1
GATE ECE 2004
MCQ (Single Correct Answer)
+1
-0.3

The equivalent inductance measured between the terminals 1 and 2 for the circuit shown in figure, is

GATE ECE 2004 Network Theory - Network Elements Question 36 English
A
L1 + L2 + M
B
L1 + L2 - M
C
L1 + L2 + 2M
D
L1 + L2 - 2M
2
GATE ECE 2002
MCQ (Single Correct Answer)
+1
-0.3

The dependent current source shown in Figure

GATE ECE 2002 Network Theory - Network Elements Question 38 English
A
Delivers 80 W
B
Absorbs 80 W
C
Delivers 40 W
D
Absorbs 40 W
3
GATE ECE 2002
MCQ (Single Correct Answer)
+1
-0.3

The differential equation for the current i(t) in the circuit of Fig. is

GATE ECE 2002 Network Theory - Network Elements Question 37 English
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
$$\frac{\operatorname d^2\mathrm i}{\operatorname d\mathrm t^2}+\;2\;\frac{\operatorname d\mathrm i}{\mathrm{dt}}\;+\;2\mathrm i(\mathrm t)\;=\sin\left(\mathrm t\right)$$
4
GATE ECE 2001
MCQ (Single Correct Answer)
+1
-0.3

The voltage e0 in Fig. is

GATE ECE 2001 Network Theory - Network Elements Question 39 English
A
2 V
B
$$\frac43\;\mathrm V$$
C
4 V
D
8 V
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