Electricity

— Physics · AS Level (7408) · Current, Circuits, Kirchhoff's Laws & Potentiometers —

★ Exam Tip: The #1 tested distinction in this chapter is ratio vs rate. Voltage, resistance, resistivity are RATIOS — read one point off a graph and divide, NEVER take a gradient. Current is a RATE — you DO take a gradient (or use area under an I-t graph if current isn't constant).

1. Charge

Charge: "A property that causes an object to experience a force in an electric field." Measured in Coulombs (C). Q=It. Can be +ve or −ve. Scalar. Like charges repel, unlike charges attract.

★ Charge is quantised: Q = Ne (e = elementary charge = 1.6×10⁻¹⁹ C, N = number of elementary charges)

Charge is a disbalance of +ve and −ve charges — e.g. +2C protons and −2C electrons together give net charge 0; remove the electrons and only +2C protons remain as net charge.

2. Current

Current: Rate of flow of charge. Scalar, measured in Amperes (A). I = dQ/dt = ΔQ/Δt || ΔQ = I×Δt

Gradient of a Q-t graph = Current. If current is CHANGING, use the area under an I-t graph to get ΔQ. If current is CONSTANT, ΔQ=I×Δt directly (simple rectangle area).

★ Conventional vs actual current: We follow CONVENTIONAL current — flow of +ve charges. In actuality, current is the flow of negative charges/electrons. If the examiner explicitly asks for the true direction, follow that (electron flow); otherwise always default to conventional current in diagrams and calculations.

Why conventional current stuck: early theory was built assuming positive-charge flow (before electrons were discovered). Since flipping the assumed direction doesn't change any calculated numeric answer, the "wrong" convention was kept rather than rewriting everything.

3. Voltage: Potential Difference vs EMF

Voltage is measured in volts. Both PD and EMF share the formula V=W/Q (a ratio — read one point off a graph, never a gradient).

Potential Difference	Electromotive Force (EMF)
Work done per unit charge moving that charge from ONE POINT TO ANOTHER in a circuit. OR: work done per unit charge converting electrical energy to other forms.	Work done per unit charge moving that charge AROUND THE WHOLE CIRCUIT. OR: work done per unit charge converting chemical energy to electrical energy.

Shotgun analogy — voltage vs current, "what kills": current is like the bullet doing the damage; voltage is like the shooter initiating and driving it. Current cannot exist without voltage driving it ("never any current for zero volts"). They always act together — voltage supplies the energy, current is the flow that disrupts the body's own electrical (nervous) signals.

4. Ohm's Law

"The potential difference across a component is directly proportional to the current, given the temperature remains constant." (Temperature constant ⟹ resistance constant, since R depends on temperature via lattice vibrations.)

V ∝ I ⟹ V = IR ⟹ R = V/I (ratio)

5. Resistance & Resistivity

Resistance: Hindrance to the flow of charge. Scalar, measured in Ohms (Ω). R=V/I (ratio). Caused by lattice vibrations — electrons collide with vibrating positive nuclei, impeding flow.

Relationship	Physical reason
R ∝ l	Longer wire = more lattice collisions along the path
R ∝ T	Higher T → more violent lattice vibrations → more/harder collisions
R ∝ 1/A	Larger area = more "lanes" for electrons to flow around obstructions

★ CRITICAL warning — the three k's are NOT interchangeable: R∝l gives R=k₁l; R∝T gives R=k₂T; R∝1/A gives RA=k₃. These three constants are completely different — never combine them (e.g. never write R/l=R/T). Each relationship is ONLY valid when the OTHER TWO factors are held constant. If two factors change at once, use the unified R=ρl/A instead.

★ R = ρ(l/A) ρ = resistivity → constant for a material, constant for a constant temperature

R₁A₁/l₁ = ρ = R₂A₂/l₂

Copper wire comparisonWorked example

A copper wire (length l, area A) has resistance R. What is the resistance of another copper wire, length l/2, area 2A?

R=ρl/A ⟹ R₁A₁/l₁ = R₂A₂/l₂ ⟹ RA/l = R₂(2A)/(l/2) ⟹ R = 4R₂ ⟹ R₂ = R/4

"Is it voltage or current that kills?" — worked numbers: LED resistance 13Ω, human resistance 10,000Ω. At 9V: LED current ≈0.7A; human current = 9×10⁻⁴A. "It's not the voltage that kills and neither the current alone kills. Since everything has a resistance, you need a specific amount of current to kill, which can only be produced with a high enough voltage."

6. Ohmic vs Non-Ohmic Conductors (I-V Graphs)

Ohmic: V∝I, straight line, R constant

Filament bulb: R↑ as I flows (heats up, curves toward V)

Thermistor: R↓ as I flows (temp↑ → R↓)

Filament bulb: "As current flows, temperature increases, resistance therefore increases and the graph bends towards the voltage axis."

Thermistor: As temp↑, R↓ (used as a thermometer). As temp↓, R↑.

Diode: forward bias needs the 0.7V "knee voltage" to conduct; reverse breakdown at very high −ve V

Circuit Symbols

LDR: As light intensity increases, resistance decreases (and vice versa). Thermistor: As temperature increases, resistance decreases (and vice versa).

7. Series & Parallel Circuits

Parallel connection: 2 or more things connected across each other. Series connection: 2 or more things connected one after another (current has only one direction to flow).

Identifying mixed circuits — 3 worked patterns: 1) R₁ in series with (R₂∥R₃): "R₁ is in series with the combined resistance of R₂ & R₃ in parallel."
2) R₂,R₃ in series (one branch) ∥ R₁ (other branch): "R₂ and R₃ are in series. R₁ is in parallel to the combined resistance of R₂ & R₃."
3) Bridge/diamond arrangement redrawn as two parallel branches, each an internal series pair: "R₁ and R₂ are in series and R₃ and R₄ are in series. The branches are in parallel."

Series Circuits	Parallel Circuits
Current remains the SAME throughout	Voltage is the SAME across each branch
V_T = V₁+V₂+...	I_T = I₁+I₂+...
R_T = R₁+R₂+...	1/R_T = 1/R₁+1/R₂+...
Cells in series ADD: V_T=2+3=6V (increases voltage)	Equal cells in parallel: V_T stays the same, e.g. 3V (provides more current/capacity)

Derivation — series resistance:

V_T=V₁+V₂+V₃ ⟹ (I×R_T)=(I×R₁)+(I×R₂)+(I×R₃) [I cancels] ⟹ R_T=R₁+R₂+R₃

Derivation — parallel resistance:

I_T=I₁+I₂+I₃ ⟹ (V/R_T)=(V/R₁)+(V/R₂)+(V/R₃) [V cancels] ⟹ 1/R_T=1/R₁+1/R₂+1/R₃

8. Voltmeters vs Ammeters

Voltmeter	Ammeter
Measures voltage. Always connected in PARALLEL.	Measures current. Always connected in SERIES.
Ideal: ∞ resistance	Ideal: ZERO resistance
Why: R=∞ ⟹ I=V/R=V/∞=0 — so no current flows into the voltmeter, and the circuit being measured isn't disturbed	Why: R=0 ⟹ V=IR=0 — no voltage drop across it, no hindrance to current, doesn't disturb the circuit

9. Worked Circuit Exercises

Exercise 1 — series circuit, lamp turn-on + charge/workWorked example

4Ω resistor, 1Ω lamp, 3Ω resistor, 12V battery, all in series, current 1.5A.

Will a lamp needing 3V turn on? V_T=I_T R_T: 12=I_T(4+1+3) ⟹ I_T=1.5A; V(1Ω lamp)=(1.5)(1)=1.5V ⟹ lamp will NOT turn on (1.5V < 3V) Circuit runs 1hr: Q_T=I×Δt=1.5×3600=5400C; W_T=V_T×Q_T=12×5400=64,800 J Work done across the 3Ω resistor: V(3Ω)=(1.5)(3)=4.5V; W=(4.5)(5400)=24,300 J

Exercise 2 — short circuit concept, correcting a flawed circuitConceptual

John's circuit has a lamp F that never turns on when the switch closes.

Q) What is a short? "You give a zero resistance pathway" ⟹ I=V/R→0 ⟹ I=∞ — current takes the zero-resistance path, bypassing the intended component entirely.

The fix: put the ammeter directly in series with the lamp, and the voltmeter directly across just the lamp — remove any parallel path that bypasses (shorts out) the lamp.

Exercises 3–4 — combined series-parallel networksWorked examples

Two parallel branches (2Ω+1Ω vs 3Ω+4Ω) off a 36V battery: R_T=(1/3+1/7)⁻¹=2.1Ω ⟹ I_T=36/2.1=17.14A; branch currents 12A and 5.14A; branch voltages 24V, 12V, 15.4V, 20.6V — found via V=IR on each resistor using its own branch current.

A more complex 5-branch network (10V battery) is solved the same way: find R_T for each parallel section first, use V_T=I_T R_T for the total current, then apply V=IR locally within each branch using that branch's own current.

★ General method for combined networks: 1) simplify parallel sections into their equivalent resistance. 2) Find total current from the battery via V_T=I_T R_T. 3) Work OUTWARD from the battery: the total current splits at each parallel junction according to each branch's own resistance (V=IR locally), and voltmeters/ammeters read whatever is local to their position.

Exercise 5 & 6 — reading EMF/current directly from I-V graphsGraph-reading technique

Series circuit: variable resistor + filament lamp, ammeter reads 3A. I-V graph gives two curves ("resistor," "filament lamp"). At I=3A, resistor reads V=30, lamp reads V=40.

Series ⟹ same current through both, total V = sum: V_T = 30+40 = 70V (the battery's EMF)

Parallel circuit: variable resistor (4.2A) ∥ another component (3A), battery 6V. At V=6V, resistor branch reads I=3A, other branch reads I=4.2A.

Parallel ⟹ same voltage across both, total I = sum: I_T = 3+4.2 = 7.2A

Trick: read the SHARED quantity (current for series, voltage for parallel) straight off the x/y axis at the given value, then read each curve's corresponding value — no algebra needed, just careful graph reading.

10. Electrical Power & Energy

Three equivalent derivations from P=dW/dt:

V=W/Q ⟹ W=VQ; P=d(VQ)/dt = V(dQ/dt) [V const] ⟹ ① P=IV
Sub V=IR: P=I(IR) ⟹ ② P=I²R
Sub I=V/R: P=(V/R)V ⟹ ③ P=V²/R

E = IVt OR E = I²Rt OR E = V²t/R

w18qp22q6 — series circuit + resistivityStructured Q

Battery EMF 7.0V (negligible r), in series with resistor X (5.2Ω), lamp Y (6.0Ω), resistor Z (unknown, PD=1.4V).

(i) Current: V_T=V₁+V₂+V₃: 7=I(5.2)+I(6)+1.4 ⟹ I=0.5A (ii) R_Z=V/I=1.4/0.5 ⟹ R_Z=2.8Ω (iii) % efficiency battery→lamp: η=(P in lamp)/(P by battery)×100 = I²R/(IV)×100 = [(0.5)²×6]/[(0.5)×7]×100 ⟹ η=43% (iv) Filament resistivity 3.7×10⁻⁷Ωm. Find α=(cross-sectional area)/(length): R=ρl/A ⟹ A/l=ρ/R ⟹ α=3.7×10⁻⁷/6 ⟹ α=6.2×10⁻⁸ m

s18qp21q6 — filament lamps in parallel + resistivity ratioStructured Q

Battery EMF 4.5V (negligible r), lamps P&Q in parallel, resistor R in series with the pair. I_P=0.15A. I-V graph given for both lamps.

(i) Both lamps share the same PD (parallel). At that shared V (2.7V from graph): I_P=0.15A, I_Q=0.09A ⟹ I_T=0.24A (ii) R_P=2.7/0.15=18Ω; R_Q=2.7/0.09=30Ω; R=(V_T−2.7)/I_T=(4.5−2.7)/0.24 ⟹ R=7.5Ω (iii) Diameter of P's wire = 2× diameter of Q's wire, same resistivity. Ratio l_P/l_Q: R=ρl/A ⟹ l=AR/ρ=πd²R/4ρ; l_P/l_Q = 4×(R_P/R_Q) = 4×(18/30) ⟹ ratio=2.4

(iv) Lamp Q's wire breaks. Effect on lamp P's resistance? "The voltage across lamp P increases. As the voltage increases, there is a greater current through P and therefore its resistance also increases."

w19qp22q6 — diode + resistor circuitStructured Q

I-V graph given for resistor X and a diode. Circuit: cell EMF E, resistor Y in parallel with (diode in series with X). Cell current 9.3mA, diode current 7.5mA.

(i) Diode resistance at V=0.60V: R=V/I=0.6/(7.5×10⁻³) ⟹ R=80Ω (ii) As V rises 0.60→0.75V, diode resistance decreases (confirmed: R drops from 80Ω to 56Ω at 0.7V/12.5mA) (c)(i) Reading the diode's I-V graph at 7.5mA gives ≈0.6V across it; E = 0.6+0.3 (extra drop across X) ⟹ E=0.90V (ii) Current through Y = 9.3−7.5=1.8mA; V=IR: 0.90=(1.8×10⁻³)R_Y ⟹ R_Y=500Ω (iii) Power in diode: P=IV=(7.5×10⁻³)(0.6) ⟹ P=4.5×10⁻³ W (iv) New cell EMF 0.50V: reading the graph at V=0.50V for the diode ⟹ current≈2.5mA

11. Deriving I = nAve

Electron density (n): Number of electrons per unit volume, constant for a material [unit: m⁻³].

Derivation:

N (number of electrons) = n×V = n×A×l
Q=Ne ⟹ Q=n×A×l×e
I=dQ/dt = d(nAle)/dt = nAe×(dl/dt) [n,A,e constant, dl/dt=drift velocity v]

★ I = n×A×v×e (v = drift velocity of electrons)

w18qp21q6 — heater wires, Kirchhoff + drift velocity ratioStructured Q

Heater: wires X(2.4Ω), Y(1.2Ω) in parallel, 9.0V supply (negligible r), variable resistor R, voltmeter reads 6.0V.

(i) R_XY=(1/2.4+1/1.2)⁻¹=0.80Ω; I=V/R=6/0.8=7.5A; R=V/I=3/7.5 ⟹ R=0.40Ω (ii) Power in X: P=V²/R=6²/2.4 ⟹ P=15W (iii)1. A_X=3×A_Y, same ρ,n. Ratio l_X/l_Y: R=ρl/A ⟹ R₁A₁/l₁=R₂A₂/l₂ ⟹ (2.4)(3A)/l_X=(1.2)(A)/l_Y ⟹ ratio=6.0 2. Ratio of drift velocities v_X/v_Y: I=nAve ⟹ v=I/(nAe); v_X/v_Y=(I_X/I_Y)×(1/3) [A_X=3A_Y]; I_X=2.5A, I_Y=5A ⟹ ratio≈0.17

w17qp22q6 — damaged wire, I=nAve applied to varying cross-sectionStructured Q

(i) Total charge of free electrons in a wire: Q = nAle (ii) I=dQ/dt=d(nAle)/dt=nAe(dl/dt) ⟹ I=nAve — shown

Wire damaged: resistivity unchanged, diameter reduces from d (at X) to 0.69d (at Y) over 3.0mm. Current 0.50A.

(i) Ratio v_Y/v_X: v=I/(nAe) ⟹ v_Y/v_X = A_X/A_Y = 1/0.69² ⟹ ratio=2.1 (ii)1. Main wire (X) has resistance/length 1.7×10⁻²Ωm⁻¹. Resistance/length at Y: (R/l)_Y = (R/l)_X × A_X/A_Y = 1.7×10⁻² × (1/0.69²) ⟹ ≈0.036 Ωm⁻¹ 2. Power dissipated: P=I²R = (0.5)²×(3.6×10⁻²×3×10⁻³) ⟹ P≈2.7×10⁻⁵ W

(iii) Diameter further decreases, I constant. "As the diameter decreases, the cross sectional area decreases and so the resistance increases. The power will also increase as P=I²R [with I constant]."

w18qp22q7 — sketching drift-speed vs position (tapering wire)Structured Q — graph sketch

Wire XY, diameter varies linearly: d at X, d/2 at Y. Current I is the same throughout (charge conservation).

I=nAve, I,n,e const ⟹ v∝1/A ⟹ v∝1/d² v_X×d² = v_Y×(d/2)² ⟹ v_Y = v_X×d²/(d/2)² = 4v_X

Sketch: a non-linear, increasingly steep rising curve from v_x at X to 4v_x at Y (matches v∝1/d² as d shrinks linearly).

12. Internal Resistance

The small resistance r drawn with each battery to account for power losses through it.

Derivation:

V_T=I_T R_T; E=I(R+r); E=IR+Ir; E=V_t+Ir

★ V_t = E − Ir (y=mx+c form: gradient=−r, y-intercept=E)

V_t vs I: gradient=−r, y-intercept=E, x-intercept=E/r. (Green: EMF halved, r doubled)

Note: Power delivered to the external resistor R is MAXIMUM when r=R.

Choosing R₁ vs R₂ for the same power P₂ (either side of the peak): the power loss through internal resistance is ↓P=↓I²r — since a LARGER R (R₂) means a SMALLER current, it wastes LESS power in the internal resistance even though the useful power P₂ delivered is the same either way. Prefer the larger resistance (smaller current) option.

s19qp21q6 — full internal resistance structured QStructured Q with graph

(a) Why is V < E? "Energy is wasted in the internal resistance." (b) V=E−Ir

(c)(i) V-I graph, y-intercept=2.8: "Upon linearising V=E−Ir, we can see that the e.m.f. can be given by the y-intercept."

(ii) gradient=−r=(0.7−2.8)/(1.5−0) ⟹ r=1.4Ω (d)(i) Battery stores 9.2kJ, V adjusted to 2.1V. From graph at V=2.1, I=0.5A; R=V/I=2.1/0.5 ⟹ R=4.2Ω (ii) Electrons through battery in 1.0s: Q=It=0.5×1=0.5C; Q=Ne ⟹ N=3.1×10¹⁸ (iii) Time for energy to drop to 1.6kJ: P=ΔW/Δt ⟹ Δt=(9.2−1.6)×10³/(0.5×2.8) ⟹ Δt=5.4×10³ s

w20qp23q6 — R∝1/A² derivation, R vs 1/I graph linearizationStructured Q

(b) Show R∝1/A² (volume V constant as wire extends): R=ρl/A; V=lA ⟹ l=V/A; R=ρ(V/A)/A=ρV/A² ⟹ R∝1/A² (ρ,V constant) (c) Kirchhoff's 2nd law: R=(E/I)−r [from V_t=E−Ir and R=V_t/I]

(d) R vs 1/I graph given: linearized form R=E(1/I)−r, gradient=E, y-intercept=−r.

(i) Power in R at I=2.0A: 1/I=0.5 ⟹ R=5.4Ω (from graph); P=I²R=(2)²×5.4 ⟹ P=22W (ii)1. y-intercept=−r=−0.6 ⟹ r=0.6Ω 2. gradient=E=(5.4−(−0.6))/(0.5−0) ⟹ E=12V

13. Kirchhoff's Laws

Kirchhoff's First Law (KCL): Sum of incoming currents at a junction = sum of outgoing currents. Based on conservation of charge.

★ ΣI_incoming = ΣI_outgoing

Three worked KCL junctionsWorked examples

1) In: 1A,5A,12A. Out: 3x. ⟹ 1+5+12=3x ⟹ x=6A 2) In: 0.5A,2.5A. Out: x ⟹ x=3A. [Separate junction] In: 4A,9A,3A. Out: y ⟹ y=16A 3) In: 7A,8A,3A. Out: xA (all treated as one signed equation): 7+8+3+x=0 ⟹ x=−18A — negative result tells us our marked direction was WRONG (it's actually incoming, not outgoing)

Kirchhoff's Second Law (KVL): Sum of EMFs = sum of PDs in a closed loop. "The sum of voltages in a closed loop is always zero." Based on conservation of energy.

+E − IR = 0 ⟹ E = IR

★ Sign-convention method (5 steps):

Make any ONE battery your reference battery and mark its + and − terminals.
For any resistors/energy-consuming devices, make the leg closer to the battery's + terminal "+".
Choose a direction of looping: ACW or CW.
As you move through a component: + → − treat the voltage as +ve; − → + treat it as −ve.
Write the equation down and equate it to zero.

★ Multi-component rule: for multiple components, choose only ONE reference battery and use it to mark polarity of ALL resistors. If you have more than one battery, you MUST use KVL (can't just add EMFs directly).

Three worked KVL loops (single → multiple batteries)Worked examples

1 battery, 2Ω+4Ω: +12−I(4)−I(2)=0 ⟹ 12=6I ⟹ I=2A 1 battery + extra drop, 4Ω+3Ω: +12−3I−6−4I=0 ⟹ 6=7I ⟹ I=6/7 A 3 batteries (6V,3V,2V) + three resistors (2Ω,1Ω,2Ω) in one loop: +6−2I−3−1I−2I+2=0 ⟹ 5=5I ⟹ I=1A

w17qp21q5 — three-cell circuit, KCL+KVL across 2 loopsStructured Q

Rectangle WXYZ with cells E₃,E₁ and a diagonal branch (R₂,E₂). Currents I₁,I₂,I₃.

(a) KCL: I₃ = I₁ + I₂ (b) Loop WXYZW (ref E₃, CW): I₃R₃ + I₃R₄ + I₁R₁ = E₁+E₃ (c) Loop YZWY (ref E₁, CW): E₂−E₁ = I₂R₂−I₁R₁

s19qp22q5 — two batteries in one loop (no external R)Structured Q — combines internal r found earlier

5.6V battery (r=2.5Ω, found from an earlier part) connected to a second battery (7.2V, r=3.5Ω), no external resistor.

+7.2 − I(3.5) − I(2.5) − 5.6 = 0 ⟹ 1.6=6I ⟹ I=0.27A

14. Potentiometers

Potentiometer: An electrical device used to measure the EMF of a battery. A driver cell (known EMF) drives current through a uniform wire; a galvanometer (used as a sensitive ammeter) finds the "balance point" where no current flows in the unknown-battery branch.

★ Why this works: V_t=E−Ir. At balance, I=0 ⟹ V_t = E exactly — the terminal voltage equals the EMF precisely when no current flows. This is why a potentiometer measures EMF directly, unlike a voltmeter which always draws some current (however small) and therefore always reads slightly less than the true EMF.

Since R∝l and V∝R (constant driver current) ⟹ ★ V ∝ l (voltage along the wire is proportional to length — the basis of every potentiometer calculation)

Basic balance-point exampleWorked example

Driver: 12V across a 1m wire. Balance point at 62.5cm.

12V:1m = x:0.625m ⟹ x = E = 7.5V

★ Progressive variants — where an added resistor matters: the position of an extra resistor in the circuit determines whether it affects the balance-point calculation at all.

Variant A — resistor in the GALVANOMETER branch (does NOT affect the answer)Worked example

A 5Ω resistor sits in the unknown-battery branch (in series with the galvanometer). At balance, I=0 in this whole branch.

Since I=0, there is ZERO voltage drop across EVERY component in that branch — including the 5Ω resistor. The ratio is completely unaffected: 2V:1m = 1.5V:XP ⟹ XP=0.75m

Rule: any resistor purely in the galvanometer/unknown-battery branch never affects the balance point, since I=0 there means V=IR=0 for every series component in it.

Variant B — resistor in SERIES with the driver battery (DOES affect the answer)Worked example

Now a 1.0Ω resistor sits in series with the 2V driver battery itself, before the wire (wire resistance 5Ω).

This resistor carries the full driver current, so it DOES matter: V_T=I_T R_T: 2=I_T(1+5) ⟹ I_T=1/3 A Voltage across just the WIRE (excluding the added 1Ω): V_wire=I×R_wire=(1/3)(5)=5/3 V Use this REDUCED voltage (not the full 2V) for the ratio: (5/3)V:1m = 1.5V:XP ⟹ XP=0.9m

★ Key distinction: a resistor in the driver loop (carrying the driving current) reduces the effective voltage available across the wire — recompute V_wire first via V=IR before setting up the length ratio. A resistor in the galvanometer branch (Variant A) carries no current at balance and can be ignored entirely.

Variant C — resistor changes the LOCAL loop's own currentWorked example

Same setup as Variant B (1Ω in the driver loop, wire=5Ω, driver=2V), unknown battery 1.5V/0.8Ω.

V_T=I_T R_T: 1.5=I_T(0.8+1) ⟹ I_T=5/6 A [local loop reasoning for the unknown battery's own branch] Cross-check via V_t=E−Ir: V_t=1.5−(5/6)(0.8)=5/6V ✓ Ratio: 2V:1m = (5/6)V:XP ⟹ XP=0.417m

s21qp21q5 — full potentiometer structured QStructured Q

Battery EMF 4.0V, r=0.35Ω, wire XY (0.90Ω) + fixed resistor R. PD across XY = 1.8V.

(i) Current in XY: V=IR: 1.8=I(0.90) ⟹ I=2.0A (ii) Electrons in 45s: Q=It=2(45)=90C; Q=Ne ⟹ N=5.6×10²⁰ (iii) R: V_T=I×R_T: 4=2(0.35+0.90+R) ⟹ R=0.75Ω (c)(i) Cell EMF 1.2V, balance at XP=0.30m. Total length L: 1.8V:L = 1.2V:0.30m ⟹ L=0.45m

(ii) R replaced with a GREATER resistance. Effect on P's position? "As R increases, the voltage XY decreases and therefore the pointer will have to be moved to the right to once again give a voltage of 1.2V."

w21qp23q6 — resistivity→resistance, wire cross-section comparisonStructured Q

(a) R=ρl/A=(5×10⁻⁷)(2)/(3.3×10⁻⁷) ⟹ R≈3.0Ω — shown (b)(i) EMF 1.50V, PD across XY=1.20V: V=IR: 1.2=I(3) ⟹ I=0.40A (ii) V_t=E−Ir: 1.2=1.5−0.4r ⟹ r=0.75Ω (c) Balance at XP=1.4m (wire length 2.0m): 1.2V:2m = E:1.4m ⟹ E=0.84V

(d) Wire replaced by a SECOND wire, same length & material, SMALLER cross-sectional area. Is the new XP shorter, longer, or the same? "As the area decreases, the resistance of the wire will increase and so the potential difference across it will also increase. Length XP will have to be made shorter to get the same 0.84V."

✓ Quick Recap — Full Chapter Cheat-Sheet

★ Ratio vs Rate: V, R, ρ are RATIOS (read a point, never a gradient). Current is a RATE (gradient of Q-t, or area under I-t if not constant).
Q=It=Ne. V=W/Q for both PD and EMF. V=IR (Ohm's law, temperature/R constant).
R=ρl/A. ★ The three separate proportionality constants (R∝l, R∝T, R∝1/A) are NEVER interchangeable and each needs the OTHER TWO factors held constant.
Series: I same, V adds, R adds. Parallel: V same, I adds, 1/R adds.
Voltmeter: parallel, ∞ resistance. Ammeter: series, 0 resistance.
P=IV=I²R=V²/R. I=nAve.
V_t=E−Ir. At I=0 (open circuit or potentiometer balance), V_t=E exactly. Power to external R is max when r=R.
KCL: ΣI_in=ΣI_out. KVL: ΣEMF=ΣPD in a loop — use ONE reference battery, mark polarities, loop consistently, +→− is positive.
★ Potentiometer: V∝l along the wire. A resistor in the GALVANOMETER branch never matters (I=0 there at balance). A resistor in the DRIVER loop DOES matter — recompute the effective V across the wire first.