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1.4. The Source‐Free Parallel RLC Circuits
Now let us look at parallel forms of RLC networks
Assume the initial current and voltage to be:
 Applying KCL at the top node gives;
 takig derivative with respect to t and dividing by C results in;
1.4. The Source‐Free Parallel RLC Circuits
s
From this, we can find the roots of the characteristic equation to be:
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1.4. The Source‐Free Parallel RLC Circuits
As in last time, there are three scenarios to consider.
Overdamped Case (∝
negative and real Critically Damped Case (∝
real and equal
1.4. The Source‐Free Parallel RLC Circuits
Under Damped Case (∝
arecomplex
To get the values for the constants, we need to know v(0) and dv(0)/dt.
canbedeterminedfrominitialconditions;
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1.4. The Source‐Free Parallel RLC Circuits
 The voltage waveforms will be similar to those shown for the series network.
 Note that in the series network, we first found the inductor current and then solved for the rest from that.
 Here we start with the capacitor voltage and similarly, solve for the other variables from that.
Example 1.5. In the parallel circuit of fig. , find assuming 0
5 , 0
0,
1 for t>0, 10
.
Consider these cases; R=1.923 Ω, R=5 Ω, R=6.25 Ω. İf R=1.923 Ω
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Example 1.5. Since (∝
in this case, overdamped. The roots of the characteristic equation are;
Apply initial conditions to get and At t =0
Example 1.5. Must be differeantiated
At t =0
With
and the solution gets… 4
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Example 1.5. When R=5 Ω
remains 10;
Since α=
, the responce is
Critically damped…
Apply initial conditions to get and Example 1.5. Must be differeantiated
With
and the solution gets… 5
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Example 1.5. İn the last case R=6.25 Ω…
Solution is…
Response for three degrees of damping
1.5. Step Response of Series RLC Circuits
 Now let us consider what happens when a DC voltage is suddenly applied to a second order circuit.
 Consider the circuit shown. The switch closes at t=0.
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1.5. Step Response of Series RLC Circuits
This equation has two compenants;
This is similar to the response for the source free version of the series circuit, except the variable is different.
Natural response;
Forced response;
The solution to this equation is a combination of transient response and steady state
1.5. Step Response of Series RLC Circuits
 İf we set 0,
contains only natural response (
 (t) can be expressed as three conditions;
 The forced response is the steady‐state or final value of  The final value of the capacitor voltage is the same as the source voltage .
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1.5. Step Response of Series RLC Circuits
Thus the complete solutions for the three conditions of damping are:
The variables A1 and A2 are obtained from the initial conditions, v(0)
and dv(0)/dt.
Example 1.6. For the circuit in Fig., find for t>0. Consider these cases:
R=5Ω, R=4Ω,R=1Ω.
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Example 1.6. Example 1.6. 
is the forced response or steady‐state response.
 It is final value of the capacitor voltage.

=24 V.
For t>0 , current i, 9
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Example 1.6.  Finally,  but we have to solve i(t)…
1.6. Step Response of Parallel RLC Circuits
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Example 1.7. Solution:
 For t<0, the switch is open
 The circuit partitioned into two independent subcircuits.
Capacitor voltage equal the voltage of 20Ω resistor.
Example 1.7.  For t>0, the switch is closed  We have parallel RLC circuit.
 The voltage source is off or short‐circuited.
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Example 1.7. The final value of I…
Using initial conditions we get and
Example 1.7. From i(t) we obtain v(t)…
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