of energy dissipation in this oscillator, due to the resistor? You have$$\langle P \rangle = R|I_0|^2 \langle \cos^2{(\omega t - \phi_Z)} \rangle = \frac{1}{2}R |I_0|^2 = \frac{R|V_0|^2}{2(R^2 + (\omega L - \frac{1}{\omega C})^2)}$$At resonance, $\omega_0 L - \frac{1}{\omega_0 C} = 0$. Half-power thus occurs when $\omega_h L - \frac{1}{\omega_h C} = \pm R$, or in other words when$$\omega_h = \mp \Gamma + \sqrt{\Gamma^2 + \omega_0^2}$$with $\Gamma := R/2L$. Then, you may identify the full-width half-power as $\delta \omega = \frac{R}{L}$, which results in a quality factor $Q = \frac{\omega_0}{\delta \omega} = \frac{1}{R} \sqrt{\frac{L}{C}}$.

In other words, complex numbers allow you to fairly easily analyse key properties of oscillating circuits, and oscillating systems in general.

Thanks for clarifying what is happening,
Can I please get some time to digest what is written in post although it is written in very simple but I am new to complex numbers and I think it will take me a little bit of time to understand it.

 

[Charles Link]

Perhaps I can offer a very simple description of the ac circuit problem:
The circuit is driven by a sinusoidal voltage of known amplitude. It is known that in each component of the circuit, the voltage and current will also be sinusoidal in time at that same frequency, but we need to compute the amplitude and phase of each of the sinusoidal waveforms across each component. The approach is similar to ohm's law for a DC circuit, but involves using complex impedances instead of simple resistances.

 

[nucl34rgg]

Now only thing I am not able to understand is the last step where the denominator is being square rooted.

The denominator was squared and then square rooted. He is expressing the magnitude.