Electricity & Charge Basics

Question 1

(a) The ampere is one coulomb of charge passing a point per second: $1\,\text{A} = 1\,\text{C/s}$, i.e. $I = q/t$. (1) $I = V/R = 12/4.8 = 2.5\,\text{A}$. (2)

(b) $q = It = 2.5 \times 60 = 150\,\text{C}$. (2) $N = q/e = 150 / (1.602\times10^{-19}) = 9.363\times10^{20}$ electrons $\approx 9.36\times10^{20}$. (2)

(c) $P = VI = 12 \times 2.5 = 30\,\text{W}$. (2) $P = I^2R = 2.5^2 \times 4.8 = 6.25\times4.8 = 30\,\text{W}$. Agreement confirms consistency. (2)

(d) Copper is a metal: overlapping/partially-filled conduction band gives a sea of delocalised free electrons, so a small field drives large current. Plastic (insulator) has a large band gap and no free carriers, so electrons stay bound. (2) Conventional current flows from + to − terminal (12 V high side to 0 V) through the heater; electron flow is opposite — electrons drift from − to + terminal. (2)

(e) $t = 8\times3600 = 28800\,\text{s}$; $E = Pt = 30\times28800 = 864000\,\text{J} = 8.64\times10^5\,\text{J}$. (1) In kWh: $E = 30\,\text{W} \times 8\,\text{h} = 240\,\text{Wh} = 0.24\,\text{kWh}$. (1) Power (W) is rate of energy transfer per second; energy (J) is the accumulated total = power × time. (1)

Question 2

(a) A DC signal has constant polarity/value with time; $v(t)$ is AC — it periodically reverses sign. Frequency $f = 50\,\text{Hz}$, period $T = 1/f = 0.02\,\text{s}$, peak value $V_{pk}=5\,\text{V}$. (3)

(b) $i(t) = C\frac{dv}{dt} = C\cdot 5\cdot(2\pi\cdot50)\cos(2\pi\cdot50\,t)$. (2) $i(t) = 100\times10^{-6}\times5\times100\pi\cos(100\pi t) = 0.05\pi\cos(100\pi t)$ A. (1) Peak current $I_{pk}= 0.05\pi \approx 0.157\,\text{A}$. (1) Current leads voltage by $90^\circ$ (cosine vs sine). (1)

(c) Farad: capacitance of a body that holds 1 coulomb per volt, $C=q/V$, so $1\,\text{F}=1\,\text{C/V}$. (2) Henry: inductance producing 1 volt for a current change of 1 A/s, $v_L=L\,di/dt$, so $1\,\text{H}=1\,\text{V·s/A}$. (2)

(d) Work to move charge $dq$ across potential $v$ where $v=q/C$: $E=\int_0^Q v\,dq = \int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C}$. (2) With $Q=CV$: $E = \frac{(CV)^2}{2C} = \tfrac12 CV^2$. (2)

(e) $E = \tfrac12 (100\times10^{-6})(5)^2 = \tfrac12\times10^{-4}\times25 = 1.25\times10^{-3}\,\text{J} = 1.25\,\text{mJ}$. (2)

Question 3

(a) Series: $R_{tot}=R_1+R_2+R_3 = 100+220+330 = 650\,\Omega$. (1) $I = V_s/R_{tot} = 9/650 = 0.013846\,\text{A} \approx 13.85\,\text{mA}$. (1) $V_k = I R_k$: $V_1 = 1.3846\,\text{V}$, $V_2 = 3.0462\,\text{V}$, $V_3 = 4.5692\,\text{V}$. (1) Sum $= 9.000\,\text{V}$ ✓ (KVL). (1)

(b) (function correct 4, both checks 2)

def series_circuit(Vs, resistors):
    Rtot = sum(resistors)
    I = Vs / Rtot
    drops = [I * R for R in resistors]
    assert abs(sum(drops) - Vs) < 1e-9          # KVL check
    P = Vs * I
    assert abs(P - sum(I**2 * R for R in resistors)) < 1e-9  # power check
    return I, drops, P
 
# series_circuit(9, [100,220,330]) -> (0.013846..., [1.384..,3.046..,4.569..], 0.1246..)

(c) (1 each) Battery: DC EMF source, measured in volts (V). Resistor: opposes current, ohms ($\Omega$). Ammeter: in series, measures current in amperes (A), ideally zero resistance. Voltmeter: across (parallel to) the resistor, measures potential difference in volts (V), ideally infinite resistance.

[
  {"claim":"Q1 current 2.5 A and electrons ~9.363e20", "code":"I=Rational(12)/Rational(48,10); q=I*60; N=q/1.602e-19; result = (I==Rational(5,2)) and abs(N-9.363e20)<1e18"},
  {"claim":"Q1 power equal both formulas =30 W", "code":"I=Rational(5,2); result = (12*I==30) and (I**2*Rational(48,10)==30)"},
  {"claim":"Q2 cap energy at 5V = 1.25 mJ", "code":"E=Rational(1,2)*100e-6*5**2; result = abs(E-1.25e-3)<1e-12"},
  {"claim":"Q3 total current and drops sum to Vs", "code":"Rtot=100+220+330; I=Rational(9,Rtot); drops=[I*R for R in (100,220,330)]; result = (sum(drops)==9) and abs(float(I)-0.0138461538)<1e-9"}
]