Electricity Flashcards
Hader
What exactly is a potential difference?
The difference in the amount of energy that charge carriers have between two points in a circuit
A student has set up a parallel circuit with two loops. Each loop has two 10Ω resistors. Points P and Q lie on the wire before the loops join back together again.
The student uses a voltmeter to find the potential difference between P and Q. What would the voltmeter reading be? Explain your answer.
0 - each loop has the same resistance (and initial potential difference because they’re parallel) so have had the same voltage drop. This means there’s no potential difference between points P and Q.
power supply symbol
two filled-in dots with a gap between them
voltage/power rating meaning
the maximum p.d./power that a device can operate at safely
threshold voltage meaning
A value of p.d., above which allows current to flow easily in a diode
potential difference meaning
The work done per unit charge.
1V = ? (alternative units)
1 J/C
Ideal voltmeters/ammeters meaning
Ideal voltmetes have infinite resistance so no current flows through them - to measure the p.d. all the current should pass through the resistor and none through the voltmeter
Ideal ammeters have no resistance so have no p.d. across them
emf meaning
the terminal potential difference (p.d. across the power supply e.g. cell) when no current flows
the amount of electrical energy provided by a power supply per coulomb of charge passing through (ε = E / Q)
What happens in a wire when a complete circuit is formed with a cell?
when a p.d. is induced, electrons move towards the positive charge
How do you calculate the resistance for a non-ohmic conductor from an IV graph?
Read off the values for current and p.d. then use the resistance formula
Explain why an IV graph of a filament lamp looks like that.
- As the current increases, its temperature also increases
- This causes the particles in the metal to vibrate faster (with a greater amplitude)
- This makes it harder for the charge-carrying electrons to travel, reducing the rate of electron movement
- This increases the bulb’s resistance as the current can’t flow as easily
Why do filament bulbs’ IV graphs plateau?
As the current and so temeprature increases, the resistance increases, which decreases the current.
What does a V-I graph for a filament lamp look like?
look online
quadrant 1 = x2 graph
quadrant 3 = -x2 graph
forward bias meaning
The direction in a diode which allows current to flow
reverse bias meaning
The direction in a diode where the resistance is high so the current is small
resistivity
in terms of a) the concept and b) the formula
a) A property of a material that describes how much it opposes the flow of electric current through it.
b) The resistance of a 1m length of a material with a 1m2 cross-sectional area.
What is the A in the resistivity formula?
The cross-sectional area - usually a circle or rectangle
Why might you use a power supply when investigating the resistivity of an electrical component?
The power supply provides a constant potential difference
Why might you use flat metal electrodes at the ends of putty when investigating its resistivity?
The electrode simproves connection (they’re good electrical conductors so current can pass from one end of the putty to the other more easily)
charge carrier meaning
A particle which carries an electric charge (i.e. electrons)
semiconductor meaning
A material which conducts electricity (but not as well as metals because it has fewer charge carriers). When energy is transferred to it, its temperature increases so it can release more charge carriers and its resistance decreases
Give three examples of semiconductors.
LDRs, thermistors and diodes
NTC thermistor meaning
Negative temperature coefficient - as the temperature increases, the component’s resistance decreases.
Explain why thermistors’ resistance increase as the temperature decreases.
As the temperature decreases, the electrons lose energy so can’t escape their atoms as easily. This means there are fewer charge carriers available, so the resistance increases.
How do you investigate the resistance of a thermistor?
- Set up circuit with a power supply, thermistor and ammeter in series with a voltmeter in parallel
- Control the temperature by placing the waterproof thermistor in a water bath (e.g. pouring boiling water into a beaker with a thermistor inside to fully cover it)
- Measure water temperature with a thermometer and current with the ammeter (as the p.d. needs to be kept constant)
- Let the water cool. For every drop in 5°C record the p.d. current and temperature
- Use R = V / I to calculate the resistance of the thermistor at each temperature recorded
risk assessment for investigating the resistance of a thermistor
Components and thermistor that come into contact with the water bath must be waterproof.
superconductor meaning
A property of certain materials which
have zero resistivity and so sero resistance below a critical temperature (this depends on the material)
Explain two superconductor example applications.
- strong electromagnets for accelerators - superconductors can have high currents flowing through them without great loss of power, producing very strong magnetic fields (e.g. for medicine & Maglev trains)
- efficient electric/power cables - reduces energy/power loss
Give two disadvantages of using superconducting wires.
- Very expensive
- Difficult to cool the material to below its critical temperature
How do you find the resitivity of a material?
- Measure the diameter of the test wire three times with a micrometer to calculate the mean in metres. Then halve this to get the radius
- Calculate the cross-sectional area using A = πr2
- Set up a circuit with an open switch, length of wire clamped with crocodile clips at the 0cm end of a meter ruler, a flying lead, an ammeter in series and a voltmeter across the wire in parallel
- Attach the flying lead to the end of the test wire and measure its length with the ruler
- Close the switch and measure the current through the circuit and the p.d. across the wire
- Open the switch
- Use R = V / I to calculate the resistance
- Reapeat twice more and calculate the mean resistance at this length of wire
- Move the flying lead to 10cm below the end of the test wire and repeat to calculate the average resistance for each length of test wire
- Plot a graph of resistance against length
- Draw a line of best fit . The gradient = R/l = ρ / A
- Therefore resistivity = the gradient * the cross-sectional area of the wire
Suggest one way, other than using a switch, to keep the temperature constant when investigating the resistivity of a wire.
Use a small current - for example by including a resistor in the circuit.
formula for resistance, energy, time and current
E=I2Rt
formula for p.d., resistance, time and energy
E=V2Rt
formula for time, energy, current and p.d.
E=VIt
potential divider formula
Vout = R2 / (R1 + R2) * Vs
where:
Vout = the output p.d.
R1 = the resistance of the resistor(s) in the main series circuit
R2 = the resistance of the resistor that you’re measuring the p.d. over; the potential divider
Vs = the p.d. supply
Kirchoff’s first law (+ give an example of a time this applies)
The total current entering a point is equal to the total current leaving that point e.g. a series circuit setup
Kirchoff’s second law
In a closed circuit, the total of the electromotive forces is equal to the total of the potential differences
What is the unit of emf?
volts
symbol representing emf
epsilon - ε
In parallel circuits, what is the total of the emfs equal to?
The p.d. in each loop (as they’ve got the same p.d.)
One loop of a parallel circuit has a resistance of 4 ohms while another has 6 ohms. The total current that flows though the battery is 5A. What current flows through each loop?
The currents are shared in the inverse ratio of the resistance – a smaller resistance has a bigger current
4 ohms loop => 5 x (4/10) = 2A
6 ohms loop => 5 x (6/10) = 3A
idk if this is even how you work it out anymore
effective resistance
the single resistor which replaces two or more resistors which gives the same effect on the circuit by allowing the same amount of current to flow (i.e. a resistor that has the total resistance of all the resistors)
internal resistance formula not on formula sheet (with labels)
ε = Ir + V
r = internal resistance (resistance across the power supply)
V = terminal p.d.
internal resistance
A resistance to the flow of current inside a cell due to the chemicals it is composed of; the resistance in a power source when electrons collide with atoms inside it and transfer (and so lose) energy.
terminal p.d.
The p.d. across the terminals of a cell when a current flows
difference between EMF and terminal p.d.
EMF = the p.d. across the terminals of a cell when no current flows
terminal p.d. = the p.d. across the terminals of a cell when a current flows
How do you find the EMF from a graph of p.d. against current?
y-intercept
How do you find the internal resistance from a graph of p.d. against current?
-gradient (modulus of p.d. / current)
What happens to the resistance of an ohmic conductor if you double the potential difference across it? Give a reason for your answer.
Nothing — the resistance of an ohmic conductor is constant (if the physical conditions it’s under are constant).
Sketch a V-I graph for a lamp.
Positive cubic graph
Sketch the characteristic V-I graph of a diode.
Like a limiting factor graph which plateaus in the first quadrant, very high gradient in the third quadrant
Sketch the characteristic I-V graph of a diode.
Low current in the reverse direction (there is a small gradient in the third quadrant) but increasing gradient in the first quadrant
A student is trying to measure the resistivity of some putty, which has a resistance of 20 ohms. They use a variable resistor with a maximum resistance of 10 ohms. Is there an issue? Explain your answer.
Yes because the resistance of the variable resistor is only half of the putty - the variable resistor should have a greater resistance.
State and explain whether a filament lamp is an ohmic or non−ohmic conductor up to its working power.
non-ohmic conductor
current is not (directly) proportional to voltage OR resistance is not constant
Lamps P and Q are parallel to each other. Lamp Q melts so that it no longer conducts. Explain how the brightness of lamp P changes.
The resistance of lamp P increases so the share of the potential difference across P increases, making lamp P glow brighter.
Cells A and B are in parallel with each other while cells C and D are in series. Given that the cells are identical, which pair of cells will go flat first? [3]
Cells C and D
The charge/current passing through them is greater (double) - the current splits between cells A and B
The energy given to charge passing through cells A and B per second is double/more than in cells C and D (need to use the phrase per second to get the final mark)
A lamp is connected to a 12V dc supply. State and explain the change, if any, to the final current through the lamp if it is connected to the same supply with another similar lamp in parallel. [2]
Current through the lamps stays the same (1)
as both lamps are connected directly to the supply (1)
When a lamp is initially switched on, its current peaks then decreases to its normal current. State and explain why a filament lamp is most likely to fail as it is switched on.
The resistance is initially low (which leads to this peak current). This means there’s a sudden rapid change in temperature.
Explain why bulbs are placed in parallel. [2]
So that the bulbs are connected directly across the power supply/battery, meaning that if one lamp blows, the others are still on.
Three types of lights and their power ratings are listed below.
Tail light 8.0W
Sidelight 5.0W
Headlight 60W
State and explain which lamp filament has the least resistance.
Headlight - it has the greatest power rating (so more charge can flow through the headlight than the other lights per second ig?)
A wire is stretched to twice its original length by a process that keeps its volume constant. If the resistivity of the metal of the wire remains constant, show that the wire’s resistance quadruples.
contant volume ->
l1 * 2 = l1
and
A1 / 2 = A2
therefore new R = (p * 2l) / (1/2 A)
R new = 4R
Explain, in terms of electron movement, why the resistance of the filament lamp
changes as the voltage increases. [3]
- As voltage increases the current increases
- More collisions (per second) between the (conduction) electrons and the lattice ions / vibration of the lattice/wire/metal ions increases
- (Rate of) vibration of the lattice ions increases, causing a greater number of collisions per second and so causing increased resistance
For full marks, you must use either the phrase “rate of” or “per second”
A series circuit has a cell with emf ε and internal resistance r, a current of I and two resistors. The potential difference across R1 is V1 and the pd across R2 is V2.
Explain how the law of conservation of energy applies to this circuit.
You should consider the movement of one coulomb of charge around the circuit. [2]
(1 C of) the charge gains ε J on passing through cell
OR
energy transferred (by 1 C) in R1 is V1 (J)
OR
energy transferred (by 1 C) in R2 is V2 (J)
OR
energy transferred (by 1 C) in r is Ir (J) (1)
ε = IR1 + IR2 + Ir = V1 + V2 + Ir (1)
(for second mark, you need either the middle or last expressions)
Energy transferred may be written as work done
It says energy in the question, so use the word energy in your answer!
What is made when a voltmeter is connected in parallel to a variable resistor?
A potential divider
Oh no! A variable resistor has three manufacturing faults. What effect on the rate of increase in p.d. for each single movement of the sliding contact will there be when there’s:
a) a thinner layer of conducting material between A and P?
b) a scratch at Q?
c) a conducting connector laid over the conducting layer at R?
Explain why.
a) greater rate of increase in p.d. as there’s less resistance
b) short steep increase (large rate of increase) in p.d. as, again, there’s less resistance
c) no increase in p.d. (i.e. the p.d. reaches its maximum when the sliding contact reaches R) as the current would just flow through the conducting connector instead (this has a lower resistance than the conducting layer of the variable resistor)
Resistors X and Y are connected in series with a 6.0 V battery of negligible internal resistance.
X has resistance R and Y has resistance 0.5R.
A voltmeter of resistance R is connected across Y.
What is the reading on the voltmeter?
Let R = 2Ω:
Resistance of Y and voltmeter together in parallel = 2/3Ω
Total resistance = 2/3 + 2 = 8/3Ω
V = IR -> I through series circuit = V / R = 6 / (8/3) = 9/4A
V across resistor Y and voltmeter = IR = 9/4 * (2/3) = 1.5V
I think this is correct
A circuit consists of a cell, a thermistor in one loop with ammeter 1 and a fixed resistor with ammeter 2 in a parallel circuit.
The cell has a constant electromotive force and negligible internal resistance. Readings from the two ammeters are taken.
Describe what happens to the current in each ammeter when the temperature of the thermistor decreases.
Temp decrease -> R of thermistor increases -> reading on A1 decreases
The p.d. across the resistor and resistance of the fixed resistor remain constant, so the reading on A2 is unchanged
https://www.youtube.com/watch?v=sLJwAV9WhSQ
A circuit is set up so a 1.5V cell with negligible internal resistance is connected to an ammeter in series, a voltmeter and 20Ω resistor in parallel and a second loop with two 10Ω resistors.
What are the readings on the ammeter and voltmeter?
The voltmeter has infinite resistance so no current flows through the first loop.
This means the circuit where current flows is essentially just the cell, ammeter and two 10Ω resistors - the total resistance is therefore 20Ω.
Using I = V / R, I = 1.5 / 20 = 0.075A
The voltmeter is connected in parallel with the cell, so the p.d. reading is 1.5V
Which is equivalent of the ohm?
A JC-2s-1
B Js
C JC-2s
D Js-1
V = IR, R = V / I
E = VQ, V = E / Q
Q = It, I = Q / t
R = (E / Q) / I = (E / Q) / (Q / t)
Ω = (JC-1) / (C-1s-1)
= JC-2s -> C
Lamps A and C are identical with a resistance of 6Ω and operating voltage of 6.0V, and lamps B and D are identical with a resistance of 3Ω and operating voltage of 3.5V.
Lamps A and B are in series with each other while lamps C and D in series - these pairs of lamps are in parallel with each other. A 9V battery is in parallel with the pair of lamps A & B and the pair of lamps C & D.
Determine which two lamps are operating at their normal brightness.
The proportion of the resistance of a lamp is proportional to the potential difference across the lamp
Therefore the p.d. across A & C is 6Ω/9Ω * 9V = 6V, the p.d. across B & D is 3Ω/9Ω * 9V = 3V
so the two lamps which have potential difference across them equal to their operating voltage is A & C, so these are the two lamps working at their normal brightness
