Charging and Discharging Capacitors

Question 1

what equipment do you need for setting up a circuit in order to find the capacitance of a capacitor

Answer 1

• a cell with known pd
• a capacitor
• a resistor
• a voltmeter
• a switch

Question 2

how does the circuit need to be set up

Answer 2

• a series circuit is made with the cell and capacitor
• a voltmeter is placed in parallel to the capacitor
• and a resistor is also placed in parallel to the capacitor, but not across the voltmeter
• the switch would be on the junction where the parallel circuits are made

Question 3

what do you firstly need to do to the capacitor when carrying out this experiment

Answer 3

charge it up

Question 4

do you charge it with the resistor connected or not and why

Answer 4

• with the resistor not connected

- so the capacitor charges up instantly

Question 5

therefore how do you charge up the capacitor

Answer 5

• disconnect the resistance from the circuit

- hold the switch down for a long enough time for the capacitor to charge (its instant anyway)

Question 6

having measured the initial pd across the capacitor, what you do do after it is fully charged up, before you start taking values of anything

Answer 6

• connect the resistor back into the circuit

- disconnect the circuit from the cell

Question 7

what would be observed the moment you disconnect the circuit from the cell

Answer 7

• the pd across the capacitor will drop quickly

- and continue to drop but at a gradually slower rate throughout

Question 8

what values are you taking down

Answer 8

• the value the pd is at (in interval form)

- the time it takes for the pd to decrease by that set interval every time

Question 9

simply speaking, what would you use the pd values for later on (the next deck) and why

Answer 9

• you would ln your results

- so you can draw a graph

Question 10

a charged capacitor is disconnected from its circuit and placed into another one with only a bulb. how does the discharge of the capacitance initially go and what is the current like

Answer 10

• there would be a rush of electrons and therefore current
• because the discharge is as high as it can be
• in other words, the current starts at its maximum

Question 11

why does the discharge initially occur at such a quick rate in the beginning

Answer 11

• because the pd on the capacitor is at its maximum

- so the force exerted on the electrons by the electric field it produces is also at its maximum

Question 12

what happens to the current as time goes on

Answer 12

• the current would gradually decrease

- until it eventually got to 0

Question 13

why does the current decrease like so

Answer 13

• because some electrons have discharged
• the pd across the capacitor decreases
• so therefore the electric force exerted on the electrons decreases
• this slows down the flow of electrons, therefore slowing down current
• (as current is the rate of flow of charge)

Question 14

what equation simply describes / supports this occurrence

Answer 14

a decrease in V in I = V / R means a decrease in I

Question 15

what observation change would you see in respects to the light bulb because of all this

Answer 15

it would get dimmer

Question 16

what would graphs of discharge current, pd across capacitor and charge in capacitor against time look like in this DISCHARGING event

Answer 16

• they would all look like radioactive half life graph

- a (decreasingly) exponential decrease in all of them over time

Question 17

what do i mean by a decreasingly exponential decrease over time

Answer 17

• the line is going downwards

- but the gradient of the line gradually decreases

Question 18

what are the two simple solutions for increasing the time the lightbulb is on, given the same power supply

Answer 18

• store more charge on the capacitor

- decrease the rate at which the capacitor discharges

Question 19

why would storing more charge on the capacitor increase the bulbs light lifespan

Answer 19

• firstly, you would need a bigger capacitor for this as capacitors usually fully charge anyway
• but with this, it means for the same rate of discharge, charge/current will be able to flow around the circuit for longer
• therefore applying a current to the bulb for longer

Question 20

why does decreasing the rate at which the capacitor discharges increase the light lifespan

Answer 20

• slower rate = longer time charge/current is flowing around circuit for a given initial charge
• therefore applying a current to the bulb for longer

Question 21

how would you decrease the rate at which the capacitor discharges

Answer 21

increase the resistance of the bulb or just add more resistance

Question 22

what is the equation for the time constant

Answer 22

T = RC

Question 23

what are each of those variables

Answer 23

• T (s) = time constant, which is actually a backwards J not just a T
• R (ohms) = resistance
• C (farads) = capacitance

Question 24

what does the time constant allow you to do

Answer 24

work out the rate of discharge of a capacitor

Question 25

without going balls deep into the maths, what does the time constant also tell you

Answer 25

the time it takes for the current to fall to 37% of its starting value

Question 26

what would graphs of charging current, pd across capacitor and charge in capacitor against time look like for a CHARGING event

Answer 26

• the charging current graph would look like a decreasingly exponential decrease graph
• the pd across capacitor would look like a x^1/2 graph
• the charge on capacitor would look like a x^1/2 graph