Graphs of charge and discharge
Capacitor charge and discharge graphs are exponential curves.
If a larger value of capacitance were used with the same value of The opposition in an electrical component to the movement of electrical charge through it. Resistance is measured in ohms. in the above circuit it would be able to store more charge. As a result, it would take longer to charge up to the supply voltage during charging and longer to lose all its charge when discharging.
If a larger value of resistance were used with the same value of capacitance in the above circuit, then a smaller current would flow, therefore it would take longer for the capacitor to charge up and longer for it to discharge.
Question
When the capacitor is charged current flows on either side of the capacitor. Do electrons cross the gap to allow the current on the other side of the circuit?
No, the electron flow from the negative terminal is absorbed on the lower plate. Electron flow to the positive supply terminal comes from the upper plate. No electrons cross the gap, which is often an insulating material.
Question
In the circuit shown below, the capacitor is initially uncharged.
The switch is now closed.
What is the initial reading on the ammeter?
\(E= 9.0V\)
\(I= ?\)
\(R= 1000\Omega\)
\(I= \frac{E}{R} = \frac{{9.0}}{{1000}} = 9 \times {10^{ - 3}}{\rm A}\)
Question
What is the charge stored on the capacitor when the p.d. across the resistor is \(3.0V\)?
\(E=p.d.\,across\,capacitor\,+\,p.d.\,across\,resistor\)
\(9.0=p.d.\,across\,capacitor+3.0\)
\(p.d.\,across\,capacitor=6.0V\)
\(C= 12 \times {10^{ - 6}}F\)
\(Q= ?\)
\(V = 6.0V\)
\(C= \frac{Q}{V}\)
\(Q = CV\)
\(=12 \times {10^{ - 6}} \times 6.0\)
\(=7.2 \times {10^{ - 5}}C\)
Question
Give two effects produced by increasing the value of the resistor.
1. Capacitor will take longer to charge up to \(9.0 V\)
2. Initial current will be lower than \(9 \times{10^{-3}}{ \rm A}\)