# Objective Questions on Circuit Theory | Page – 4

1. 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.

2. 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.

3. In active filter which element is absent

In active filter inductor is absent which are bulky and expensive at lower frequency.

4. 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.

5. 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.

6. 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.

7. 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.

8. 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 × 103 = 100 KV

9. 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 Isc = I ×(100 / % X) = 50 ×(100 / 20 ) = 250 Amp

10. A electric current in a circuit is given by I(s) = (2s+8) / (s2 + 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) / (s2 + 4s + 12) = 2 Amp.
Power dissipated is [i(0 + )]2 × 5 = 20 Watts.

11. 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 = √(22 + 2/2 + 22 × 2/2) = 3 A.

12. 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

f0 = 1 / [ 2π√(L × C)] and f1 = 1 / [ 2π√(2L × 2C)], f0 / f1 = 2, f1 = f0 / 2.

13. 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 .

14. 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

Vrms = √(12 + 42 / 2 + 22 / 2) = √11 V.
P = V2/ R = 11 W.

15. For the resonance circuit ω0 = 105, Q = 50, R = 400 Ω the value of C is-

ω0 / R = 50, L = 50 × 400 / 105 = 0.2, C = 1/ ω02 × L = 1/(0.2) × 1010.

16. An RLC series circuit resonates at a frequency wr the ratio of wr L/R = 10 the variable frequency voltage applied to the circuit is 20 sin (ω t + π/3)the voltage measured across the capacitance

Voltage across inductor VL = VXQ. Where V is rms of applied voltage and Q factor of the coil.
Vm = 20 V, ωr L / R = Q = 10
VL = (20 / √2) × 10 = 200 / √2 V.

17. 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 ⇒ I2 × R = 1000.
52 × R = 1000 ⇒ R = 1000 / 25 = 40 ohm.

18. 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.

19. What is the relation between line voltage & phase voltage in case of delta connection?

Line voltage is equal to phase voltage in case of delta connection.

20. 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.