Sunday, May 20, 2012

ECIL Previous year Papers

ECIL (Electronics Corporation Of India Ltd.) GET Examination conducted for the recruitment of the  engineers in the various deparments.
 Previous papers of this Exam is given below...

Electronic and communication 2010
1.The current I in the given network.
a) 1A
b) 3A
c) 5A
d) 7A

2.For the Delta‐ Wye transformation in given figure, the value of the resistance R is.
a) 1/3 ohms
b) 2/3 ohms
c) 3/2 ohms
d) 3 ohms

3.In the given network, the Thevenin’s equivalent as seen by the load resistance Rl is
a) V=10 V, R= 2ohms
b) V=10V, R=3 ohms
c) V=15V, R= 2ohms
d) V=15V, R=3 ohms

4.The current I in a series R‐L circuit with R=10 ohms and L=20mH is given by i=2sin500t A. If v is the voltage across the R‐L combination then i
a) lags v by 45 degree
b) is in‐phase with v
c) leads v by 45
d) lags v by 90

5.In thr given network, the mesh current I and the input impedance seen by the 50 V source, respectively, are
a) 125/13 A and 11/8 ohms
b) 150/13 A and 13/8 ohms
c) 150/13 A and 11/8 ohms
d) 125/13 A and 13/8 ohms

6.A voltage sourcehaving a source impedance Z = R + jX can deliver maximum Average power to a load impedance Z, when
a) Z = R + jX
b) Z = R
c) Z = jX
d) Z = R –jX

7.In the given circuit, the switch S is closed at t=0. Assuming that there is no initial Charge in the capacitor, the current i(t) for t>0 is
a) V/R e^ (‐2t/RC)
b) V/R e^ (‐t/RC)
c) V/2R e^ (‐2t/RC)
d) V/2R e^ (‐t/RC)
Click here for full document

No comments:

Post a Comment