Charging and Discharging Capacitors Flashcards
what equipment do you need for setting up a circuit in order to find the capacitance of a capacitor
- a cell with known pd
- a capacitor
- a resistor
- a voltmeter
- a switch
how does the circuit need to be set up
- 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
what do you firstly need to do to the capacitor when carrying out this experiment
charge it up
do you charge it with the resistor connected or not and why
- with the resistor not connected
- so the capacitor charges up instantly
therefore how do you charge up the capacitor
- disconnect the resistance from the circuit
- hold the switch down for a long enough time for the capacitor to charge (its instant anyway)
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
- connect the resistor back into the circuit
- disconnect the circuit from the cell
what would be observed the moment you disconnect the circuit from the cell
- the pd across the capacitor will drop quickly
- and continue to drop but at a gradually slower rate throughout
what values are you taking down
- the value the pd is at (in interval form)
- the time it takes for the pd to decrease by that set interval every time
simply speaking, what would you use the pd values for later on (the next deck) and why
- you would ln your results
- so you can draw a graph
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
- 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
why does the discharge initially occur at such a quick rate in the beginning
- 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
what happens to the current as time goes on
- the current would gradually decrease
- until it eventually got to 0
why does the current decrease like so
- 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)
what equation simply describes / supports this occurrence
a decrease in V in I = V / R means a decrease in I
what observation change would you see in respects to the light bulb because of all this
it would get dimmer
what would graphs of discharge current, pd across capacitor and charge in capacitor against time look like in this DISCHARGING event
- they would all look like radioactive half life graph
- a (decreasingly) exponential decrease in all of them over time
what do i mean by a decreasingly exponential decrease over time
- the line is going downwards
- but the gradient of the line gradually decreases
what are the two simple solutions for increasing the time the lightbulb is on, given the same power supply
- store more charge on the capacitor
- decrease the rate at which the capacitor discharges
why would storing more charge on the capacitor increase the bulbs light lifespan
- 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
why does decreasing the rate at which the capacitor discharges increase the light lifespan
- slower rate = longer time charge/current is flowing around circuit for a given initial charge
- therefore applying a current to the bulb for longer
how would you decrease the rate at which the capacitor discharges
increase the resistance of the bulb or just add more resistance
what is the equation for the time constant
T = RC
what are each of those variables
- T (s) = time constant, which is actually a backwards J not just a T
- R (ohms) = resistance
- C (farads) = capacitance
what does the time constant allow you to do
work out the rate of discharge of a capacitor
without going balls deep into the maths, what does the time constant also tell you
the time it takes for the current to fall to 37% of its starting value
what would graphs of charging current, pd across capacitor and charge in capacitor against time look like for a CHARGING event
- 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
