The capacitor is initially uncharged. the switch is moved to position 1 first for 1.5s and then after 1.5s the switch is moved to position 3 and remains there for 3.5s.
What is the maximum instantaneous energy stored in the capacitor during the period from t = 0 to t = 5 s?

how do i go about solving this problem. Does the capacitor reach its max voltage at 1.5s since in position 3 the capacitor is being discharged.

 

Given the circuit setup when the switch is in position 1, you should be able to determine the length of time it takes for the capacitor to charge. It should be a function of the resistor and capacitor in the circuit.

Hint: look at your title...

 

it doesnt reach 5 time constant. the switch is moved to position 3 after 1.5s so what i get for the voltage across the capacitor after 1.5s is 1.2 x 10^-5 V. since the capacitor will be discharged when in position 3 the maximum voltage will hence be 1.2x10^-5 V.
which gives Energy= (0.5)(5)(1.2x10^-5)^2.

is this correct?

 

That does appear to be correct, although I did not verify the actual voltage number. The components give a rather large time constant, which results in a rather small voltage charge in 1.5s. The instantaneous is given by Energy = 0.5*C*V^2, which you have calculated.

 

I strongly suspect that the capacitance is supposed to be 5µF and not 5F. Look at the space between the '5' and the 'F'; I think they intended to go back and insert a 'µ' manually. But, you pretty much have to go with what you've got.

The only issue I have with your solution is that you won't track units:

Energy= (0.5)(5F)(1.2x10^-5V)^2 = 30pJ.

 

I just blogged the formulas for RC time constants. My blog #3