1
GATE ECE 2003
+2
-0.6
The current flowing through the resistance R in the circuit in figure has the form P cos 4t, where P is
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 2003
+2
-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
A
$$10\cos \left( {t + {{55}^0}} \right) + 10\,\cos \left( {2t + {{10}^0} + {{\tan }^{ - 1}}\,2} \right)$$
B
$$10\cos \left( {t + {{55}^0}} \right) + 10\sqrt {{3 \over 2}} \,\cos \left( {2t + {{55}^0}} \right)$$
C
$$10\cos \left( {t - {{35}^0}} \right) + 10\cos \left( {2t + {{10}^0} - {{\tan }^{ - 1}}\,2} \right)$$
D
$$10\cos \left( {t - {{35}^0}} \right) + 10\sqrt {{3 \over 2}} \,\cos \left( {2t - {{35}^0}} \right)$$
3
GATE ECE 2002
+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
A
$$90\angle32.44^\circ$$
B
$$80\angle32.44^\circ$$
C
$$80\angle-32.44^\circ$$
D
$$90\angle-32.44^\circ$$
4
GATE ECE 2001
+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
A
Fig. (i)
B
Fig. (ii)
C
Fig. (iii)
D
Fig. (iv)
EXAM MAP
Medical
NEETAIIMS