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.

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

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

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

INITIAL CONDITIONS

2. INDUCTOR The expression for current through the

inductor is given by

INITIAL CONDITIONS

Hence if i(0-)=0A , then i(0+)=0ASo we can visualize inductor as a open circuit at t=0+

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

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

INITIAL CONDITIONS

3. CAPACITOR The expression for voltage across the

capacitor is given by

INITIAL CONDITIONS

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

INITIAL CONDITIONS

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

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.

INITIAL CONDITION

EXAMPLE-1 : In the network shown in the figure the switch is closed at t=0. Determine i, (di/dt) and (d2i/dt2) at t=0+ .

At t=0- , the switch is

Closed. Due to which

il(0-)=0A

Vc(0-)=0V

INITIAL CONDITION

At t=0+ the circuit is

From the circuit

il(0+)=0A

Vc(0+)=0V

INITIAL CONDITION• Writing KVL clockwise for the circuit

Putting t=0+ in equation (2)

INITIAL CONDITION

• Differentiating equation (1) with respect to time

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) & (d2i/dt2) at t=0+

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

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

INITIAL CONDITION

INITIAL CONDITION

THANK YOU