1. Transient Analysis
Transient Analysis

2. The Growth of current in R-L CircuitvR vL i
When switch S just switches to a : t=0 , i=0 , VL =0
Then current start flowing with the rate of di/dt thus a voltage develop in inductor given by

3. i vR vL Using Voltage Kirchoff’s law Divided by R
Where I = V/R = steady current
 = L/R = time constant

4. Rearrange the equation
Integrate both sides we have
Then the solution is
(*)
Apply the initial state t=0 and i=0
Substitute in (*) , we have Or
Voltage
Substitute t = L/R

5. vL, i V
V/R
vL, i t

6. Decay analysis for R-L circuitvL vR i
When switch S is switched to b iL will begin to decay in a reverse direction with the rate of di/dt thus a voltage develop in inductor given by

7. Using Voltage Kirchoff’s lawOr
Rearrange
where  = L/R
Integrate
The solution
Apply initial condition t=0 , i=I
Or we have

8. Substitude  = L/R and IR = V ; we have
Voltage
Substitude  = L/R and IR = V ; we have
vL = - V e-t/
vL, i
V/R + - V vL i t

9. Example 1 For network as in figure
Determine the mathematical expressions for the variation of the current in the inductor following the closure of the switch at t=0 on to position 1
When the switch is closed on to position 2 at t=100ms, determine the new expression for the inductor current and also for the voltage across R;
Plot the current waveforms for t=0 to t=200ms.

10. (a) For the switch in position 1, the time constant is
Therefore
(b) For the switch in position 2, the time constant is
Therefore
The voltage drop

11. At time t=0 till t=40ms i follows the equation
At time t=40ms till t=100ms i is saturated which is equaled to 1
At time t>100ms i follows the equation

12. Example 2
For the network in example 1, the switch is closed on to position 1 as before but it is closed on to position 2 when t=10ms. Determine the current expressions and hence plot the current waveforms
When switch is switched on to position 1 , i follows
At t=10ms the magnitude of i is
When switch is switched on to position 2 , i

13. Example 3
For network in the Figure , the switch is closed on to position 1 when t=0 and then moved to position 2 when t=1.5ms. Determine the current in the inductor when t=2.5ms.

14. First we simplify the supply of the circuit using Norton Theorem
The simplified circuit is

15. Switch at position 1
Current Maximum magnitude
At time t=1.5ms
Switch at position 2
t2=2.5ms- t1=2.5ms-1.5ms=1ms

16. Energy stored
To neutralize the induced e.m.f which represents the power absorbed by magnetic field a voltage source is needed. The energy produced by the voltage and current developed is stored in the form of magnetic field. This is given by
and
Average energy absorbed
Total energy absorbed

17. The Growth of current in R-C CircuitvC vR i
When switch S just switches to a : t=0 , q=0 , VC =0
Then voltage start develops across the capacitor with the rate of dVC/dt thus a current flowing in the circuit is given by

18. vR i vC Using Voltage Kirchoff’s law V = vR + vC Or V – vC = vR = iR
Substitute i we have Or
Where t =RC

19. Rearrange we have
Integrate
Then we have
At t=0 we have vC=0 thus A=ln V
Substitute A we have Or

20. Voltage development and current decay in serial RC circuit
Substitute VC in the current formulation vC i
vC, i V R t
Voltage development and current decay in serial RC circuit

21. Decay analysis for R-C circuitvC vR i
When the switch is connected to b, vC = V. The accumulated charges in C now start to discharge via R. The initial current iI is given by:
In the process of discharging, voltage across C, vC is decaying. When charges are full discharged, vC = 0 and the final current iF = 0.

22. Using voltage‘s Kirchoff law
vC = -vR
Therefore
But
Substitute we have
Rearrange
And substitute
RC = 
Then we have

23. Integrate
We get
A = ln V
If t=0 ; vC=V , then we have
Substitute A we have
We get or

24. Decaying of Voltage and currrent in RC circuit
For current we have t
vC, i V
-V/R vC i
Decaying of Voltage and currrent in RC circuit

25. Example 4
A capacitor is made of two pieces of aluminium foil separated by a piece of paper of 0.2mm thick and having a relative permittivity of r = 5. Determine the area of aluminium required to produce a capacitance of 1200 pF? Assume that the permittivity of the free space is o ,8.854 x F/m.
C = pF = x F
d = 0.2 mm = 2 x 10-4 m
r = 5; dan o = x F/m
A = ?

26. Example 5
Two capacitors, C1 = 1200 pF and C2 = 2200 pF are connected in parallel. The combination is connected to an a.c voltage source VS = 155 V having a series resistor RS. After the voltage across the capacitor reach a steady state, the connection of capacitor to the battery is disconnected and discharged via a resistor RP.
Draw the circuit.
Calculate Rs so that the energy stored in capacitor reaches 25 mJ in 250ms during capacitor charging
Calculate Rp , after 1ms discharging, the voltage drop is 20% from its steady state.
Calculate the energy stored in capacitor after 1.5ms discharging.

27. (1) RS RP C1
1200 pF C2
2200 pF VS
155 V

28. (2) Total capacitance Given Voltage across cap. or During charging
rewrite or
then

29. (3)
During discharging
Given that vc drop 20% from its steady voltage , vC= 0.8VS or

30. (4)
After 1.5ms discharge