Question

A 1.0μF capacitor with an initial stored energy of 0.50 J is discharged through a 1.0MΩ resistor. (a) What is the initial charge on the capacitor? (b) What is the current through the resistor when the discharge starts? Find an expression that gives, as a function of time t, (c) the potential difference V C ​ across the capacitor, (d) the potential difference V R ​ across the resistor, and (e) the rate at which thermal energy is produced in the resistor.

Answer 1

Answer:

A) q_o = 0.001 C

B) I = 0.001•e^(-t)

C) V_c = 1000e^(-t)

D) V_r = 1000e^(-t)

E) P = e^(-2t) watts

Explanation:

A) We are given;

Initial stored energy; U_o = 0.5 J

Capacitance; C = 1.0μF = 1 × 10^(-6) F

To find the charge, we will use the formula for energy in capacitors which is given by;

U = q²/2C

Thus, since we are dealing with initial energy, U is U_o and q is q_o

Making q the subject, we have;

q_o = √2CU_o

q_o = √(2 × 1 × 10^(-6) × 0.5)

q_o = 0.001 C

B) The charge as a function of time is expressed as;

q = q_o•e^(-t/RC)

Now the current is gotten by differentiating the charge function. Thus;

I = (q_o/RC)•e^(-t/RC)

Where;

R is Resistance = 1.0MΩ = 1 × 10^(6) Ω

C is capacitance = 1 × 10^(-6) F

(q_o/RC) is the initial current = 0.001/(1 × 10^(6) × 1 × 10^(-6))

(q_o/RC) = 0.001 A

Thus;

I = 0.001•e^(-t/(1 × 10^(6) × 1 × 10^(-6)))

I = 0.001•e^(-t)

C) Formula for potential difference across the capacitor is;

V_c = IR

I = 0.001•e^(-t)

R = 1 × 10^(6) Ω

Thus;

V_c = 1 × 10^(6) × 0.001•e^(-t)

V_c = 1000e^(-t)

D) Potential difference across the resistor will be the same as that across the capacitor because the resistor is connected in parallel to the capacitor.

Thus;

V_r = V_c = 1000e^(-t)

E) rate at which thermal energy is produced is basically the power.

Thus;

P = (V_r)²/R

P = (1000²e^(-2t))/1 × 10^(6)

P = e^(-2t) watts