# How can this circuit be simplified in the first moments after the circuit breaker closes?

The book states that in this circuit

after the transient the branch with the capacitor can be ignored while the branch with the inductance can be considered constituted only by a resistance. This seems to me simple and intuitive. However, it is also claimed that as soon as the switch is closed, in the first moments, the opposite happens: we must assume that the branch with the inductance is as if it were cut while the branch with the capacitor is as if it consisted only of the resistance $$R_3$$. This seems much more problematic to me and I don't know how to justify it.

## Edit: analysis that doesn't work

If $$i_1$$ is the current through the $$C$$ branch, and $$i_2$$ is the current through the $$L$$ branch, then differential equations of loops are $$$$-R_1 (i_1 + i_2) - i_2 R_2 - L \frac{d i_2}{dt} + V_g = 0$$$$ $$$$-R_3 i_1 - \frac{1}{C} \int_0^t i_1 dt + L \frac{d i_2}{dt} + i_2 R_2 = 0$$$$ By deriving this one we have $$$$-R_3 \frac{d i_1}{dt} - \frac{1}{C} i_1 + L \frac{d^2 i_2}{dt^2} + R_2 \frac{d i_2}{dt} = 0$$$$ After transient, all time derivatives are zero, so the first and the third gives $$$$i_1 = 0$$$$ $$$$i_2 = \frac{V_2}{R_1+R_2}$$$$ and this is not a surprise. In the first moments after closing are currents to be zero, not their derivatives. So the first gives $$$$\frac{di_2}{dt} = \frac{V_g}{L}$$$$ $$i_2 (0) = 0$$ so constant integration is zero and we have $$$$i_2 = \frac{V_g}{L} t$$$$ By substituting into the third and reasoning in a similar manner we have $$$$i_1 = \frac{R_2}{R_3} \frac{V_g}{L} t = \frac{R_2}{R_3} i_2$$$$ I would have expected for example $$i_1 \propto t$$ and $$i_2 \propto t^2$$, in that way the idea $$i_2=0$$ at the first moment would be justified. Insted I found the two corrent are simply proportional in the first moments. I don't know reactance, impedance, etc., ok I should study them, but, anyway, I study only for fun and I would simply to know why my try to answer my question doesn't work. If these steps are meaningfull or it is some stupid error.

• Maybe you should search about self inductance and how a capacitor charges because this is a basic problem about these 2 concepts Commented May 22, 2022 at 12:35

In a transient problem like this, to determine the initial conditions for the problem after an ideal switch changes state, we treat inductors as current sources and capacitors as voltage sources. The value of these sources is whatever current was flowing through the inductor, and whatever voltage was found across the capacitor, prior to the switch changing state.

This is because the current through an inductor cannot change instantly, and the voltage across a capacitor cannot change instantly. So as far as the changes in the instant when the switch closes, the inductor has an infinite impedance and the capacitor has zero impedance.

In this case, since the inductor had 0 current and the capacitor had 0 voltage before the switch changed, these become equivalent to an open circuit for the inductor and a short circuit for the capacitor.

Ammending the correct answer of Photon. This has to do with the capacitive and inductive electric reactance in your presented circuit. Capacitors resist sudden changes in their voltage (including a change from zero value) and solenoid inductors a sudden change in their current passing though them (including a change from zero value):

$$X_{C}=-\frac{1}{\omega C}=-\frac{1}{2 \pi f C}$$

Capacitive reactance for a given capacitance $$C$$ value, reduces with the increase in frequency of the signal.

$$X_{L}=\omega L=2 \pi f L$$

Inductive reactance for a given inductance $$L$$ value, increases with the increase in frequency of the signal.

We can deduce from the above equations that for $$f=0$$, thus a d.c. steady-state for a d.c. circuit, $$X_{L}$$ becomes ideally zero and $$X_{C}$$ infinite (or strictly mathematically speaking as unspecified). Therefore, in a d.c. circuit in steady-state the solenoid behaves as a short-circuit and the capacitor as an open-circuit. In an a.c. circuit $$f≠0$$ in steady-state operation the corresponding reactance of each component are calculated from the above equations in Ohm ($$Ω$$) units.

In particular for a d.c. circuit with a very fast switch-on time when first powered up a very fast signal transient will be generated thus with a very short period similar to a spike signal that effectively corresponds to a very high frequency a.c. signal. As shown from the steady-state reactance equations above this will result during this transient signal, to a relative very large reactance value for the solenoid in your circuit and a relative very low reactance value for the capacitor.

Thus, the capacitor behaves as a short-circuit and the solenoid as an open-circuit during power up of the circuit.

So, your equivalent circuit during power-up moment, can be simplified ideally to the R1 and R3 connected in series with your voltage source and R2, L2 and C3 components as non-existing in the circuit.

• why my edit try to explain doesn't work? @Markoul11
– user291161
Commented May 25, 2022 at 23:50
• @Haumea Maybe you should use standard Maxwell equations in your analysis of the circuit as pointed out also previously describing the self inductance of the solenoid and displacement current on the capacitor. Commented May 26, 2022 at 11:56