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 𝑑 2 𝑑𝑡 2 𝑑 𝑑𝑡 s 𝑠 2 From this, we can find the roots of the characteristic equation to be:

1.4. The Source-Free Parallel RLC Circuits As in last time, there are three scenarios to consider. Overdamped Case (∝> 𝝎 𝟎 ) 𝑠 1 𝑎𝑛𝑑 𝑠 2 negative and real Critically Damped Case (∝= 𝝎 𝟎 ) 𝑠 1 𝑎𝑛𝑑 𝑠 2 real and equal

1.4. The Source-Free Parallel RLC Circuits Under Damped Case (∝< 𝝎 𝟎 ) 𝑠 1 𝑎𝑛𝑑 𝑠 2 are complex To get the values for the constants, we need to know v(0) and dv(0)/dt. 𝐴 1 𝑎𝑛𝑑 𝐴 2 can be determined from initial conditions;

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 𝑣(𝑡) for t>0, assuming 𝑣 0 =5 𝑉, 𝑖 0 =0, 𝐿=1𝐻 𝑎𝑛𝑑 𝐶=10 𝑚𝐹. Consider these cases; R=1.923 Ω, R=5 Ω, R=6.25 Ω. İf R=1.923 Ω

Example 1.5. At t =0 Since (∝> 𝝎 𝟎 ) in this case, overdamped. The roots of the characteristic equation are; Apply initial conditions to get 𝑨 𝟏 and 𝑨 𝟐 At t =0

Example 1.5. At t =0 Must be differeantiated With 𝑨 𝟏 and 𝑨 𝟐 the solution gets…

Example 1.5. When R=5 Ω 𝜔 0 remains 10; Since α= 𝜔 0 , the responce is Critically damped… Apply initial conditions to get 𝑨 𝟏 and 𝑨 𝟐

Example 1.5. Must be differeantiated With 𝑨 𝟏 and 𝑨 𝟐 the solution gets…

Example 1.5. İn the last case R=6.25 Ω… Solution is… Response for three degrees of damping Cevap yanlıs bak

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.

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. The solution to this equation is a combination of transient response and steady state Natural response; 𝒗 𝒏 (𝒕) Forced response; 𝒗 𝒇 (𝒕)

1.5. Step Response of Series RLC Circuits İf we set 𝑉 𝑠 =0, 𝑣(𝑡) contains only natural response ( 𝑣 𝑛 ) 𝑣 𝑛 (t) can be expressed as three conditions; The forced response is the steady-state or final value of 𝑣(𝑡) The final value of the capacitor voltage is the same as the source voltage 𝑉 𝑠 .

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 𝑣 𝑡 𝑎𝑛𝑑 𝑖(𝑡) for t>0. Consider these cases: R=5Ω, R=4Ω,R=1Ω.

Example 1.6.

Example 1.6. 𝑣 𝑓 is the forced response or steady-state response. It is final value of the capacitor voltage. 𝑣 𝑓 =24 V. Take derivative of v(t) For t>0 , current i,

Example 1.6. Finally, but we have to solve i(t)…

1.6. Step Response of Parallel RLC Circuits The same treatment given to the parallel RLC circuit yields the same result.

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.

Example 1.7. The final value of I… Using initial conditions we get 𝑨 𝟏 and 𝑨 𝟐

Example 1.7. From i(t) we obtain v(t)…