Use the differential equation approach to find v(t) for t>0...

Key Concepts

Initial Conditions

The solution of the differential equation requires initial conditions, which usually include the initial voltage across a capacitor or the initial current through an inductor. These conditions are critical for fully determining the particular solution to the circuit's transient response.

Transient Response

The transient response of a circuit refers to how the circuit variables (such as voltage or current) change from an initial state to a steady state following a disturbance, such as a switch action. It is typically characterized by exponential functions derived from the solutions of the differential equations.

Time Constant

The time constant is a parameter that quantifies the rate at which the transient response decays or grows. It is determined by the values of the resistive and reactive components in the circuit and directly influences the exponential behavior of the circuit’s response.

Kirchhoff's Laws

Kirchhoff's voltage and current laws are fundamental in circuit analysis. They are used to set up the equations that describe the conservation of energy and charge within a circuit, which ultimately lead to the formulation of a differential equation governing the circuit's behavior.

Differential Equations in Circuit Analysis

In circuits with energy storage elements such as capacitors and inductors, the dynamic behavior is modeled through differential equations. These equations capture the time-dependent relationships between voltages and currents, allowing for the analysis of transient responses.

Transcript

00:01 Question number is 7 .3.

00:08 Here according to the given we will redraw circuit for our own convenience and the circuit is here r1 is equal to 6 kilo on node r3 is equal to 6 kilo home and like this here here here there is a capacitor c is equal to 50 micro -ferradi this note to this like this plus minus voltage source of 12 volt t is equal to zero and here is r2 is equal to 6 kilo home and here is an other resistance r4 is equal to 6 kilo hom and this circuit is closed here while plus voltage source plus minus b0 into t is this one okay before we begin our analysis we know that resistors r3 r4 resistors are 3 or 3 and r4 are in parallel so we first reduce the network to dead like this another diagram r1 which is 6 kilo -oom this is the resistance and like this negative v1 positive plus vc negative here is the capacitor c c is equal to 50 microferradi another resistance r2 is equal to 6 pheom like this and here like this t is equal to 0 this one and here is the voltage source 12 volt source which were combined at this one and here plus negative v0 into t remains same okay now that the network has simplified we begin our analysis so according to this one we will draw an other diagram r1 is equal to six kilo form potential difference v1 and 2 plus minus v1 and 2 plus minus capacitor potential difference and this and this this voltage source plus minus 12 volt and it joins here here r is equal to 3k and this comes here here is r2 which is also 6 kilo hound now now we will this is a close switch and we will now draw an open switch.

04:57 R1 is equal to 6 kilo resistance and this one negative positive resistance r2 plus minus v0 at the approaches to z plus plus side and other resistance voltage source closed circuit now this one here the circuit is open plus minus 12 volt source and it combines here this all here this is is equal to 3k and this r2 is equal to 6k and finally we will get 6k and this open one r t h this one right this 3k resistance 6k resistance and this here we have simplified our diagram as we could now we will rename this as a this as b figure this as c figure this is as d figure and this is as d figure and this is is as if we have.

06:55 Now here is the first step.

07:14 Step 1 v0 into t is equal to k1 plus k2 into e raise to per minus t over top...

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