Non Steady State: Difference between revisions
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To better understand this, let’s take a look at a simple RC circuit consisting of a battery Emf, a resistor of resistance R, and a capacitor with a capacitance C. | To better understand this, let’s take a look at a simple RC circuit consisting of a battery Emf, a resistor of resistance R, and a capacitor with a capacitance C. | ||
[[File:Wiki circuit.jpg]] | |||
Assume that the capacitor initially has a charge of Q=0. While the switch remains open at location Y, there is no current flowing through the circuit. | Assume that the capacitor initially has a charge of Q=0. While the switch remains open at location Y, there is no current flowing through the circuit. | ||
Revision as of 14:31, 30 November 2015
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 Emf, a resistor of resistance R, and a capacitor with a capacitance C.
Assume that the capacitor initially has a charge of Q=0. While the switch remains open at location Y, there is no current flowing through the circuit.
The moment the switch closes at location X, 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. 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.
Discharging a Capacitor
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