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24.09.2013 1.4. The Source‐Free Parallel RLC Circuits Now let us look at parallel forms of RLC networks Assume the initial current and voltage to be: Applying KCL at the top node gives; takig derivative with respect to t and dividing by C results in; 1.4. The Source‐Free Parallel RLC Circuits s From this, we can find the roots of the characteristic equation to be: 1 24.09.2013 1.4. The Source‐Free Parallel RLC Circuits As in last time, there are three scenarios to consider. Overdamped Case (∝ negative and real Critically Damped Case (∝ real and equal 1.4. The Source‐Free Parallel RLC Circuits Under Damped Case (∝ arecomplex To get the values for the constants, we need to know v(0) and dv(0)/dt. canbedeterminedfrominitialconditions; 2 24.09.2013 1.4. The Source‐Free Parallel RLC Circuits The voltage waveforms will be similar to those shown for the series network. Note that in the series network, we first found the inductor current and then solved for the rest from that. Here we start with the capacitor voltage and similarly, solve for the other variables from that. Example 1.5. In the parallel circuit of fig. , find assuming 0 5 , 0 0, 1 for t>0, 10 . Consider these cases; R=1.923 Ω, R=5 Ω, R=6.25 Ω. İf R=1.923 Ω 3 24.09.2013 Example 1.5. Since (∝ in this case, overdamped. The roots of the characteristic equation are; Apply initial conditions to get and At t =0 Example 1.5. Must be differeantiated At t =0 With and the solution gets… 4 24.09.2013 Example 1.5. When R=5 Ω remains 10; Since α= , the responce is Critically damped… Apply initial conditions to get and Example 1.5. Must be differeantiated With and the solution gets… 5 24.09.2013 Example 1.5. İn the last case R=6.25 Ω… Solution is… Response for three degrees of damping 1.5. Step Response of Series RLC Circuits Now let us consider what happens when a DC voltage is suddenly applied to a second order circuit. Consider the circuit shown. The switch closes at t=0. 6 24.09.2013 1.5. Step Response of Series RLC Circuits This equation has two compenants; This is similar to the response for the source free version of the series circuit, except the variable is different. Natural response; Forced response; The solution to this equation is a combination of transient response and steady state 1.5. Step Response of Series RLC Circuits İf we set 0, contains only natural response ( (t) can be expressed as three conditions; The forced response is the steady‐state or final value of The final value of the capacitor voltage is the same as the source voltage . 7 24.09.2013 1.5. Step Response of Series RLC Circuits Thus the complete solutions for the three conditions of damping are: The variables A1 and A2 are obtained from the initial conditions, v(0) and dv(0)/dt. Example 1.6. For the circuit in Fig., find for t>0. Consider these cases: R=5Ω, R=4Ω,R=1Ω. 8 24.09.2013 Example 1.6. Example 1.6. is the forced response or steady‐state response. It is final value of the capacitor voltage. =24 V. For t>0 , current i, 9 24.09.2013 Example 1.6. Finally, but we have to solve i(t)… 1.6. Step Response of Parallel RLC Circuits 10 24.09.2013 Example 1.7. Solution: For t<0, the switch is open The circuit partitioned into two independent subcircuits. Capacitor voltage equal the voltage of 20Ω resistor. Example 1.7. For t>0, the switch is closed We have parallel RLC circuit. The voltage source is off or short‐circuited. 11 24.09.2013 Example 1.7. The final value of I… Using initial conditions we get and Example 1.7. From i(t) we obtain v(t)… 12 24.09.2013 10 9 8 7 6 5 4 3 2 1 0 -1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 13