3 Electric circuits

PD

energy transferred between two points in a circuit per unit charge

EMF

work done per unit charge by the power supply or cell, converting energy into electrical potential energy of the charges

current

rate of flow of charge

derive the equation linking current with the number of electrons N flowing past a point in a time Δt with the current

The charge flow ΔQ = Ne where e is the magnitude of the charge on each electron, e=1.6×10^-19.

I=Ne/Δt

Explain how to calculate the number of electrons in a certain amount of charge

Number of electrons = total charge/charge on one electron.

N=Q/1.6×10^-19

Resistance

ratio of pd to current

Derive the formula for the ratio of currents down parallel branches of a circuit

the pd across each branch is the same, V₁=V₂.

V=IR so I₁R₁=I₂R₂.

So I₁/I₂=R₂/R₁.

Or, the ratio of the currents is the reciprocal of the ratio of the resistances.

Derive the formula for resistors in series

V_{total =} V₁ + V₂ + V₃ due to energy conservation.

V=IR so I_totalR_total = I₁R₁ + I₂R₂ + I₃R₃.

I_total = I₁= I₂ = I₃ due to charge conservation.

IR_total = IR₁ + IR₂ + IR₃.

R_total = R₁ + R₂ + R₃.

Derive the formula for resistors in parallel

I_total = I₁ + I₂ + I₃ due to charge conservation.

I = V/R so V_total/R_total = V₁/R₁ + V₂/R₂ + V₃/R₃.

V_total = V₁ = V₂ = V₃ = V due to energy conservation

V/R_total = V/R₁ = V/R₂ = V/R₃

1/R_total = 1/R₁ + 1/R₂ + 1/R₃

Calculate the effective resistance between P and Q

kirchoff’s pd law

The sum of the pd drops is equal to the sum of the emfs around a closed loop within a circuit. Due to conservation of energy

Kirchoff’s current law

The sum of the currents into a junction is equal to the sum of the currents out of the junction. Due to conservation of charge

A student wants to determine the resistance of the resistor shown in the circuit. Student A claims that both the voltmeter and ammeter should be ‘ideal’ in order for an accurate measurement using this circuit. Student B says it is only the ammeter that needs to be ‘ideal’. Explain who is correct.

To determine the resistance of the resistor, the student would use R=V/I where V is the pd across the resistor and I is the current through the resistor.

An ideal ammeter has 0 resistance.

If the ammeter had non-zero resistance then there would be pd across it.

Therefore the pd measured by the voltmeter would be the sum of the pd across the ammeter and the pd across the resistor. This would be problematic as the resistance calculation requires only the pd across the resistor.

An ideal voltmeter has ∞ resistance.

If the voltmeter had non-infinite resistance then it would draw some current and reduce the total resistance of the circuit, increasing the overall current.

However, this would not affect the resistance determination, as the ammeter would still be measuring the genuine current through the resistor.

So student B is correct.

A student wants to determine the resistance of a resistor shown in the circuit. Student A claims that both the voltmeter and ammeter should be ‘ideal’ in order for an accurate measurement using this circuit. Student B says that it is only the voltmeter that needs to be ‘ideal’. Explain who is correct.

To determine the resistance of the resistor, the student would use R=V/I where V is the pd across the resistor and I is the current through the resistor.

An ideal voltmeter has ∞ resistance.

If the voltmeter had non-infinite resistance then it would draw some current (and increase the total current in the circuit as the total R of the circuit would decrease).

This would mean that the ammeter reading would give the sum of the current through the resistor and the current through the voltmeter.

This would mean the resistance calculation would not be possible as this should only use the current through the resistor, which is not measured here.

If the ammeter had non-zero resistance, then the total R of the circuit would increase and there would be pd across it.

Neither of those consequences are problematic though, as the voltmeter still only reads the pd across the resistor, and the current reading is still the same current passing through the resistor.

So the ammeter can have a non-zero resistance and student B is correct.

Lost volts

the energy per unit charge transferred to the internal resistance of a cell

terminal pd

terminal pd, V=EMF − Ir where r is the internal resistance of the cell and I is the current through the cell.

Explain why connecting an ideal voltmeter directly around a cell allows us to measure the emf directly.

The ideal voltmeter has infinite R.

So the current is approximately 0.

So as terminal pd = EMF − lost volts, V = E − Ir.

The lost volts is 0 and the terminal pd = EMF.

A teacher connects a circuit as shown. Student A says the readings on each voltmeter should be the same. Student B says they should be different. Explain who is correct.

Student A is correct

Because the wires we assume to have0 resistance, therefore there is no pd dropped across the wires.

So, as the sum of the EMF = the sum of pd drops

Both voltmeters read the terminal pd, V = EMF − Ir

A student connects a circuit as shown. Explain what happens to the reading on the voltmeter as the R of the variable resistor is decreased.

The voltmeter reads the terminal pd, V.

V = EMF − Ir

As the R of the variable resistor is decreased, the total R in the circuit decreases

Meaning the current I = EMF/R_total will increase.

So the lost volts, Ir will increase.

And as the EMF is fixed, terminal pd will decrease.

Explain why a battery will get hot when the current passing through it is high

If the current is high, the lost volts Ir will be high.

SO there will be a large amount of energy per unit charge transferred to the internal R of the cell.

This energy is transferred to the resistor ions, causing them to vibrate with higher amplitudes, increasing the temperature.

Draw the circuit diagram required to determine the emf and internal resistance of a cell

Describe how to perform an experiment to determine the EMF and internal R of a cell

Set up a series circuit with an ammeter in series with a variable resistor and cell.

A voltmeter should be connected across a load resistor. This will measure terminal pd (this is identical to connecting the voltmeter across the cell).

Measure current using the ammeter.

Change the current by changing the resistance of the variable R and measure the current and pd again for a range of currents.

Plot current on x axis, terminal pd on the y axis.

V = −r I + EMF

y = mx + c

Gradient of line of best fit graph = −r, y-intercept = EMF.

To determine the internal resistance multiply the gradient by −1

Explain the significance of the x and y intercepts of this graph

The x intercept is when the terminal pd = 0.

This means that all of the EMF is being transferred to the internal R of the cell as ‘lost volts’.

No energy is being transferred to the load R and the current cannot continue to increase.

The y-intercept is when the EMF = the terminal pd.

This is when the current = 0 and therefore no energy is being transferred to the internal resistor.

lost volts, Ir = 0.

Power

rate of transfer of energy (P=E/t)

Derive power equations from P=VI

a) P=VI

b) V=IR

substituting b into a, P = IRxI = I²R

c) I = V/R

substituting c into a, P = V x V/R = V²/R

Explain which bulb is brighter in each case.

In circuit A, the pd V across each bulb is the same.

Therefore, as P = V²/R, the bulb with the lower resistance will have a higher power.

Therefore the 20Ω bulb is brighter in A.

In circuit B, the current I through each bulb is the same.

There fore as P = I²R, the bulb with the higher R will have a higher power.

So the 40Ω bulb is brighter in B.

The cell has internal resistance.

i) How does the pd across A and B before an extra light bulb is connected compare with that after an extra light bulb is connected?

ii) What can we say about the brightness of each light bulb after an extra light bulb is connected to the circuit (assuming the brightness of the bulb increases with the energy transferred)?

i) When the extra light bulb is connected in parallel, the total R of the circuit decreases.

This means the total current in the circuit increases, as I_total = EMF/R_total and the EMF is fixed.

So the lost volts, Ir, increases.

This means the terminal pd across AB decreases, as V_terminal = EMF − Ir.

ii) The power across AB is given by P = V²/R where V is the pd across AB and R is the resistance of the bulb between A and B.

As the terminal pd decreases as extra bulbs are added (because lost volts increase due to increase in current caused by the decrease in total R), the power across AB will decrease.

And therefore the brightness will decrease as extra bulbs are added.

Electrical power is transmitted by a cable of resistance 1.2Ω. The power transmitted from the power station is 2000MW and the transmission pd is 1200kV. Calculate the efficiency of this process.

Transmission pd is not the pd across the cable.

current I_transmission = P_transmission/_transmission = 2×10⁹/1.2×10⁶ = 1667A

Power delivered to the cable = I²R = 1667² × 1.2 = 3.334×10⁶W = power wasted

Efficiency = useful power output/total power input

Useful power output = 2000MW − 3.33MW = 1997MW

Efficiency = 1997/2000 = 0.9985

99.9% efficient

Name and define each of the terms in the equation I=nqvA

I is current - rate of flow of charge

n is the number density of free charge carriers - the number of charge carriers per unit volume (n=N/V)

q is the charge on an electron, 1.6×10^-19C

v is the drift velocity of electrons - the mean speed of electrons due to an applied pd.

A is the cross sectional area of the material carrying current.

Derive I=nqvA

There are N electrons moving with a drift velocity v along a wire of length l and cross sectional area A.

The current I = ΔQ/Δt →equation 1

ΔQ = Nq where q is the charge on the electron.

v=l/Δt so Δt = l/v

substituting into equation 1:

I=ΔQ/Δt = Nqv/l →equation 2

There is a number density n of electrons n=N/V, so N=Vn=Aln

substituting into equation 2:

I = Nqv/l

=Alnqv/l

=nqvA

Draw a diagram of a circuit you could use to determine the R of a wire

Explain how to determine the R of a component from a VI graph

Choose a point on the IV graph and identify the current and pd values.

R=V/I. Note that this is NOT the gradient of the line, as R is not ΔV/ΔI

Draw the IV graphs and the VI graphs for an ohmic conductor of high resistance (a), an ohmic conductor of low resistance (b), a filament lamp, a thermistor and a diode.

Explaion the IV relationship for a filament bulb

As current increases, the electrons collide more frequently with lattice resistor ions, transferring more energy to them per second OR as pd increases, electrons gain more energy and transfer more energy to lattice resistor ions.

This causes the resistor ions to oscillate with greater amplitudes so the temperature of the resistor increases.

This causes more frequent collisions between the electrons and the lattice ions, limiting the drift velocity.

This causes the current, I=nqvA to increase by a smaller factor than the pd (as n, q, A all fixed).

So, as resistance is the ratio of pd to current, if pd increases by a greater factor than current does, the ratio R=V/I increases.

Explain the resistance behaviour of a thermistor as temperature increases (due to external factors)

As temperature increases, the electrons in the thermistor are provided with more energy.

This means they can be promoted from the valence band of the semiconductor to the conduction band.

As they are free and able to conduct, the number density of free charge carriers, n increases.

Thus, current, I=nqvA will increase by a greater factor than the pd.

As resistance is the ratio of pd to current, R=V/I, as current can be higher for the same pd as this higher temperature, the resistance decreases.

Explain the IV relationship for a thermistor (here, the temperature increases due to the current flowing through the thermistor so we must explain this step first).

As current increases, the electrons collide more frequently with lattice resistor ions, transferring more energy overall to them per second OR as pd increases, electrons gain more energy and transfer more energy to lattice resistor ions.

This causes the resistor ions to oscillate with greater amplitudes so the temperature of the resistor increases.

As temperature increases, electrons can gain energy and so more electrons are promoted into the conduction band of the thermistor, so n, the number density of free charge carriers increases.

Thus, current, I=nqvA will increase by a greater factor than the pd.

So, as resistance is the ratio of pd to current, if current increases by a greater factor than pd does, the ratio R=V/I decreases.

Explain the resistance, light intensity behaviour for an LDR

Long answer:

A higher intensity means a greater number of photons incident on the LDR per second, per m².

So as intensity increases, more photons incident per second means more electrons gain energy from the incident photos per second (note that each electron still gains the same amount of energy from each photon, as photon energy is proportional to frequency not intensity).

More electrons are promoted into the conduction band of the LDR per second, so n, the number density of free charge carriers increases.

Current increases.

So, as resistance is the ratio of pd to current, if pd remains constant as current increases, resistance = V/I decreases.

Short answer:

Higher intensity of EM radiation means more electrons gain energy from photons.

Increasing the number density of free charge carriers

Increasing current, reducing resistance.

Explain the shape of the IV graph shown

This is a diode. It allows current to flow in one direction only.

The diode conducts when the pd is above the threshold voltage, around 0.7V.

For negative pds (in the reverse bias direction) the resistance is very high, but at a very large negative potential difference the diode will conduct in the reverse direction.

Derive the formula for potential dividers.

Describe what is meand by a potential divider circuit and draw a circuit diagram to illustrate this.

A potential divider circuit is a simple circuit involving a fixed supply potential difference that is shared between 2 resistors.

This can either be two separate resistors as shown in A or a rheostat or potentiometer as shown in B.

Describe what will happen to the voltmeter reading in the circuit if the resistance of the variable resistor increases

There is a fixed supply pd and the other resistor is of fixed resistance.

When the variable resistor’s resistance increases, the voltmeter reading increases.

This is because the pd is shared in the ratio of the resistances.

Describe how the voltmeter reading in the circuit changes as the slider is moved from the bottom of the resistor to the top

At the bottom of the resistor, the voltmeter reading would be 0V.

At the top of the resistor, the voltmeter reading would be equal to the supply pd.

Pd is proportional to the length of the wire.A

A student moves the slider along the potentiometer. Explain why pd is proportional to the length of the wire.

Resistance is proportional to length for a constant cross sectional area and resistivity, R=ρl/A.

The current is fixed as the total resistance in the circuit is unchanged as the slider is moved.

V is directly proportional to R as V=IR.

So V∝R∝I

Explain which of these circuits would be better for plotting a graph of current against potential difference for plotting an IV graph for the component.

The right hand circuit would be better as it achieves a range of pds and currents from 0V to the EMF of the cell, whereas the left hand circuit has a minimum pd and current that is not close to 0.

For the right hand circuit, the pd across the component can vary from 0V (when the slider is on the left) tot he maximum pd of the power supply (when the slider is on the right)

and the current through the component will be 0A when the slider is on the left and maximum when the slider is on the right.

For the left hand circuit, the component would always receive a share of the fixed total pd, and the current through it would never be 0.

Minimum pd across the component in the left circuit would be V = EMF x R_component/R_total, and minimum current would be EMF/R_total

so you could not plot an IV graph that starts close to zero for the left hand circuit.

Example graph for left hand circuit (left) and example graph for right hand circuit (right)

Resistivity

The resistance of a unit cube

Describe what resistivity depends on

Resistivity is independent of dimensions of the wire - it is unique to an individual material only and depends on temperature.

Describe how the resistivity of a thermistor changes with temperature

Resistivity of a metal filament increases with temperature (as number density of charge carriers, n increases)

Describe how the resistivity of a metal filament changes with temperature

Resistivity of a metal filament increases with temperature (as drift velocity, v decreases due to more frequent collisions between electrons and ions)

Describe how to perform an experiment to determine the resistivity of a wire

Measure the resistance of a wire using an ohmmeter for different values of the length of the wire.

Measure diameter of wire using micrometer (in three places along the wire, taking an average).

Calculate the cross-sectional area of the wire using A=π(d/2)².

Plot a graph of resistance against length, the gradient of the line of best fit will be the resistivity/cross sectional area.

Multiply the gradient by cross sectional area to get the resistivity.