Question

5-mu or micro F capacitor is charged to 110 V and is then connected across a 600-ohm resistor. What is the initial charge on the capacitor? C * [2.5 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] What is the initial current just after the capacitor is connected to the resistor? 0.1833333333 A * [2.5 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0.1833333333 OK What is the time constant of this circuit? 0.003 s * [2.5 points] 2 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0.003 OK How much charge is on the capacitor after 6 ms?

Answer 1

Answer: a) Qo=550 *10^-6 C; b) Io=0.18 A; c) τ=3 ms

Explanation: In order to explain this problem we have to consirer the capacitor expresion given by :

C=Q/V then Q=C*V

the initial charge of the capacitor is:

Qo= 5*10^-6*110=550 μC

When the capacitor is connected to a resistor the stored charge is consume at the resitance as a current (dQ/dt):

The charge as a function of time is given by;

Q(t)= Qo*e(-t/τ) where τ is the equal R*C and represent the time constant.

the we have that the current is:

I=dQ/dt= (Qo/τ)*e(-t/τ)=(Qo)/(R*C)*e(-t/R*C)

Io=(Qo/τ)=(550*10^-6)/(600*5*10^-6)=0.18 A