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This lesson deals with the use of either capacitors or inductors (not both at the same time) in single-time-constant circuits. This usually means that the circuit consists of a single inductor or capacitor, several resistors, and one or more switches. Such a circuit is shown below.

Assume that the switch has been open for a long time, and that at t = 0 the switch is closed. Intuitively, you should expect that before the switch is closed, the capacitor will charge up to 50 V. This is so because, if you wait long enough, the current in the 4000 Ω resistor will decay to zero, and there will be no voltage across the resistor. Then, by KVL, the capacitor and battery voltages must be equal.

After the switch is thrown, things get more complicated: The capacitor will begin to drain through the 1000 Ω resistor, but current will continue to be provided by the battery. You might guess that the capacitor voltage would decay, but wouldn't go all the way to zero, and you would be right.

Click on this link: http://micro.magnet.fsu.edu/electromag/java/timeconstant/ to see a demonstration of the RC time constant.

Below is a systematic approach to solving problems like this. Read it carefully, then we'll apply it to the above problem.

This technique works for first order systems with constant sources.

For simplicity, assume t_{i} = 0, where t_{i}
is the initial time, *i.e.* the time when the switch is
closed (or opened).

1. Write down:

i_{C} = C(dv_{C}/dt)

v_{L} = L(di_{L}/dt)

i(t) = i_{f} + (i_{0} - i_{f})e^{-t/τ}

v(t) = v_{f} + (v_{0} - v_{f})e^{-t/τ}

2. For 0^{-}

- Recall that "steady state" has been reached.
- Therefore, v
_{C}= constant, i_{L}= constant - Therefore, i
_{C}= 0, v_{L}= 0 - Redraw the circuit, open circuiting capacitors (because i
_{C}= 0) and short circuiting inductors (because v_{L}= 0) - Apply KVL and KCL to find unknown voltages and currents,
particularly v
_{C}or i_{L}.

3. For 0^{+}

- Recall that you can't change v
_{C}or i_{L}instantaneously. - Therefore, v
_{C}(0^{+}) = v_{C}(0^{-}) and i_{L}(0^{+}) = i_{L}(0^{-}) - Apply KCL and KVL to find v
_{0}and i_{0}.

4. For infinity

- Repeat 2.
- Calculate v
_{f}or i_{f}.

5. Finally

- Redraw circuit, removing C or L, shorting voltage sources, and opening current sources.
- Calculate R
_{TH}"looking into" the terminals where you took out C or L. - Calculate τ = R
_{TH}C or τ = L/R_{TH}. - Plug values into i(t) or v(t) equation.
- Simplify.

1. Write down:

i_{C} = C(dv_{C}/dt)

v(t) = v_{f} + (v_{0} - v_{f})e^{-t/τ}

2. For 0^{-}

- Recall that "steady state" has been reached.
- Therefore, v
_{C}= constant - Therefore, i
_{C}= 0 - Redraw the circuit, open circuiting capacitors (because i
_{C}= 0)

- Apply KVL to find the unknown voltage.

This is easy:

-50 + v_{4000}+ v = 0

But v_{4000}= 0 because it caries no current, so

v = v(0^{-}) = 50 V

3. For 0^{+}

- Recall that you can't change v
_{C}instantaneously. - Therefore, v
_{C}(0^{+}) = v_{C}(0^{-})

v(t = 0) = v_{0}= 50 V

4. For infinity

- Repeat 2.

By the voltage divider formula,

v = 50×1000/(4000 + 1000) = 10 V

- Calculate v
_{f}.

v_{f}= v = 10 V

5.

- Redraw circuit, removing C, shorting voltage sources.

- Calculate R
_{TH}"looking into" the terminals where you took out C.

R_{TH}= 4000 × 1000/(4000 + 1000) = 800 Ω - Calculate τ = R
_{TH}C.

τ = 800 × 20×10^{-6}= 16×10^{-3 }s = 16 ms = .016 s - Plug values into v(t) equation.

v(t) = v_{f}+ (v_{0}- v_{f})e^{-t/τ}= 10 + (50 - 10)e^{-t/.016}V - Simplify.

v(t) = 10 + 40e^{-t/.016}V

Here's what a graph of the capacitor voltage looks like:

Check out this first order circuit explanation and example: http://people.clarkson.edu/~svoboda/eta/plots/FOC.html (created by James A. Svoboda of Clarkson University).

At this time you should complete
Tutorial 8
on **first order circuits**.