1
GATE ECE 2003
MCQ (Single Correct Answer)
+2
-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
A
(0.18 + j 0.72)
B
(0.46 + j 1.90)
C
-(0.18 + j 1.90)
D
-(0.192 + j 0.144)
2
GATE ECE 2002
MCQ (Single Correct Answer)
+2
-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 ZL (in ohm) is approximately GATE ECE 2002 Network Theory - Sinusoidal Steady State Response Question 57 English
A
$$90\angle32.44^\circ$$
B
$$80\angle32.44^\circ$$
C
$$80\angle-32.44^\circ$$
D
$$90\angle-32.44^\circ$$
3
GATE ECE 2001
MCQ (Single Correct Answer)
+2
-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 GATE ECE 2001 Network Theory - Sinusoidal Steady State Response Question 31 English 2 GATE ECE 2001 Network Theory - Sinusoidal Steady State Response Question 31 English 3 GATE ECE 2001 Network Theory - Sinusoidal Steady State Response Question 31 English 4 GATE ECE 2001 Network Theory - Sinusoidal Steady State Response Question 31 English 5
A
Fig. (i)
B
Fig. (ii)
C
Fig. (iii)
D
Fig. (iv)
4
GATE ECE 2000
MCQ (Single Correct Answer)
+2
-0.6
In Fig., the steady state output voltage corresponding to the input voltage $$\left( {3 + 4\sin \,\,100\,t} \right)$$ $$V$$ is GATE ECE 2000 Network Theory - Sinusoidal Steady State Response Question 32 English
A
$$3 + {4 \over {\sqrt 2 }}\sin \left( {100\,t - {\pi \over 4}} \right)\,\,V$$
B
$$3 + 4\sqrt 2 \sin \left( {100\,t - {\pi \over 4}} \right)\,\,V$$
C
$${3 \over 2} + {4 \over {\sqrt 2 }}\sin \left( {100\,t + {\pi \over 4}} \right)\,\,V$$
D
$$3 + 4\sin \left( {100\,t - {\pi \over 4}} \right)\,\,V$$
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