Laplace transform analysis gives
The Laplace transform of the derivative of the signal f(t) = e^{t} u(t) is
The derivative of function f(t) is
f'(t) = e^{t}δ (t) + u (t) (e^{t})
= e^{t} u(t) + e^{t}δ(t)
For a series R  C circuit excited by a dc voltage of 20 V, and with timeconstant τ seconds, the voltage across the capacitor at time t = τ is given by
V_{c}(t) = V(1  e^{t/τ}) volts
V = 20 volt (given)
At t = τ, V_{c}_{t = τ} = 20 (1  e^{1}) volts
The Laplacetransform equivalent of a given network will have 7/5 F capacitor replaced by 5
An RC circuit has a capacitor C = 2 μF in series with a resistance R = 1 MΩ The time of 6 secs will be equal to
Time constant of series RC circuit is
τ = RC = 1 x 10^{6} x 2 x 10^{6}
= 2 secs
Hence, time of 6 seconds will be equal to 3τ = 3 times of time constant.
The Laplace transform of the gate function shown below is
From the given waveform, we have
∴
Assuming initial condition to be zero, the current i(t) in the circuit shown below is
The given circuit in Laplace domain is shown below.
Here
(Using current divider theorem)
or, i(t) = 0.25 e^{05t} u(t)
A 10 volts step voltage is applied across a RC series circuit at t = 0 having R = 100 Ω, and C = 100 μF.
The values of i(t) and di(t)/dt at t = 0 are respectively given by
At t = 0, capacitor acts as shortcircuit.
So,
Time constant of given circuit is
τ = RC = 100 x 100 x 10^{6}
= 10^{2} sec = 0.01 sec
With zero initial condition, current in the series RC circuit is
Hence,
di/dt = 10 e^{100t}
i(0^{+}) = 0.1 A
If the input voltage V_{1}(t) = 10 e^{2t} V then, the output voltage V_{2}(t) for the circuit shown below is
Given circuit is a lowpass filter having transfer function
When, V_{1}(t) = 10e^{2t}
In the circuit shown below, the switch 'S' is opened at t = 0. Prior to that, switch was closed.
The current i(t) is at t = 0^{+ }is
For t = 0^{}, switch Sis closed. Hence, in steady state capacitor will act as open circuit.
Thus, V_{0}(0^{}) = V_{C}(0^{+}) = 2 volts
The circuit at t = 0^{+} is shown below where,
cananitor ants as a constant voltage source.
Using source transformation,
and
Hence,
A rectangular pulse of duration T and magnitude A has the Laplace transform
∴
The time constant of the circuit shown below is
R_{eg} across inductor
(5 A C.S open circuited)
∴ Time constant of RL circuit is
Assertion (A): Laplace transform is preferred for solving networks involving higher order differential equations.
Reason (R): The classical method for solving differential equations of higher order is quite cumbersome.
Assertion (A): In a purely capacitive circuit, the current wave is more distorted than the voltage wave.
Reason (R): The harmonics in the current wave are increased in proportion to their frequency numbers.
The time constant of the circuit shown below is
For finding time constant of the circuit 5 A current source is open circuited and 2 V voltage source is short circuited.
C_{eq }= C
∴
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