Circuits & networks - Initial Conditions of Resistor, Inductor & Capacitor

1. INITIAL CONDITIONS : WHY TO STUDY
• Differential Equations written for a network
may contain arbitrary constants equal to the
order of the differential equations.
• The reason for studying initial conditions is to
find the value of arbitrary constants that
appear in the general solution of differential
equations written for a given network.

2. INITIAL CONDITIONS
• In Initial conditions, we find the change in selected variables
in a circuit when one or more switches are moved from open
to closed positions or vice versa.
t=0-
indicates the time just before changing
the position of the switch
t=0 indicates the time when the position of
switch is changed
t=0+
indicates the time immediately after
changing the position of switch

3. INITIAL CONDITIONS
• Initial condition focuses solely on the current
and voltages of energy storing elements
(inductor and capacitor) as they will
determine the circuit behavior at t>0.
• PAST HISTORY OF THE CIRCUIT WILL SHOW UP
AS THE CAPACITOR VOLTAGES AND INDUCTOR
CURRENTS

4. INITIAL CONDITIONS
1. RESISTOR
 The voltage current relation of an ideal resistance is
V=R*I
 From this equation it can be concluded that the
instantaneous current flowing through the resistor
changes if the instantaneous voltage across it
changes & vice versa
 The past voltage or current values have no effect on
the present or future working of the resistor i.e.. It’s
resistance remains the same irrespective of the past
conditions

5. INITIAL CONDITIONS
2. INDUCTOR
 The expression for current through the
inductor is given by

6. INITIAL CONDITIONS
Hence if i(0-
)=0A , then i(0+
)=0A
So we can visualize inductor as a open
circuit at t=0+

7. INITIAL CONDITIONS
• If i(0-
)=I0 , then i(0+
)=I0 i.e. the inductor can be
thought as a current source of I0 as shown

8. INITIAL CONDITIONS
FINAL CONDITIONS :
 From the basic relationship
V= L*(di/dt)
We can state that V=0 in steady state conditions at t= as
(di/dt)=0 due to constant current

9. INITIAL CONDITIONS
3. CAPACITOR
 The expression for voltage across the
capacitor is given by

10. INITIAL CONDITIONS
If V(0-
)=0V , then V(0+
)=0V indicating the
capacitor as a short circuit

11. INITIAL CONDITIONS
If V(0-
)= V volts, then the capacitor can be
visualized as a voltage source of V volts

12. INITIAL CONDITIONS
• Final Conditions
The current across the capacitor is given by the equation
i=C*(dv/dt)
which indicates that i=0A in steady state at t=
due to capacitor being fully charged.

13. INITIAL CONDITION
EXAMPLE-1 : In the network shown in the figure
the switch is closed at t=0. Determine i, (di/dt)
and (d2
i/dt2
) at t=0+
.
At t=0-
, the switch is
Closed. Due to which
il(0-
)=0A
Vc(0-
)=0V

14. INITIAL CONDITION
At t=0+
the circuit is
From the circuit
il(0+
)=0A
Vc(0+
)=0V

15. INITIAL CONDITION
• Writing KVL clockwise for the circuit
Putting t=0+
in equation (2)

16. INITIAL CONDITION
• Differentiating equation (1) with respect to time

17. INITIAL CONDITION
 Example 2: The position of switch was changed from
1 to 2 at t=0. Steady State was achieved when the
switch was at position 1. Find i, (di/dt) & (d2
i/dt2
) at
t=0+

18. INITIAL CONDITION
At t=0-
, the circuit is shown in figure
The inductor is in steady state so it is
assumed to be shorted.
So the current through it is
il(0-
)=20/10=2A
Vc(0-
)=0V

19. INITIAL CONDITION
So at t=0+
, the switch is at position 2
Here the Inductor behaves as a current source
of 2A. The circuit is shown below
il(0+
)=2A
Vc(0+
)=0V

20. INITIAL CONDITION

21. INITIAL CONDITION

22. THANK YOU