1
AIEEE 2010
MCQ (Single Correct Answer)
+4
-1
In a series $$LCR$$ circuit $$R = 200\Omega$$ and the voltage and the frequency of the main supply is $$220V$$ and $$50$$ $$Hz$$ respectively. On taking out the capacitance from the circuit the current lags behind the voltage by $${30^ \circ }.$$ On taking out the inductor from the circuit the current leads the voltage by $${30^ \circ }.$$ The power dissipated in the $$LCR$$ circuit is
A
$$305$$ $$W$$
B
$$210$$ $$W$$
C
$$zero$$ $$W$$
D
$$242$$ $$W$$
2
AIEEE 2010
MCQ (Single Correct Answer)
+4
-1
In the circuit shown below, the key $$K$$ is closed at $$t=0.$$ The current through the battery is
A
$${{V{R_1}{R_2}} \over {\sqrt {R_1^2 + R_2^2} }}$$ at $$t=0$$ and $${V \over {{R_2}}}$$ at $$t = \infty$$
B
$${V \over {{R_2}}}$$ at $$\,t = 0$$ and $${{V\left( {{R_1} + {R_2}} \right)} \over {{R_1}{R_2}}}$$ at $$t = \infty$$
C
$${V \over {{R_2}}}$$ at $$\,t = 0$$ and $${{V{R_1}{R_2}} \over {\sqrt {R_1^2 + R_2^2} }}$$ at $$t = \infty$$
D
$${{V\left( {{R_1} + {R_2}} \right)} \over {{R_1}{R_2}}}$$ at $$t=0$$ and $${V \over {{R_2}}}$$ at $$t = \infty$$
3
AIEEE 2009
MCQ (Single Correct Answer)
+4
-1
An inductor of inductance $$L=400$$ $$mH$$ and resistors of resistance $${R_1} = 2\Omega$$ and $${R_2} = 2\Omega$$ are connected to a battery of $$emf$$ $$12$$ $$V$$ as shown in the figure. The internal resistance of the battery is negligible. The switch $$S$$ is closed at $$t=0.$$ The potential drop across $$L$$ as a function of time is :
A
$${{12} \over t}{e^{ - 3t}}V$$
B
$$6\left( {1 - {e^{ - t/0.2}}} \right)V$$
C
$$12{e^{ - 5t}}V$$
D
$$6{e^{ - 5t}}V$$
4
AIEEE 2007
MCQ (Single Correct Answer)
+4
-1
In an $$a.c.$$ circuit the voltage applied is $$E = {E_0}\,\sin \,\omega t.$$ The resulting current in the circuit is $$I = {I_0}\sin \left( {\omega t - {\pi \over 2}} \right).$$ The power consumption in the circuit is given by
A
$$P = \sqrt 2 {E_0}{I_0}$$
B
$$P = {{{E_0}{I_0}} \over {\sqrt 2 }}$$
C
$$P=zero$$
D
$$P = {{{E_0}{I_0}} \over 2}$$
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