GATE ECE 2002 | GATE ECE Year Wise Previous Years Questions

GATE ECE 2002

MCQ (Single Correct Answer)

-0.3

In figure, the switch was closed for a long time before opening at t = 0. the voltage V_x at t = 0⁺ is GATE ECE 2002 Network Theory - Transient Response Question 49 English

GATE ECE 2002 Network Theory - Transient Response Question 49 English

25 V

50 V

-50 V

0 V

GATE ECE 2002

MCQ (Single Correct Answer)

-0.6

If the 3-phase balanced source in Fig. delivers 1500 W at a leading power factor of 0.844, then the value of Z_L (in ohm) is approximately GATE ECE 2002 Network Theory - Sinusoidal Steady State Response Question 57 English

GATE ECE 2002 Network Theory - Sinusoidal Steady State Response Question 57 English

$$90\angle32.44^\circ$$

$$80\angle32.44^\circ$$

$$80\angle-32.44^\circ$$

$$90\angle-32.44^\circ$$

GATE ECE 2002

MCQ (Single Correct Answer)

-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

$$2\;\frac{\operatorname d^2\mathrm i}{\operatorname d\mathrm t^2}+\;2\;\frac{\operatorname d\mathrm i}{\mathrm{dt}}\;+\;\mathrm i(\mathrm t)\;=\sin\left(\mathrm t\right)$$

$$\frac{\operatorname d^2\mathrm i}{\operatorname d\mathrm t^2}+\;2\;\frac{\operatorname d\mathrm i}{\mathrm{dt}}\;+\;2\mathrm i(\mathrm t)\;=\cos\left(\mathrm t\right)$$

$$2\;\frac{\operatorname d^2\mathrm i}{\operatorname d\mathrm t^2}+\;2\;\frac{\operatorname d\mathrm i}{\mathrm{dt}}\;+\;\mathrm i(\mathrm t)\;=\cos\left(\mathrm t\right)$$

$$\frac{\operatorname d^2\mathrm i}{\operatorname d\mathrm t^2}+\;2\;\frac{\operatorname d\mathrm i}{\mathrm{dt}}\;+\;2\mathrm i(\mathrm t)\;=\sin\left(\mathrm t\right)$$

GATE ECE 2002

MCQ (Single Correct Answer)

-0.3

The Fourier transform F $$\left\{ {{e^{ - t}}u(t)} \right\}$$ is equal to $${1 \over {1 + j2\pi f}}$$. Therefore, $$F\left\{ {{1 \over {1 + j2\pi t}}} \right\}$$ is

$${e^f}u(f)$$

$${e^{ - f}}$$ u(f)

$${e^f}u(f)$$

$${e^{ - f}}$$u(-f)

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