GATE ECE 2012 | GATE ECE Year Wise Previous Years Questions

GATE ECE 2012

MCQ (Single Correct Answer)

-0.3

In the circuit shown below, current through the inductor is

GATE ECE 2012 Network Theory - Network Elements Question 34 English

$$\frac2{1\;+\;\mathrm j}\mathrm A$$

$$\frac{-1}{1\;+\;\mathrm j}\mathrm A$$

$$\frac1{1\;+\;\mathrm j}\mathrm A$$

0 A

GATE ECE 2012

MCQ (Single Correct Answer)

-0.6

If V_A - V_B = 6 V then V_C - V_D is

GATE ECE 2012 Network Theory - Network Elements Question 18 English

-5 V

2 V

3 V

6 V

GATE ECE 2012

MCQ (Single Correct Answer)

-0.6

Assuming both the voltage sources are in phase the value of R for which maximum power is transferred from circuit A to circuit B is GATE ECE 2012 Network Theory - Network Theorems Question 21 English

GATE ECE 2012 Network Theory - Network Theorems Question 21 English

$$0.8\Omega$$

$$1.4\Omega$$

$$2\Omega$$

$$2.8\Omega$$

GATE ECE 2012

MCQ (Single Correct Answer)

-0.6

With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed:
(i) 1Ω connected at port B draws a current of 3 A
(ii) 2.5Ω connected at port B draws a current of 2 A
GATE ECE 2012 Network Theory - Network Theorems Question 20 English

GATE ECE 2012 Network Theory - Network Theorems Question 20 English

For the same network, with 6 V dc connected at port A, 1Ω connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is

6 V

7 V

8 V

9 V

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