Question

The values of the components in a simple series RC circuit containing a switch and an initially uncharged capacitor (see figure below) are C = 0.70

Answer 1

q = Q I(1-e^-t/RC)-------------------------------------1

where RC is called timeconstant and denoted by T

so T = RC

similarly we get current variation as

i = io(e^-t/Rc)--------------------------------------------2

we can get the eqns of charge and current while discharging as

q = Qe^-t/RC -------------------------------------------------3

i = io e^-t/RC----------------------------------------------4

we can get the eqvation for Voltage as

V = Vo(1-e^t/RC) while charging-------------------------------------5

V= voe^-t/RC while discharging-----------------------------------------6

a. Q=CV

Q = 0.7uF * n10 = 7uC

so q = Q(1-e^-t/RC)

q = 7 * (1-e^-1.3/(2.5*10^6*0.7*10^-6)

q = 3.669 uC

b. current i = io(e^-t/RC)

iintial currrnet i = 10/2.5 = 4 uA

so i = 4* e^-1.3/(2.5*0.7)

i = 1.903 uA

so power across ressitor P = i^2 R = 1.903*1.903 *10^-12 * 2.5*10^6

P = 9.053*10^-6 Watts

power acrooss capacitor = enrgy/time = 0.5qV/time = 0.5 * 3.669*10^-6 * 10/1.3 = 1.411 *10^-6 J

power P = vi = 10 * 1.903 = 19.03*10^-6 Watts

Answer 2

Answer 3

formulas required are

oe R

so the charge at time 1.3 s = Q=CV

= 0.70* 10* e^[-1.3/2.5*.70] = 3.34

b] energy diddipated in the resistor = I[t]^2 *R =0.02* 10^-6 J

2] in the capacitor = .5 C [V[t]]^2= .08*10^-6 J

3] delivered by the battery

= energy stored by the capacitor + energy lost by the resistor= .10 * 10^-6 J