The switch in the figure has been closed for a very long time.
Part A: What is the charge on the capacitor?
Part B: The switch is opened at t= 0s. At what time has the charge on the capacitor decreased to 29% of its initial value?
Revelant knowledge
The capacitor is an electronic device that stores electricity through the form of an electronic field. It is an non-active electronic device equipped with 2 connections.
The result of a capacitor can be described by the term capacitance. There is some form of capacitance between electrical conductors that are in close proximity to each other in an circuit the capacitor is a device that is designed to increase the capacitance of an electrical circuit. The capacitor was initially referred to as a condenser. The name and its synonyms are still used extensively in a variety of languages, however seldom in English The most important exception is condenser microphones which are also referred to as capacitor microphones.
This problem can be solved using Ohm’s law, series resistance, Ohm’s law, expression of charge on capacitor and charge on capacitor after t
Calculate the equivalent resistance first, then calculate current in circuit. Next calculate potential difference. Finally calculate charge across capacitor. In part (B), calculate time required to charge capacitor to 29\% .
This is the equivalent resistance to series resistance.
{R_{{\rm{eq}}}} = {R_1} + {R_2}
Here {R_1}
Ohm’s law
V = IR
Here, I
The capacitor’s charge is:
Q = CV
Here Q
After t, the charge on the capacitor is $0.
Q = {Q_{{\rm{max}}}}{e^{\frac{{ – t}}{{RC}}}}
Here, 29\% 0
The capacitor behaves as an open circuit when the switch is closed for a long time, and there is no current flowing from it.
Below is an image that shows the circuit condition at 29\% 1
Use the series resistance formula to calculate equivalent resistance.
The series resistance equivalent is:
{R_{{\rm{eq}}}} = {R_1} + {R_2}
Here {R_1}
Substitute 29\% 4
29\% 5
Use Ohm’s law to calculate the current.
Ohm’s law
V = IR
Here, I
Modify the equation V = IR
29\% 9
Substitute {R_{{\rm{eq}}}} = {R_1} + {R_2}0
{R_{{\rm{eq}}}} = {R_1} + {R_2}1
Now, calculate the voltage across {R_{{\rm{eq}}}} = {R_1} + {R_2}2
{R_{{\rm{eq}}}} = {R_1} + {R_2}3
The maximum capacitor capacity is now
The capacitor’s charge is:
{R_{{\rm{eq}}}} = {R_1} + {R_2}4
Here 29\% 0
Substitute {R_{{\rm{eq}}}} = {R_1} + {R_2}6
{R_{{\rm{eq}}}} = {R_1} + {R_2}7
Below is an image of the circuit at {R_{{\rm{eq}}}} = {R_1} + {R_2}8
Calculate the equivalent resistance {R_{{\rm{eq}}}} = {R_1} + {R_2}8
The series resistance equivalent is:
{R_{{\rm{eq}}}} = {R_1} + {R_2}
Here {R_1}
Substitute {R_1}2
{R_1}3
Now, calculate the capacitor charge after t
After t, the charge on the capacitor is $0.
Q = {Q_{{\rm{max}}}}{e^{\frac{{ – t}}{{RC}}}}
Here, 29\% 0
Modify the equation Q = {Q_{{\rm{max}}}}{e^{\frac{{ – t}}{{RC}}}}
{R_1}9
Substitute V = IR0
V = IR1
Convert second to millisecond
V = IR2Ans Part A
The charge on the capacitor is V = IR3Part A
The time required is V = IR4