1
GATE ECE 2002
+1
-0.3

The differential equation for the current i(t) in the circuit of Fig. is A
$$2\;\frac{\operatorname d^2\mathrm i}{\operatorname d\mathrm t^2}+\;2\;\frac{\operatorname d\mathrm i}{\mathrm{dt}}\;+\;\mathrm i(\mathrm t)\;=\sin\left(\mathrm t\right)$$
B
$$\frac{\operatorname d^2\mathrm i}{\operatorname d\mathrm t^2}+\;2\;\frac{\operatorname d\mathrm i}{\mathrm{dt}}\;+\;2\mathrm i(\mathrm t)\;=\cos\left(\mathrm t\right)$$
C
$$2\;\frac{\operatorname d^2\mathrm i}{\operatorname d\mathrm t^2}+\;2\;\frac{\operatorname d\mathrm i}{\mathrm{dt}}\;+\;\mathrm i(\mathrm t)\;=\cos\left(\mathrm t\right)$$
D
$$\frac{\operatorname d^2\mathrm i}{\operatorname d\mathrm t^2}+\;2\;\frac{\operatorname d\mathrm i}{\mathrm{dt}}\;+\;2\mathrm i(\mathrm t)\;=\sin\left(\mathrm t\right)$$
2
GATE ECE 2002
+1
-0.3

The dependent current source shown in Figure A
Delivers 80 W
B
Absorbs 80 W
C
Delivers 40 W
D
Absorbs 40 W
3
GATE ECE 2001
+1
-0.3

The voltage e0 in Fig. is A
2 V
B
$$\frac43\;\mathrm V$$
C
4 V
D
8 V
4
GATE ECE 2000
+1
-0.3

In the circuit of Fig., the value of the voltage source E is A
-16 V
B
4 V
C
-6 V
D
16 V
GATE ECE Subjects
Signals and Systems
Network Theory
Control Systems
Digital Circuits
General Aptitude
Electronic Devices and VLSI
Analog Circuits
Engineering Mathematics
Microprocessors
Communications
Electromagnetics
EXAM MAP
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