topic 3 - electric circuits

Define potential difference

Energy transferred (work done) between two points per unit charge. V = W/Q.

Define emf

Work done per unit charge by a power supply, converting energy into electrical potential energy of charges.

Define current

Rate of flow of charge. I = ΔQ/Δt.

Derive I = Ne/Δt

Charge flow ΔQ = Ne (e = 1.6×10⁻¹⁹ C). So I = ΔQ/Δt = Ne/Δt.

Calculate number of electrons from charge

N = Q / e, where e = 1.6×10⁻¹⁹ C.

Define resistance

Ratio of potential difference to current. R = V/I.

Derive ratio of currents in parallel branches

V same across branches: I₁R₁ = I₂R₂ → I₁/I₂ = R₂/R₁.

Derive resistors in series

V_total = V₁+V₂+V₃. V=IR, I constant → R_total = R₁+R₂+R₃.

Derive resistors in parallel

I_total = I₁+I₂+I₃. I=V/R, V constant → 1/R_total = 1/R₁+1/R₂+1/R₃.

Kirchhoff's voltage law

Sum of pd drops = sum of emfs around closed loop (energy conservation).

Kirchhoff's current law

Sum of currents into junction = sum of currents out (charge conservation).

Define lost volts

Energy per unit charge transferred to internal resistance of a cell: Ir.

equation for terminal pd, defining symbols

V = EMF − Ir, where r = internal resistance, I = current.

Measuring EMF with ideal voltmeter

o The ideal voltmeter has infinite resistance o So the current is approximately zero
o So as terminal pd = EMF – lost volts,
o V=E–Ir

o The lost volts is zero
o And terminal pd is equal to EMF

Effect of decreasing variable resistor on voltmeter reading

o The voltmeter reads the terminal potential difference, V
o V=EMF–Ir
o As the resistance of the variable resistor is decreased, the total resistance in the circuit

decreases
o Meaning the current, I = EMF / (Rtotal) will increase o So the lost volts, Ir will increase
o And as the EMF is fixed
o Terminal pd will decrease

Why battery gets hot at high current

o If the current is high, the lost volts, Ir will be high
o So there will be a large amount of energy per unit charge transferred to the internal

resistance of the cell
o This energy is transferred to the resistor ions
o Causing them to vibrate with higher amplitudes, increasing the temperature

Circuit to determine EMF and internal resistance

Experiment to determine EMF and internal resistance

Significance of x-intercept (V=0)

o The x intercept is when the terminal potential difference is zero
o This means that all of the EMF is being transferred to the internal resistance of the cell as ‘lost

volts’
o no energy is being transferred to the load resistance and the current cannot continue to

increase

Significance of y-intercept (I=0)

the y-intercept is when the EMF is equal to the terminal potential difference
o this is when the current is zero and therefore no energy is being transferred to the internal

resistor
o lost volts, Ir is zero

Define power

Rate of transfer of energy. P = E/t.

Derive P = I²R and P = V²/R from P = VI

P = VI. Substitute V = IR → P = I²R. Substitute I = V/R → P = V²/R.

which bulb is brighter in each case

In circuit A, the potential difference, V, across each bulb is the same
o Therefore, as P = V2/R, the bulb with the lower resistance will have a higher power o Therefore the 20Ω bulb is brighter in A
o In circuit B, the current, I, through each bulb is the same
o Therefore, as P = I2R, the bulb with the higher resistance will have a higher power o So the 40Ω bulb is brighter in B

Efficiency of power transmission (cable resistance 1.2 Ω, 2000 MW, 1200 kV)

I = P_transmit / V_transmit = 2×10⁹ / 1.2×10⁶ = 1667 A. Power wasted = I²R = 1667² × 1.2 ≈ 3.33 MW. Useful output = 2000 − 3.33 = 1997 MW. Efficiency = 1997/2000 = 99.9%.

Define terms in I = nqvA

I = current (rate of flow of charge); n = number density of free charge carriers; q = charge on electron (1.6×10⁻¹⁹ C); v = drift velocity; A = cross-sectional area.

Derive I = nqvA

I = ΔQ/Δt = (Nq)/(l/v) = Nqv/l. N = nV = nAl → I = nAlqv/l = nqvA.

Circuit to determine resistance of a wire

Determine resistance from V-I graph

Choose a point, R = V/I. (Not gradient — gradient = 1/R only for ohmic conductor if graph passes through origin.)

IV graphs: ohmic conductor (high R)

IV graphs: ohmic conductor (low R)

IV graph: filament lamp

IV graph: thermistor (self-heating)

Increasing V → I increases more rapidly (resistance decreases).

IV graph: diode

Very low current in reverse bias; forward bias: negligible current until ~0.7 V, then steep increase.

VI graph: ohmic conductor (high R)

Straight line through origin, steep slope (V vs I).

VI graph: ohmic conductor (low R)

Straight line through origin, shallow slope (V vs I).

VI graph: filament lamp

Increasing I → V increases but curve steepens (resistance increases).

VI graph: thermistor (self-heating)

Increasing I → V increases but curve flattens (resistance decreases).

VI graph: diode

Very low I in reverse bias; forward bias: negligible I until ~0.7 V, then steep rise in I for small V increase.

Explain IV relationship for 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

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

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

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

o 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

Resistance behaviour of thermistor as temp increases due to external factors

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

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

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

o Thus, current, I= nqvA will increase by a greater factor than the potential difference.

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

IV relationship for thermistor (self-heating)

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

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

o 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.

o Thus, current, I= nqvA will increase by a greater factor than the potential difference.

o 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

Resistance-light intensity behaviour for LDR

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

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

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

o Current increases

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

Explain shape of diode IV graph

This is a diode o It allows current to flow in one direction only o The diode conducts when the pd is above the threshold voltage, around 0.7V o For positive potential differences above 0.7V the resistance is very low o 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 potential divider formula

Potential divider circuit

Circuit sharing fixed supply pd between two resistors (fixed + variable, or potentiometer).

Effect of increasing variable resistor (circuit a)

There is a fixed supply potential difference and the other resistor is of fixed resistance

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

o This is because the potential difference is shared in the ratio of the resistances

Effect of moving slider on potentiometer (circuit b)

At the bottom of the resistor, the voltmeter reading would be 0V o At the top of the resistor, the voltmeter reading would be equal to the supply potential difference o Potential difference is proportional to the length of the wire

Why pd ∝ length in potentiometer

resistance is proportional to length for a constant cross sectional area and resistivity, R=ρl/A o The current is fixed as the total resistance in the circuit is unchanged as the slider is moved o V is directly proportional to R as V=IR o So V α R α l

Better circuit for plotting IV graph

Right-hand circuit (potential divider with slider): V and I can vary from 0 to max. Left circuit has non-zero minimum.

Define resistivity

Resistance of a unit cube. ρ = RA/l.

What resistivity depends on

Material and temperature. Independent of dimensions.

Resistivity of thermistor with temperature

Resistivity of a thermistor decreases with temperature (as number density of charge carriers, n increases

Resistivity of metal filament with temperature

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

Experiment to determine resistivity of wire

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

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

o Calculate the cross sectional area of the wire using A = π (d/2)2 . o Plot a graph of resistance against length, the gradient of the line of best fit will be the resistivity / cross sectional area.

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

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(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)?

*(b) We usually assume that ammeters have negligible resistance and voltmeters have infinite resistance. The determination of ε and r is not affected by using an ammeter with non-negligible resistance but is affected by using a voltmeter with a low resistance. Explain why.