GATE ECE 2003 | Sinusoidal Steady State Response Question 38 | Network Theory

GATE ECE 2003

MCQ (Single Correct Answer)

-0.6

An input voltage $$v(t)$$ $$ = 10\sqrt 2 \,\,\cos \,\,\left( {t + {{10}^0}} \right) + 10\sqrt 5 \,\,\cos \left( {2t + {{10}^0}} \right)\,\,V$$ is applied to a series combination of resistance $$L = 1H$$. the resulting steady - state current $$i(t)$$ in ampere is

$$10\cos \left( {t + {{55}^0}} \right) + 10\,\cos \left( {2t + {{10}^0} + {{\tan }^{ - 1}}\,2} \right)$$

$$10\cos \left( {t + {{55}^0}} \right) + 10\sqrt {{3 \over 2}} \,\cos \left( {2t + {{55}^0}} \right)$$

$$10\cos \left( {t - {{35}^0}} \right) + 10\cos \left( {2t + {{10}^0} - {{\tan }^{ - 1}}\,2} \right)$$

$$10\cos \left( {t - {{35}^0}} \right) + 10\sqrt {{3 \over 2}} \,\cos \left( {2t - {{35}^0}} \right)$$

GATE ECE 2003

MCQ (Single Correct Answer)

-0.6

The current flowing through the resistance R in the circuit in figure has the form P cos 4t, where P is GATE ECE 2003 Network Theory - Sinusoidal Steady State Response Question 56 English

GATE ECE 2003 Network Theory - Sinusoidal Steady State Response Question 56 English

(0.18 + j 0.72)

(0.46 + j 1.90)

-(0.18 + j 1.90)

-(0.192 + j 0.144)

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 2001

MCQ (Single Correct Answer)

-0.6

When the angular frequency $$\omega $$ in Fig. is varied from $$0$$ to $$\infty $$ the locus of the current phasor $${{\rm I}_2}$$ is given by GATE ECE 2001 Network Theory - Sinusoidal Steady State Response Question 31 English 1