GATE ECE 2002 | GATE ECE Year Wise Previous Years Questions

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GATE ECE 2002

MCQ (Single Correct Answer)

-0.6

In the network of Figure, the maximum power is delivered to R_L if its value is

GATE ECE 2002 Network Theory - Network Theorems Question 26 English

16 Ω

(40/3) Ω

60 Ω

20 Ω

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

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

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

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

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

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