Final answer:
The capacitor voltage after charging for one time constant will be approximately 6.32V, and the stored energy at this time is 40% of the maximum energy.
Step-by-step explanation:
The energy stored in a capacitor is given by the formula E = (1/2)CV², where E is the energy, C is the capacitance, and V is the voltage across the capacitor. In this scenario, the capacitor is charged through a resistor for a time equal to one time constant. The time constant is given by the formula τ = RC, where R is the resistance and C is the capacitance. After charging for one time constant, the voltage across the capacitor will be approximately 0.632 times the maximum voltage. So, if the power supply is set to 10V, the capacitor voltage after one time constant will be approximately 6.32V.
The stored energy at this time can be calculated using the formula E = (1/2)CV². The maximum energy that can be stored in the capacitor is obtained when it is fully charged, so the maximum energy is Emax = (1/2)CV²max. As a fraction of the maximum energy, the stored energy at this time is E/Emax = (6.32V)²/(10V)², which simplifies to 0.4 or 40%.