| Outline: |
| Capacitor Review |
| Inductor Review |
| DC Steady State |
| Switches |
| Analysis of circuits with switches and storage elements |
Capacitor Review
A Capacitor is an element which stores charge. It is comprised of two
conducting plates sepparated by a non-conducting material called a dielectric.
For every + unit charge put on one plate, there is an equal - unit charge on the
the other plate. Thus the entire capacitor is charge neutral. Since the + and - charges are
separated by a dielectric there is an electric field (or voltage) across the capacitor.
If positive charge is put on one plate then work is done because like-charges repel each other.
Recall that voltage is related to work done on charge by the equation V=J/C.
The more charge on the capacitor's plates, the more work had to be done to put the charge there,
and the higher the voltage across the capacitor.
In this class all capacitors will be considered linear and will obey the following
charge-voltage relationship:
q = CV
If this equation is differentiated we get:
dq/dt = C(dV/dt)
or i = C(dV/dt)
The capacitor on the left is charging, therefore the voltage is increasing.
The capacitor on the right is actually discharging, therefore the voltage is decreasing.
Note the sign difference.
the capacitor behaves like an 'open circuit' since no current flows.
for the charge to leave the plate. Therefore charge on the capacitor is continuous.
Since V = q/C then voltage across a capacitor is also continuous.
(i.e. The voltage across a capacitor can never change instantaneously.)
However the current can change instantaneously since i = C dV/dt.
This is because current is the rate of charge moving over time.
Energy stored in a capacitor is:
E = 1/2 CV2
Using the above concepts, let's analyze the following circuit:
| This circuit has both a switch and a capacitor: The switch opens at t=0 The switch is closed for t<0 and is open for t>0. This can be seen by inspecting the switch's arrow direction. We will not try to analyze the circuit at t=0 since the circuit's state at t=0 is unkonwn. Instead we will look at the circuit at t=0- (the time right before the switch moves) and t=0+ (the time right after the switch moves). In this problem it is given that V4 = 8V at t=0-. With this information find the following things at t=0- and t=0+:
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| Here are snapshots of the circuit at t=0- and t=0+: _______________________________________________________ t = 0-:
At t=0- the switch is closed. Since V4 = 8V and and the capacitor is in parallel with V4, then VC = 8V at t=0-. To determine i1 we need to find the voltage across the horizontal 4 ohm resistor. To find this, we will apply KVL: -20V + 4i1 + 8V = 0 4i1 = 20 - 8 = 12V i1 = 12/4 = 3A To find the current in the capacitor we have to apply KCL at the marked node: i1 = i + iC 3A = 2A + iC iC = 1A dV/dt = iC/C = 1A/(.25F) = 4V/s _______________________________________________________ t = 0+:
At t=0+ the switch is open. Since the voltage across a capacitor can notchange instantly, then VC(0-) = VC(0+) Therefore VC(0+) = 8V Since the switch is now open no current flows through the horizontal resistor, so i1 = 0 To find the current in the capacitor we have to apply KCL at the marked node: i1 = i + iC 0A = 2A + iC iC = -2A dV/dt = iC/C = -2A/(.25F) = -8V/s dV/dt is negative, therefore for t>0 the capacitor is discharging. The energy stored in the capacitor is being absorbed by the resistor. Eventually all the initial energy stored in the capacitor will be absorbed by the resitor. |
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Study Problems
After clicking on the following link enter 6-1 for the problem and 1 for the step:
Study Problem 6-1
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Inductor Review
An inductor is an element which stores a magnetic field. An inductor is
a wire coiled around a material called a core. The core is typically made of
a magnetic material however the core can be anything from a toilet paper roll to a piece of wood.
(The behavior of the inductor IS effected by the core material though.)
Recall from Physics that static charge sets up an electrostatic force. Also recall that
moving charge sets up a magnetic field called flux around the current-carrying wire.
If the wire is then coiled as shown below, then the flux increases linearly with every coil turn.
The flux is proportional to the current flowing through the inductor.
We can use the following equation where L is the inductance of the inductor.(L is constant)
Flux = Li
Through experimentation it was found that if the flux changes then a voltage is induced.
If the flux is constant then there is no voltage across the inductor.
V = d(Flux)/dt
V = d(Li)/dt
V = L di/dt
When the current is changing a voltage is induced across the inductor.
This voltage opposes the change in the current.
When the current stops changing and stays constant, then the voltage collapses.
Therefore when conditions are constant (unchanging) the inductor appears like a
wire or short circuit.
Current is continuous in an inductor. In other words, current can not
change instantaneously. If current could change instantaneously then an infinite
voltage would be induced across the capacitor. Infinite voltage is a practical impossibility.
Therefore current can not change instantaneously through an inductor.
Voltage across an inductor can change instantaneously.
The energy stored in an inductor is:
E = 1/2 Li2
Using the above concepts, let's analyze the following circuit:
| This circuit has both a switch and an inductor: The switch closes at t=0 The switch is open for t<0 and is closed for t>0. This can be seen by inspecting the switch's arrow direction. We will not try to analyze the circuit at t=0 since the circuit's state at t=0 is unkonwn. Instead we will look at the circuit at t=0- (the time right before the switch moves) and t=0+ (the time right after the switch moves). In this problem it is given that i6 at t=0- is 2A With this information find the following things at t=0- and t=0+:
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| Here are snapshots of the circuit at t=0- and t=0+: _______________________________________________________ t = 0-:
At t=0- the switch is open. Since we know that i6 is given to be 2A,we can apply KCL to the marked node to find iL. Σ ientering = Σ ileaving 5A = i6 + iL 5A = 2A + iL iL = 3A at t=0-. Now to find diL/dt: diL/dt = VL/L We need to find VL first: Apply KVL to the right-most loop: VL + 4iL - (2A)(6 Ohms) = 0 VL + 4(3A) - 12 = 0 VL = 12 - 12 VL = 0 diL/dt = VL/L = 0/2 = 0 _______________________________________________________ t = 0+:
iL(0-) = iL(0+) This is always true! Therefore iL(0+) = 3A Current through an inductor is continuous We also need to fine i6: Note that the 6 Ohm resistor has been shorted out by the closed switch. A resistor in parallel with a wire has zero voltage across is. If the 6 Ohm resistor has zero volts across it then the current through the 6 Ohm has to be zero due to Ohm's Law, V=IR. Thus i6(0+) = 0 Now to find diL/dt: diL(0+)/dt = VL(0+)/L We need to find VL(0+) first: VL + 4iL - 0 = 0 VL + 4(3A) = 0 VL = -12 diL/dt = VL/L = -12/2 = -6A/s |
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After clicking on the following link enter 6-2 for the problem and 1 for the step:
Study Problem 6-2
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DC Steady State
When a switch exist in a circuit, the state of the circuit
changes when the switch moves from open-to-closed or closed-to-open.
The state is defined by the voltages and currents in the circuit.
When the switch moves, the circuit diagram is altered and that is why the
voltage and current values change in the circuit.
Steady State is when all voltages and currents in the circuit
have reached a steady value (i.e. all currents and voltages
are constant - not changing). This occurs some finite amount of time
after the switch moves.
Immediately after the switch moves from open-to-closed or closed-to-open
the current and voltage values are 'transitioning' from their old values to their new
values. This period of time is called the transient response.
The transient response is the way in which currents and voltages get from their
initial value (at t=0-) to their final value (at t>0).
This will be discussed in the next chapter.
Steady State is when all voltages and currents in the circuit have settled down
to their final values. This occurs after the transient response has died out.
DC Steady State is the final state of the circuit when a DC source is present.
In DC Steady State all voltages and currents will be CONSTANT.
To find the Steady State response for a circuit assume a long time after the switch
has moved (t >> 0). This will ensure that the transient response has 'died out'.
In unchanging conditions the voltage across a capacitor is constant and the
current through an inductor is constant. Recall:
iC = Cdv/dt
VL = Ldi/dt
Since the derivatives are zero then iC=0 and VL = 0 for DC Steady State.
What does this mean???
In DC Steady State capacitors look like open circuits and inductors look like wires.
The following example will illustrate DC Steady State:
| In this circuit the switch is moving from position-a to position-b at t = 0. We can assume Steady State at t = 0- because nothing happens to the switch before t = 0 . |
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| We now look at the circuit at t = 0-. Since we are assuming Steady State, we can make the capacitor an open circuit. If the capacitor is 'open' then no current flows to the capacitor. Applying KVL we find that VC = 5V (There is no drop across the resistor.) |
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| At t = 0+ the switch is now in position-b, however the circuit is not in steady state. VC = 5V because VC(0-) = VC(0+). There is a 2V drop across the resistor, therefore there is now current in the resistor. The current comes from the fact that the capacitor is discharging. |
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| We now assume the transient response has died out. At t >> 0 there is no transient response left. Also after a long, long time (i.e. t >> 0), we know that we are in a Steady State condition again. Again we assume that the capacitor looks like an open circuit. Applying KVL we find that VC = 3V. We can now sketch the response to get curve shown to the right. The time it takes the capacitor to discharge from 5V to 3V is determined by the value of the capacitor and the value of the resistor. We will look at this in the next section. |
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After clicking on the following link enter 6-3 for the problem and 1 for the step:
Study Problem 6-3
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More on Switches
A switch moves from one position to another.
This changes the circuit at the specified time.
The direction of the arrow determines which way
the switch is moving.
| When a switch moves from one position to another it happens at the time specified on the switch. We are only interested in the time right before the switch moves and the time after the switch moves. In other words, if a switch moves at t= 0, then we consider the circuit at t=0- (right before t=0) and t=0+ (right after t=0). We do not consider the circuit at t=0 because we don't know where the switch is physically during this quick switching time.  The switch can move at times other than t=0 as shown below: |
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Case 1: Here the switch CLOSES at t=0. At t=0- the switch is open and at t=0+ the switch is closed. Case 2: The switch OPENS at t=5s. At t=5- the switch is closed and at t=5+ the switch is open. Case 3: The switch moves from position-a to position-b at t=10s. At t=10- the switch is in position-a and at t=10+ the switch is in position-b. |
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Analysis of circuits with switches and storage elements
After clicking on the following link enter 6-4 for the problem and 1 for the step:
Study Problem 6-4
After clicking on the following link enter 6-5 for the problem and 1 for the step:
Study Problem 6-5
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