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Transient Analysis Transient Analysis

The Growth of current in R -L Circuit i v. R v. L When The Growth of current in R -L Circuit i v. R v. L 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

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

Rearrange the equation Integrate both sides we have (*) Then the solution is Apply 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 Voltage Substitute = L/R Or

v L, i V v L, V/R i 0 t v L, i V v L, V/R i 0 t

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

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

Voltage Substitude = L/R and IR = V ; we have v. L = Voltage Substitude = L/R and IR = V ; we have v. L = - V e-t/ v. L , i V/R i + 0 t - v. L V

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

(a) For the switch in position 1, the time constant is Therefore (b) For (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

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

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

Example 3 For network in the Figure , the switch is closed on to 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. 5 ms. Determine the current in the inductor when t=2. 5 ms.

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

Switch at position 1 Current Maximum magnitude At time t=1. 5 ms Switch at Switch at position 1 Current Maximum magnitude At time t=1. 5 ms Switch at position 2 t 2=2. 5 ms- t 1=2. 5 ms-1. 5 ms=1 ms

Energy stored To neutralize the induced e. m. f which represents the power absorbed 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

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

i v. R v. C Using Voltage Kirchoff’s law Or V = v. R i v. R v. C Using Voltage Kirchoff’s law Or V = v. R + v. C V – v. C = v. R = i. R Substitute i we have Or Where =RC

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

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

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

Using voltage‘s Kirchoff law v. C = -v. R Therefore But Substitute we have Using voltage‘s Kirchoff law v. C = -v. R Therefore But Substitute we have Rearrange Then we have And substitute RC =

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

For current we have v. C, i V v. C 0 -V/R t i For current we have v. C, i V v. C 0 -V/R t i Decaying of Voltage and currrent in RC circuit

Example 4 A capacitor is made of two pieces of aluminium foil separated by Example 4 A capacitor is made of two pieces of aluminium foil separated by a piece of paper of 0. 2 mm thick and having a relative permittivity of r = 5. Determine the area of aluminium required to produce a capacitance of 1200 p. F? Assume that the permittivity of the free space is o , 8. 854 x 10 -12 F/m. C = 1200 p. F = 1200 x 10 -12 F d = 0. 2 mm = 2 x 10 -4 m r = 5; dan o = 8. 854 x 10 -12 F/m A = ?

Example 5 Two capacitors, C 1 = 1200 p. F and C 2 = Example 5 Two capacitors, C 1 = 1200 p. F and C 2 = 2200 p. F 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. 1. Draw the circuit. 2. Calculate Rs so that the energy stored in capacitor reaches 25 m. J in 250 ms during capacitor charging 3. Calculate Rp , after 1 ms discharging, the voltage drop is 20% from its steady state. 4. Calculate the energy stored in capacitor after 1. 5 ms discharging.

(1) RS VS 155 V RP C 1 1200 p. F C 2 2200 (1) RS VS 155 V RP C 1 1200 p. F C 2 2200 p. F

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

(3) During discharging Given that vc drop 20% from its steady voltage , v. (3) During discharging Given that vc drop 20% from its steady voltage , v. C= 0. 8 VS or

(4) After 1. 5 ms discharge (4) After 1. 5 ms discharge