Question

A 10 F capacitor is connected through a 1 M resistor to a constant potential difference of 100

V. Compute (a) the charge at time 5s (b) the current in the same instant (c) the time require for the

charge to increase from zero to 5×10-04 C (d) the final charge in the circuit (e) the relaxation time

and half-life of the circuit. (f) the time require for the charge to increase from 5×10-04 C to 7×10-04

C.

Answer 1

$c= 10\; \mu F \\ R=1 \; M \Omega \\ V_b= 100 \;V$

a)

$Q=cV_b[1 - e^{-\frac{t}{Rc}}] \\ t= 5 \;s \\ Q= 10 \times 10^{-6} \times 100 [1 -e^{-\frac{5}{10}}] \\ Q= 3.93 \times 10^{-4} \;C$

b)

Carging current at t= 5 s

$I= \frac{V_b}{R}e^{-\frac{t}{RC} } \\ = \frac{3.93 \times 10^{-4}}{5} \\ = 7.86 \times 10^{-5} \;A$

c)

$Q(t) = 5 \times 10^{-4} \;C \\ 5 \times 10^{-4}= 10 \times 10^{-6} \times 100 [1 -e^{-\frac{t}{10} }] \\ 0.5 = [1 -e^{-\frac{t}{10} }] \\ e^{-\frac{t}{10}} = 0.5 \\ -\frac{t}{10} = ln(0.5) \\ t=6.93 \;s$

d)

$Q=cV_b = 10 \times 10^{-6} \times 100 \\ Q=10^{-3} \;C$

e) The relaxation time:

$τ = RC = 10^6 \times 10 \times 10^{-6} \\ = 10 \;s$

Half-life of the circuit:

$V=V_b[1 – e^{-\frac{t}{Rc}}] \\ At\;t= t_{1/2} \; \\ V = \frac{V_b}{2} \\ 0.5 = [1 – e^{-\frac{t_{1/2}}{10}}] \\ t_{1/2} = 6.93 \;s$

f)

$Q=cV_b[1 - e^{-\frac{t}{Rc}}] \\ 7 \times 10^{-4} = 10 \times 10^{-4}[1 -e^{-\frac{t_1}{10}}] \\ t_1= 12.039 \;s \\ 5 \times 10^{-4}= 10 \times 10^{-4}[1- e^{-\frac{t_2}{10}}] \\ t_2= 6.93 \;s$

The time require for the charge to increase from $5 \times 10^{-4} \; C$ to $7 \times 10^{-4} \; C$ :

$Δt=t_1-t_2 = 5.11 \;s$