1
GATE ECE 1990
MCQ (Single Correct Answer)
+2
-0.6
The Op-Amp of fig has a very poor open loop voltage gain of 45 but is otherwise ideal. The gain of the Amplifier equals: GATE ECE 1990 Analog Circuits - Operational Amplifier Question 43 English
A
5
B
20
C
4
D
4.5
2
GATE ECE 1990
MCQ (Single Correct Answer)
+2
-0.6
The CMRR of the differential Amplifier of the fig is equal to GATE ECE 1990 Analog Circuits - Operational Amplifier Question 44 English
A
1000
B
0
C
900
D
1800
3
GATE ECE 1989
MCQ (Single Correct Answer)
+2
-0.6
Refer to Fig. GATE ECE 1989 Analog Circuits - Operational Amplifier Question 18 English
A
For $${V_i} > 0,\,\,{V_0}\, = - {{{R_2}} \over {{R_1}}}{V_i}$$
B
For $${V_i} > 0,\,\,{V_0}\, = 0.$$
C
For $${V_i} < 0,\,\,{V_0}\, = - {{{R_2}} \over {{R_1}}}{V_i}$$
D
For $${V_i} < 0,\,\,{V_0}\, = 0.$$
4
GATE ECE 1988
MCQ (Single Correct Answer)
+2
-0.6
The OP-AMP shown in fig. below is ideal. $$R\, = \,\sqrt {L/C.} $$ The phase angle between V0 and Vi at $$\omega = 1/\sqrt {LC} $$ is GATE ECE 1988 Analog Circuits - Operational Amplifier Question 19 English
A
$$\pi /2$$
B
$$\pi $$
C
$$3\pi /2$$
D
$$2\pi $$
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