
Two capacitors $ C_1 $ and $ C_2 $ are connected in parallel to a battery. The charge-time graph shows the charge on each capacitor as a function of time.
From the graph, we can see that $ C_1 $ reaches a higher charge value than $ C_2 $ as $ t \to \infty $.
Since the capacitors are connected in parallel, they have the same voltage $ V $ across them.
We know that $ Q = CV $, where $ Q $ is the charge, $ C $ is the capacitance, and $ V $ is the voltage.
Since $ C_1 $ has a higher charge $ Q_1 > Q_2 $ at the same voltage $ V $, it must have a larger capacitance $ C_1 > C_2 $.
The energy stored in a capacitor is given by $ U = \frac{1}{2} CV^2 $.
Since $ C_1 > C_2 $ and both capacitors have the same voltage $ V $, the energy stored in $ C_1 $ is greater than the energy stored in $ C_2 $, i.e., $ U_1 > U_2 $.
Thus, we have $ C_1 > C_2 $ and $ U_1 > U_2 $.
Final Answer:
The final answer is $ \ C_1 > C_2,\ U_1 > U_2 $.
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged.
Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor \( \frac{R}{\sqrt{R^2 + \omega^2 L^2}} \), where \( \omega \) is the frequency of the supply across resistor \( R \) and inductor \( L \). If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage.
In light of the above statements, choose the most appropriate answer from the options given below:
The output of the circuit is low (zero) for:

(A) \( X = 0, Y = 0 \)
(B) \( X = 0, Y = 1 \)
(C) \( X = 1, Y = 0 \)
(D) \( X = 1, Y = 1 \)
Choose the correct answer from the options given below:
