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

Answer 2

(A) The initial charge on the capacitor will be 550×10⁻⁶ columb.

(B) The value of the time constant will be 3ms.

(C) The initial current just after the capacitor is connected to the resistor will be 0.18 ampere (A).

What is capacitance?

Capacitance refers to the capacitor's capacity to hold charges. It is made up of two electrical conductors separated by a distance. The gap between the conductors can be filled by a vacuum or a dielectric,

The given data in the problem is:

C is the capacitance = 5-mu or micro farad=5×10⁻⁶

Q₀ is the initial charge on the capacitor=?

I₀ is the initial current just after the capacitor =?

τ is the time constant of this circuit=?

V is the i voltage = 110 V

(A) The initial charge on the capacitor will be 550×10⁻⁶ columb.

The charge on a capacitor is given by

$\rm Q_0=C_0V\\\\\rm Q_0=5\times10^{-6}\times110\\\\\rm Q_0=550\times10^{-6} \;coloumb$

Hence the initial charge on the capacitor will be 550×10⁻⁶ columb.

(B) The value of the time constant will be 3ms.

The time constant is defined by the product of resistance and  capacitance,

$\rm \tau=RC\\\\\rm \tau=600\times550\times10^{-6}\\\\ \tau=3\times10^{-3}=3ms$

Hence the value of the time constant will be 3ms.

(C) The initial current just after the capacitor is connected to the resistor will be 0.18 ampere (A).

Electric current is defined as the charge per unit time.

$\rm I_0=\frac{Q_0}{\tau} \\\\\rm I_0=\frac{550\times 10^{-6}}{3\times10^{-3}}\\\\\rm I_0=0.18\; A$

Hence the initial current just after the capacitor is connected to the resistor will be 0.18 ampere (A)

To learn more about the capacitance refer to the link;

https://brainly.com/question/12356566