GATE EE 2011 | GATE EE Year Wise Previous Years Questions

GATE EE 2011

MCQ (Single Correct Answer)

-0.6

Solution, the variable $${x_1}$$ and $${x_2}$$ for the following equations is to be obtained by employing the Newton $$-$$ Raphson iteration method
equation (i) $$10\,{x_2}\,\sin \,{x_1} - 0.8 = 0$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$10\,x_2^2\, - 10\,{x_2}\cos \,{x_1} - 0.6 = 0$$
Assuming the initial values $${x_1} = 0.0$$ and $${x_2} = 1.0$$ the Jacobian matrix is

$$\left[ {\matrix{ {10} & { - 0.8} \cr 0 & { - 0.6} \cr } } \right]$$

$$\left[ {\matrix{ {10} & 0 \cr 0 & {10} \cr } } \right]$$

$$\left[ {\matrix{ 0 & { - 0.8} \cr {10} & { - 0.6} \cr } } \right]$$

$$\left[ {\matrix{ {10} & 0 \cr {10} & { - 10} \cr } } \right]$$

GATE EE 2011

MCQ (Single Correct Answer)

-0.3

Circuit turn-off time of an $$SCR$$ is defined as the time

taken by the $$SCR$$ turn of

required for the $$SCR$$ current to become zero

for which the $$SCR$$ is reverse biased by the commutation circuit

for which the $$SCR$$ is reverse biased to reduce its current below the holding current

GATE EE 2011

MCQ (Single Correct Answer)

-0.6

The input voltage given to a converter is
$${V_i} = 100\sqrt 2 \,\,\,\sin \,\,\,\left( {100\pi t} \right)\,\,V$$
The current drawn by the converter is
$${i_i} = 10\sqrt 2 \,\,\,\sin \,\,\,\left( {100\pi t - {\pi \over 3}} \right)\,\, + 5\sqrt 2 $$
$$\sin \left( {300\pi t + {\pi \over 4}} \right)\,\, + \,\,2\sqrt 2 \,\,\sin \left( {500\pi t - {\pi \over 6}} \right)A$$

The input power factor of the converter is

$$0.31$$

$$0.44$$

$$0.5$$

$$0.71$$

GATE EE 2011

MCQ (Single Correct Answer)

-0.6

The active power drawn by the converter is

$$181$$ $$W$$

$$500$$ $$W$$

$$707$$ $$W$$

$$887$$ $$W$$

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