1
GATE EE 2008
MCQ (Single Correct Answer)
+2
-0.6
In the circuit shown in the figure, the value of the current $$i$$ will be given by GATE EE 2008 Electric Circuits - Network Elements Question 8 English
A
$$0.31$$ $$A$$
B
$$1.25$$ $$A$$
C
$$1.75$$ $$A$$
D
$$2.5$$ $$A$$
2
GATE EE 2008
MCQ (Single Correct Answer)
+2
-0.6
Assuming ideal elements in the circuit shown below, the voltage $${V_{ab}}$$ will be GATE EE 2008 Electric Circuits - Network Elements Question 10 English
A
$$-3$$ $$V$$
B
$$0$$ $$V$$
C
$$3$$ $$V$$
D
$$5$$ $$V$$
3
GATE EE 2008
MCQ (Single Correct Answer)
+1
-0.3
The Thevenin's equivalent of a circuit operating at $$\omega = 5\,\,rad/s\,\,{V_{oc}} = 3.71\angle - {15.9^ \circ }$$ and $${Z_0} = 2.38 - j0.667\Omega .$$ At this frequency, the minimal realization of the Thevenin's independence will have a
A
resistor and a capacitance and an inductor
B
resistor and a capacitor
C
resistor and an inductor
D
capacitor and an inductor
4
GATE EE 2008
MCQ (Single Correct Answer)
+1
-0.3
Two $$8$$ $$A$$ $$D$$ $$C$$ $$s$$ one of single slope integrating type and other of successive approximation type. Take $${T_A}$$ and $${T_B}$$ times to connect $$5V$$ analog input signal to equivalent digital output. If the input analog signal is reduced to $$2.5V$$ the approximate time taken by the two $$ADC's$$ will respectively be
A
$${T_A},\,\,{T_B}$$
B
$${\raise0.5ex\hbox{$\scriptstyle {{T_A}}$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}},\,\,{T_B}$$
C
$${T_A},\,\,{\raise0.5ex\hbox{$\scriptstyle {{T_B}}$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}$$
D
$${\raise0.5ex\hbox{$\scriptstyle {{T_A}}$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}},\,\,{\raise0.5ex\hbox{$\scriptstyle {{T_B}}$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}$$
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