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The ratio of the transfer function Io/Is is
Io/Is = (s + 4)/ (s + 4 + 3/s) = s(s + 4)/(s + 1)(s + 3).
The voltage across 200 μF capacitor is given by
The steady state voltage across capacitor is
The transformed voltage across the 60 μF capacitor is given by
The initial current through capacitor is
⇒
The current through an 4 H inductor is given by
The initial voltage across inductor is
The amplifier network shown in fig.is stable if
⇒ s^{2} + (6  2K) s + 1 = 0
(6  2K) > 0 ⇒ K < 3
Let v_{1} be the node voltage of middle node
A circuit has a transfer function with a pole s = 4 and a zero which may be adjusted in position as s= a The response of this system to a step input has a term of form Ke^{ 4t}. The K will be (H= scale factor)
A circuit has input v_{in }(t) = cos 2t u(t) V and output i_{o }(t) = 2sin 2t u(t) A. The circuit had no internal stored energy at t = 0. The admittance transfer function is
A two terminal network consists of a coil having an inductance L and resistance R shunted by a capacitor C. The poles of the driving point impedance function Z of this network are at and zero at 1. If Z(0) =1the value of R, L, C are
The current response of a network to a unit step input is
The response is
The characteristic equation is s^{2} (s^{2} + 11s + 30) = 0 ⇒ s^{2} (s + 6)(s + 5) = 0
s = 6, 5, Being real and unequal, it is overdamped.
The circuit is shown in fig.
The current ratio transfer function I_{o}/I_{S IS}
The circuit is shown in fig.
The response is
The characteristic equation is (s+1 (s+3) = 0. Being real and unequal root, it is overdamped response.
The circuit is shown in fig.
If input i_{s} is 2u(t) A, the output current i_{o} is
In the network of Fig. , all initial condition are zero. The damping exhibited by the network is
The roots are imaginary so network is underdamped
The voltage response of a network to a unit step input is
The response is
The characteristic equation is s(s^{2} + 8s + 16) = 0 , (s + 4)^{2} = 0, s = 4, 4
Being real and repeated root, it is critically damped.
The response of an initially relaxed circuit to a signal v_{s} is e^{2t }u(t). If the signal is changed to , the response would be
⇒
⇒
⇒
Consider the following statements in the circuit shown in fig.
1. It is a first order circuit with steady state value of v_{C}= 10/3, i = 5/3A
2. It is a second order circuit with steady state of v_{C }= 2 V , i = 2 A
3. The network function V(s)/I(s)has one pole.
4. The network function V(s)/I(s) has two poles.
It is a second order circuit. In steady state
It has one pole at s = 2
The singularity near to origin is pole. So it may be RC impedance or RL admittance function.
Poles and zero does not interlace on negative real axis so it is not a immittance function
The singularity nearest to origin is a zero. So it may be RL impedance or RC admittance function. Because of (D) option it is required to check that it is a valid RC admittance function. The poles and zeros interlace along the negative real axis. The residues are real and positive.
22 docs274 tests

22 docs274 tests
