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state source free rg circuit ? and explain beiefly? and prof the therom?

raju
201 Points
5 years ago
A source-free RC circuit occurs when its dc source is suddenly discon-
nected. The energy already stored in the capacitor is released to the
resistors.
Consider a series combination of a resistor and an initially charged
capacitor,  (The resistor and capacitor may be the
equivalent resistance and equivalent capacitance of combinations of re-
sistors and capacitors.) Our objective is to determine the circuit response,
which, for pedagogic reasons, we assume to be the voltage v(t) across
the capacitor. Since the capacitor is initially charged, we can assume that
at time t = 0, the initial voltage is
v(0) = V 0
with the corresponding value of the energy stored as
1 CV 0 22
Applying KCL at the top node of the circuit in
w(0) = i C + i R = 0
By definition, i C = C dv/dt and i R = v/R. Thus,
C dvv+ = 0dtRordvv+= 0dtRC
This is a first-order differential equation, since only the first derivative of
v is involved. To solve it, we rearrange the terms as
dv1=−dtvRC
Integrating both sides, we get
tln v = −+ ln ARC
where ln A is the integration constant. Thus,
vtln = −ARC
Taking powers of e produces
v(t) = Ae −t/RC
But from the initial conditions, v(0) = A = V 0 . Hence,
v(t) = V 0 e −t/RC
This shows that the voltage response of the RC circuit is an exponential
decay of the initial voltage. Since the response is due to the initial energy
stored and the physical characteristics of the circuit and not due to some
external voltage or current source, it is called the natural response of the
circuit.
The natural response of a circuit refers to the behavior (in terms of voltages and
currents) of the circuit itself, with no external sources of excitation.
The natural response is illustrated graphically in Fig.Note that at
t = 0, we have the correct initial condition as  As t increases,
the voltage decreases toward zero. The rapidity with which the voltage
decreases is expressed in terms of the time constant, denoted by the lower