Non Steady State

by Javier Rodriguez

RC circuits in non-steady state are defined by changing charges, currents, and voltages as a function of time.

In other words, Q’(t) ≠ 0

Non-steady state circuits are specifically important when analyzing the charging or discharging a capacitor. Surely, when a capacitor fully charged (Q/C = Emf) or fully discharged (Q=0), the circuit is in steady state — the derivative of charge with respect to time is 0.

Charging a Capacitor

To better understand this, let’s take a look at a simple RC circuit consisting of a battery and a capacitor with a capacitance C.

Assume that the capacitor initially has a charge of Q=0. While the switch remains open, there is no current flowing through the circuit.

The moment the switch closes, the voltage across the capacitor does not instantaneously equal the Emf of the battery. A current initially starts to flow, which allows charges to collect on the plates of the capacitor. At this moment, Q’(t) > 0, and, inger. As time passes, the charges continue to build up, which permits less current to pass through the capacitor — causing the current to decrease under an exponential decay, I’(t) < 0.

Because the voltage drop across a capacitor is directly proportional to charge, this voltage follows the same trend. To find the the exact voltage drop across the plates at any given time t, we can use the following exponential equations:

V(t) = Emf (1 - e-t/RC)

Looking at this equation, we see that as time increases, e-t/RC grows increasingly smaller. After enough time passes, this value will approximate at 0, leaving the voltage at time t to resemble:

V(t) = Emf

At this moment, the capacitor will be fully charged, and no current will flow through the circuit.

Let's say there is a location B at [math]\displaystyle{ \left(x_1, 0, 0\right) }[/math]. Now in order to find the potential difference between A and C, we need to find the potential difference between A and B and then between B and C.

The potential difference between A and B is [math]\displaystyle{ dV = V_B - V_A = -\left(E_x*\left(x_1 - 0\right) + 0*0 + 0*0\right) = -E_x*x_1 }[/math].

The potential difference between B and C is [math]\displaystyle{ dV = V_C - V_B = -\left(E_x*0 + 0*\left(-y_1 - 0\right) + 0*0\right) = 0 }[/math].

Therefore, the potential difference A and C is [math]\displaystyle{ V_C - V_A = \left(V_C - V_B\right) + \left(V_B - V_A\right) = E_x*x_1 }[/math], which is the same answer that we got when we did not use location B.

Discharging a Capacitor

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