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Physics

Capacitance

Capacitors in d.c. circuits

Charging and discharging a capacitor from a d.c. source

Capacitor in dc circuit - The circuit is powered by a battery (E). In this circuit there is a resistor (R) and ammeter and capacitor (C) in series. There is a voltmeter in parallel to the capacitor. There is a switch which switched between position '1' and '2'.

The circuit above is used to investigate the charge and discharge of a capacitor. The supply has negligible internal resistance [resistance: The opposition in an electrical component to the flow of electricity through it. Resistance is measured in ohms..

The capacitor is initially uncharged. When the switch is moved to position 1, electrons [electrons: Sub-atomic particles, with a negative charge and a negligible mass relative to protons and neutrons. move from the negative terminal of the supply to the lower plate of the capacitor. Electrons on the lower plate repel electrons from the upper plate, which then move to the positive terminal of the supply.

This movement of charge is opposed by the resistor R, so the initial current [current: Moving electric charges, for example, electrons moving through a metal wire. in the circuit is I = E/R

Charges cannot move from the lower plate to the upper plate as the plates are separated by an insulating material.

During the charging of a capacitor:

  • the charging current decreases from an initial value of E/R to zero
  • the p.d. across the capacitor plates increases from zero to a maximum value of E, when the capacitor is fully charged
  • at all times the sum of the p.d. across the capacitor and the p.d. across the resistor equals the e.m.f of the supply
  • the p.d. across the resistor (given by VR = IR) decreases from an initial value of E to zero when the capacitor is fully charged

When the switch is moved to position 2, electrons move from the lower plate through the resistor to the upper plate of the capacitor.

The movement of electrons through the ammeter [ammeter: A device used to measure electric current. is in the opposite direction to that of charging.

During the discharging of a capacitor:

  • the discharging current decreases from an initial value of -E/R to zero
  • the p.d. across the capacitor plates decreases from E to zero, when the capacitor is fully discharged
  • the p.d. across the capacitor and the p.d. across the resistor are always equal
  • the p.d. across the resistor (given by VR = IR) decreases from an initial value of E to zero when the capacitor is fully discharged

Graphs shown below have exponential curves

Graphs of charge and discharge

Graphs of charge and discharge - the voltage increases as it is charged, and decreases as it's discharged. I=E/R falls as it charges and I=-E/R increases as its getting discharged.

If a larger value of capacitance were used with the same value of resistance 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.

In a d.c. circuit a capacitor stores charge and energy. It acts as a block to d.c. and builds up a d.c. voltage.

Question
Capacitor in a dc circuit - there is a 9V battery in series with a switch, which has a capacitor, ammeter and 1kOhm resistor in series. A voltmeter is used to measure the voltage of the capacitor - this is in parallel to the capacitor. The capacitor measures 12 micro Farad .

In the circuit shown above the capacitor is initially uncharged.

The switch is now closed.

What is the initial reading on the ammeter?

Answer

E = 9.0 V

I = ?

R = 1000 \Omega

I = {E \over R} = {{9 \cdot 0} \over {1000}} = 9 \times 10^{ - 3} A

Question

What is the charge stored on the capacitor when the p.d. across the resistor is 3.0 V?

Answer

E = p.d. across capacitor + p.d. across resistor

9.0 = p.d. across capacitor + 3.0

p.d. across capacitor = 6.0 V

 

C = 12 \times 10^{-6} F

Q = ?

V = 6.0 V

C = {Q \over 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.

Answer
  1. Capacitor will take longer to charge up to 9.0 V
  2. Initial current will be lower than 9 × 10-3 A

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