Two capacitors \( C_1 \) and \( C_2 \) are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are \( U_1 \) and \( U_2 \), respectively. Which of the given statements is true?
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Capacitance affects charging time and the energy stored; a higher capacitance results in slower charging and more energy storage at the same voltage.
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 $.
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