* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Example 1.5.
Electronic engineering wikipedia , lookup
Transmission tower wikipedia , lookup
Voltage regulator wikipedia , lookup
Power MOSFET wikipedia , lookup
Fault tolerance wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Stray voltage wikipedia , lookup
Current source wikipedia , lookup
Schmitt trigger wikipedia , lookup
Alternating current wikipedia , lookup
Voltage optimisation wikipedia , lookup
Electrical substation wikipedia , lookup
Opto-isolator wikipedia , lookup
Rectiverter wikipedia , lookup
Surge protector wikipedia , lookup
Buck converter wikipedia , lookup
Mains electricity wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Flexible electronics wikipedia , lookup
1.4. The Source-Free Parallel RLC Circuits ο Applying KCL at the top node gives; ο takig derivative with respect to t and dividing by C results in; 1.4. The Source-Free Parallel RLC Circuits π2 ππ‘ 2 π 2 π ππ‘ s 1.4. The Source-Free Parallel RLC Circuits Overdamped Case (β> ππ ) π 1 πππ π 2 negative and real Critically Damped Case (β= ππ ) π 1 πππ π 2 real and equal 1.4. The Source-Free Parallel RLC Circuits Under Damped Case (β< ππ ) π 1 πππ π 2 are complex π΄1 πππ π΄2 can be determined from initial conditions; Example 1.5. In the parallel circuit of fig. , find π£(π‘) for t>0, assuming π£ 0 = 5 π, π 0 = 0, πΏ = 1π» πππ πΆ = 10 ππΉ. Consider these cases; R=1.923 β¦, R=5 β¦, R=6.25 β¦. Δ°f R=1.923 β¦ Example 1.5. Since (β> ππ ) in this case, overdamped. The roots of the characteristic equation are; Apply initial conditions to get π¨π and π¨π At t =0 Example 1.5. Must be differeantiated At t =0 With π¨π and π¨π the solution getsβ¦ Example 1.5. When R=5 β¦ π0 remains 10; Since Ξ±= π0 , the responce is Critically dampedβ¦ Apply initial conditions to get π¨π and π¨π Example 1.5. Must be differeantiated With π¨π and π¨π the solution getsβ¦ Example 1.5. Δ°n the last case R=6.25 β¦β¦ Solution isβ¦ Response for three degrees of damping 1.5. Step Response of Series RLC Circuits This equation has two compenants; Natural response; ππ (π) Forced response; ππ (π) 1.5. Step Response of Series RLC Circuits ο Δ°f we set ππ = 0, π£(π‘) contains only natural response (π£π ) ο π£π (t) can be expressed as three conditions; ο The forced response is the steady-state or final value of π£(π‘) ο The final value of the capacitor voltage is the same as the source voltage ππ . 1.5. Step Response of Series RLC Circuits ο Thus the complate solutionβ¦ Example 1.6. For the circuit in Fig., find π£ π‘ πππ π(π‘) for t>0. Consider these cases: R=5β¦, R=4β¦,R=1β¦. Example 1.6. Example 1.6. ο π£π is the forced response or steady-state response. ο It is final value of the capacitor voltage. ο π£π =24 V. For t>0 , current i, Example 1.6. ο Finally, ο but we have to solve i(t)β¦ 1.6. Step Response of Parallel RLC Circuits Example 1.7. Solution: ο For t<0, the switch is open ο The circuit partitioned into two independent subcircuits. Capacitor voltage equal the voltage of 20β¦ resistor. Example 1.7. ο For t>0, the switch is closed ο We have parallel RLC circuit. ο The voltage source is off or short-circuited. Example 1.7. The final value of Iβ¦ Using initial conditions we get π¨π and π¨π Example 1.7. From i(t) we obtain v(t)β¦