A 1.3 µF capacitor with an initial stored energy of 0.55 J is discharged through

Question

A 1.3 µF capacitor with an initial stored energy of 0.55 J is discharged through a 2.6 MO resistor.
(a) What is the initial charge on the capacitor?
1 C

(b) What is the current through the resistor when the discharge starts?
2 A

(c) Find an expression that gives, as a function of time t, the potential difference VC across the capacitor. (Use the following as necessary: t. Do not include untis in your answer.)
VC =

(d) Find an expression that gives, as a function of time t, the potential difference VR across the resistor. (Use the following as necessary: t. Do not include untis in your answer.)
VR =

(e) Find an expression that gives, as a function of time t, the rate at which thermal energy is produced in the resistor. (Use the following as necessary: t. Do not include untis in your answer.)
P =

Explanation / Answer

Capacitance of the capacitor C =1.3 µF =1.3*10^-6 F
Initial stored energy=0.55 J
Load resistance, R =2.6 MOhm =2.6*10^6 Ohms
Let
Initial charge on the capacitor be =q_0
Time constant = t
a)
Initial energy stored in the capacitor,U_i =1/2[ (q_0)^2/C)]
Initial charge on the capacitor be q_0 =sq.root(2U_iC) =1.19*10^-3 C
B)
Charge across the capacitor,q =q_0 (e^-t/t)
current i =dq/dt =(q_0/t) e^-t/t
Then, time constant of RC circuit, t =RC =3.38 s Now, initial current i_0 =q_0/t =0.35*10^-3 A C) the voltage during the discharge V_c =V_0 e^-t/RC V_c =(q_0/C) e^-t/RC =(0.91*10^3 v) e^-t/3.38 s =[(0.91*10^3 ) e^-0.29t] V D) i =[(q_0/t) e^-t/t] the voltage across the resistor V_R=iR =[(q_0/t) e^-t/t]R =(0.35*10^-3 A)(2.6*10^6 Ohms)e^-0.29t V_R=(0.91*10^3 v)e^-0.29t E) thermal energy is produced in the resistor P =i^2R where i =(q_0/t) e^-t/t P =[(q_0)^2/t^2]R e^-2t/t =[0.32 e^-0.59t] W V_c =V_0 e^-t/RC V_c =(q_0/C) e^-t/RC =(0.91*10^3 v) e^-t/3.38 s =[(0.91*10^3 ) e^-0.29t] V D) i =[(q_0/t) e^-t/t] the voltage across the resistor the voltage across the resistor V_R=iR =[(q_0/t) e^-t/t]R =(0.35*10^-3 A)(2.6*10^6 Ohms)e^-0.29t V_R=(0.91*10^3 v)e^-0.29t E) thermal energy is produced in the resistor P =i^2R where i =(q_0/t) e^-t/t P =[(q_0)^2/t^2]R e^-2t/t =[0.32 e^-0.59t] W

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