When a capacitor is fully charged, the maximum energy stored is given by (1/2)*C*V^2, the maximum charge is Q = CV

As the capacitance C = 1200* 10^-6 F and the maximum voltage V = 12V

Qmax = 1200*10^-6 * 12 = 14.40 mC

Emax = (1/2)*1200*10^-6*12^2

=> (1/2)*1200*10^-6*144

=> 86.40 mJ

When a capacitor discharges:

the residual charge after time t is: C*V*e^(-t/RC)

substituting the values we have:

the residual charge Q

=> 1200* 10^-6 * 12*e^(-68/(1200* 10^-6*50*10^3))

=> 14400*10^-6* = 4.63 mC

the residual voltage V = 12 * e^(-68/(1200* 10^-6*50*10^3))

=> 3.86 V

The residual energy E is (1/2)*V*Q

=> (1/2)*4.63*3.86

=> 8.94 mJ

The required values are:

**Qmax = 14.40 mC , Emax = 86.40 mJ, Q = 4.63 mC, V = 3.86 V and E = 8.94 mJ**