1
GATE EE 2012
MCQ (Single Correct Answer)
+1
-0.3
In the circuit shown below, the current through the inductor is GATE EE 2012 Electric Circuits - Network Theorems Question 28 English
A
$$\frac2{1+\mathrm j}\;\mathrm A$$
B
$$\frac{-1}{1+\mathrm j}\;\mathrm A$$
C
$$\frac1{1+\mathrm j}\;\mathrm A$$
D
0 A
2
GATE EE 2011
MCQ (Single Correct Answer)
+1
-0.3
In the circuit given below, the value of R required for the transfer of maximum power to the load having a resistance of 3 Ω is GATE EE 2011 Electric Circuits - Network Theorems Question 30 English
A
zero
B
3 Ω
C
6 Ω
D
infinity
3
GATE EE 2006
MCQ (Single Correct Answer)
+1
-0.3
In the figure the current source is $$1\,\,\angle \,0\,A,$$ $$R = \,1\,\,\Omega ,$$ the impedances are $${Z_C} = - j\,\,\Omega ,$$ and $${Z_L} = 2\,j\,\,\Omega .$$ The Thevenin equivalent looking into the circuit across $$X-Y$$ is. GATE EE 2006 Electric Circuits - Network Theorems Question 10 English
A
$$\sqrt 2 \,\,\angle 0,\,\,V,\,\,\left( {1 + 2j} \right)\,\,\Omega $$
B
$$2\,\,\angle {45^ \circ },\,\,V,\,\,\left( {1 - 2j} \right)\,\,\Omega $$
C
$$2\,\,\angle {45^ \circ },\,\,V,\,\,\left( {1 + j} \right)\,\,\Omega $$
D
$$\sqrt 2 \,\,\angle {45^ \circ },\,\,V,\,\,\left( {1 + j} \right)\,\,\Omega $$
4
GATE EE 2003
MCQ (Single Correct Answer)
+1
-0.3
In the Fig. $${Z_1} = 10\angle - {60^ \circ },\,\,{Z_2} = 10\angle {60^ \circ },\,$$ $${Z_3} = 50\angle {53.13^ \circ }.\,\,$$ Thevenin's impedance seen from $$X-Y$$ is GATE EE 2003 Electric Circuits - Network Theorems Question 17 English
A
$$56.6\,\,\angle {45^ \circ }$$
B
$$60\,\,\angle {30^ \circ }$$
C
$$70\,\,\angle {30^ \circ }$$
D
$$34.4\,\,\angle {65^ \circ }$$
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