1
JEE Main 2015 (Offline)
MCQ (Single Correct Answer)
+4
-1

An $$LCR$$ circuit is equivalent to a damped pendulum. In an $$LCR$$ circuit the capacitor is charged to $${Q_0}$$ and then connected to the $$L$$ and $$R$$ as shown below :

If a student plots graphs of the square of maximum charge $$\left( {Q_{Max}^2} \right)$$ on the capacitor with time$$(t)$$ for two different values $${L_1}$$ and $${L_2}$$ $$\left( {{L_1} > {L_2}} \right)$$ of $$L$$ then which of the following represents this graph correctly ?
$$\left( {plots\,\,are\,\,schematic\,\,and\,\,niot\,\,drawn\,\,to\,\,scale} \right)$$

A
B
C
D
2
JEE Main 2015 (Offline)
MCQ (Single Correct Answer)
+4
-1
An inductor $$(L=0.03$$ $$H)$$ and a resistor $$\left( {R = 0.15\,k\Omega } \right)$$ are connected in series to a battery of $$15V$$ $$EMF$$ in a circuit shown below. The key $${K_1}$$ has been kept closed for a long time. Then at $$t=0$$, $${K_1}$$ is opened and key $${K_2}$$ is closed simultaneously. At $$t=1$$ $$ms,$$ the current in the circuit will be : $$\left( {{e^5} \cong 150} \right)$$

A
$$6.7$$ $$mA$$
B
$$0.67$$ $$mA$$
C
$$100$$ $$mA$$
D
$$67$$ $$mA$$
3
JEE Main 2014 (Offline)
MCQ (Single Correct Answer)
+4
-1
In the circuit shown here, the point $$'C'$$ is kept connected to point $$'A'$$ till the current flowing through the circuit becomes constant. Afterward, suddenly, point $$'C'$$ is disconnected from point $$'A'$$ and connected to point $$'B'$$ at time $$t=0.$$ Ratio of the voltage across resistance and the inductor at $$t=L/R$$ will be equal to :
A
$${e \over {1 - e}}$$
B
$$1$$
C
$$-1$$
D
$${{1 - e} \over e}$$
4
JEE Main 2013 (Offline)
MCQ (Single Correct Answer)
+4
-1
In an $$LCR$$ circuit as shown below both switches are open initially. Now switch $${S_1}$$ is closed, $${S_2}$$ kept open. ($$q$$ is charge on the capacitor and $$\tau$$ $$=RC$$ is Capacitance time constant). Which of the following statement is correct ?
A
Work done by the battery is half of the energy dissipated in the resistor
B
$$t = \,\tau ,\,q = CV/2$$
C
At $$t = \,2\tau ,\,q = CV\left( {1 - {e^{ - 2}}} \right)$$
D
At $$t = \,2\tau ,\,q = CV\left( {1 - {e^{ - 1}}} \right)$$
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