- OLTF G(s)={s-2}/{s+2} is a
G(s)= (s - 2)/(s + 2) denotes a zero at s = 2 and a pole at s = - 2. So there exists a pole and a zero one at left half of s plane and other at right half of s plane. They denote symmetrical mirror images all pass filter.
- When compared a 1st order LPF a 2nd order LPF has
2nd order LPF low pass filter have higher cut off frequency than 1st order low pass filter.
- In active filter which element is absent
In active filter inductor is absent which are bulky and expensive at lower frequency.
- Advantage of active filter is
Active filters offers gain and it is also easy to tune. It derives low impedance loads. These are the advantages of active filter over passive.
- Disadvantages of constant k type filter
In m derived filter impedance is constant throughout passband and it is possible to get very rapid attenuation rise in stopband and beyond cut off frequency.
- Piezoelectric effect is carried out in
In piezoelectric effect when a mechanical strain is applied to one face of a suitably cut face of a piezoelectric crystal, it causes an emf to be developed in opposite surface of that piece. Reverse is also true.
- In Cauer 1 form last element in the network is
In cauer form 1 at s tends to zero, if Z(s) is zero and ω = 0 i.e for s tends to zero , the inductive path is conductive. Thus the last element is an inductor.
- If L & C are 4 mH & 0.0001 μF respectively a electric current chop of magnitude 50 Amp would induced a voltage
e = √(L / C)= 50 × √{(40 × 10
^{-3})/(0.001 × 10^{-6})} = 100 × 10^{3}= 100 KV - If the percentage reactance of an element is 20 % & the full load electric current is 50 Amp, the short circuit electric current will be
We know that I
_{sc}= I ×(100 / % X) = 50 ×(100 / 20 ) = 250 Amp - A electric current in a circuit is given by I(s) = (2s+8) / (s
^{2}+ 4s + 12). If the electric current flows through the 5 ohms resistor, find power dissipated at t = 0?i(0+) = lim(s → ∞) SI(s) or lim(s → ∞ ) s(2s + 8) / (s^{2}+ 4s + 12) = 2 Amp.

Power dissipated is [i(0 + )]^{2}× 5 = 20 Watts. - A electric current of {2 + √2sin( 214t + 30) + 2√2cos(952t + 45)} is measured with a thermocouple, 5 A full scale meter, what is the meter reading?
RMS value = √(2
^{2}+ 2/2 + 2^{2}× 2/2) = 3 A. - A circuit with resistor, inductor, capacitor in series is resonant of 50 Hz. If all the values are now doubled, the new resonant frequency is
f
_{0}= 1 / [ 2π√(L × C)] and f_{1}= 1 / [ 2π√(2L × 2C)], f_{0}/ f_{1}= 2, f_{1}= f_{0}/ 2. - Can a 250 V, 5 A single way switch be used in place of a 250 V, 15 A Switch?
No. The contact strip will melt if electric current more than 5 A flows in 250 V, 5 A rated switch .
- A periodic voltage having the fourier series v(t)= 1 + 4sinωt + 2cosωt volts is applied across a one ohm resistor. The power dissipated in the one ohm resistor is
V
_{rms}= √(1^{2}+ 4^{2}/ 2 + 2^{2}/ 2) = √11 V.

P = V^{2}/ R = 11 W. - For the resonance circuit ω
_{0}= 10^{5}, Q = 50, R = 400 Ω the value of C is-ω_{0}/ R = 50, L = 50 × 400 / 10^{5}= 0.2, C = 1/ ω_{0}^{2}× L = 1/(0.2) × 10^{10}. - An RLC series circuit resonates at a frequency w
_{r}the ratio of w_{r}L/R = 10 the variable frequency voltage applied to the circuit is 20 sin (ω t + π/3)the voltage measured across the capacitance - An ac voltage of 200 V at 50Hz is applied to a coil which draws 5 amp and dissipates 1000 W. the resistance and impedance of the coil respectively are
P = 1000 W ⇒ I
^{2}× R = 1000.

5^{2}× R = 1000 ⇒ R = 1000 / 25 = 40 ohm. - In what connection we get neutral
Star connection is a method of connecting 3 phase circuit in such a way that one end of each phase of 3 phases are connected together thus forming a common star point called neutral.
- What is the relation between line voltage & phase voltage in case of delta connection?
- A capacitor of 50 microfarad is connected in series with a resistance of 120 ohms. If above circuit is connected across 240 V,50 Hz 1-φ supply find capacitive reactance?
X
_{c}= 1/ ( 2πfC ) = 1 / ( 2 × 3.14 × 50 × 50 × 10^{-6}= 63.7 ohms.

Design with by SARU TECH