Noesis
Interleaved — Phase 1

Hardware interleaved practice

printable — key stays hidden on paper

Instructions: Solve each problem showing your work. Watch carefully — consecutive problems draw from different topics, so identify the correct law or definition before computing. Use SI units. Marks in [brackets].

  1. A wire carries a steady current of 2 A2\text{ A}. How much charge passes a cross-section in 30 s30\text{ s}, and how many electrons does that represent? (electron charge =1.602×10−19 C= 1.602\times10^{-19}\text{ C}) [3]

  2. A 9 V9\text{ V} source connects to two resistors in series, R1=3 kΩR_1 = 3\text{ k}\Omega and R2=6 kΩR_2 = 6\text{ k}\Omega. Find the voltage across R2R_2 using the voltage divider rule. [3]

  3. A material has loosely bound electrons in a partially filled conduction band and its conductivity increases when doped. Classify it (conductor / insulator / semiconductor) and briefly justify. [2]

  4. A 12 V12\text{ V} battery drives a 4 Ω4\ \Omega resistor. Compute the current, the power dissipated (two different formulas), and the energy delivered over 10 s10\text{ s}. [4]

  5. At a circuit node, three currents flow in: 3 A3\text{ A}, 1.5 A1.5\text{ A}, and IxI_x; two flow out: 4 A4\text{ A} and 2 A2\text{ A}. Apply KCL to find IxI_x. [2]

  6. An RC circuit has R=10 kΩR = 10\text{ k}\Omega and C=47 μFC = 47\ \mu\text{F}, charging from 0 V0\text{ V} toward 5 V5\text{ V}. Find the time constant τ\tau, and the capacitor voltage at t=τt = \tau. [3]

  7. Find the equivalent resistance between the terminals of this network: a 6 Ω6\ \Omega resistor in series with a parallel combination of 4 Ω4\ \Omega and 12 Ω12\ \Omega. [3]

  8. State the direction of conventional current relative to electron flow in a wire, and explain in one sentence why the two conventions differ. [2]

  9. A single loop contains a 10 V10\text{ V} source and three resistors dropping 3 V3\text{ V}, 4 V4\text{ V}, and V3V_3. Apply KVL to find V3V_3. [2]

  10. A source of EMF 20 V20\text{ V} with internal (Thevenin) resistance Rth=5 ΩR_{th} = 5\ \Omega drives a load RL=15 ΩR_L = 15\ \Omega. Find the load current and load voltage, then give the Norton equivalent (current source value and parallel resistance). [4]

Answer keyMark scheme & solutions

Q1 — Subtopic 1.1.4 (current/ampere) + 1.1.1 (coulomb/electron). Why: "current × time" signals charge; then convert charge to electron count. Q=It=2×30=60 CQ = It = 2 \times 30 = 60\text{ C}. n=Q/e=60/1.602×10−19=3.75×1020n = Q/e = 60 / 1.602\times10^{-19} = 3.75\times10^{20} electrons.

Q2 — Subtopic 1.2.5 (voltage divider). Why: Series resistors + "voltage across one" = divider, not Ohm's law alone. V2=VR2R1+R2=9⋅69=6 VV_2 = V\dfrac{R_2}{R_1+R_2} = 9\cdot\dfrac{6}{9} = 6\text{ V}.

Q3 — Subtopic 1.1.2 (conductors/insulators/semiconductors). Why: Definition-recall; keyword "doped" + "partially filled band" points to semiconductor. Semiconductor. Its conductivity lies between conductor and insulator, and doping (adding impurities) controllably changes carrier concentration — characteristic of semiconductors like silicon.

Q4 — Subtopic 1.1.6 (Ohm's Law), 1.1.7 (power), 1.1.13 (energy vs power). Why: Multiple requested quantities force distinguishing P=VIP=VI vs P=I2RP=I^2R and W=PtW=Pt. I=V/R=12/4=3 AI = V/R = 12/4 = 3\text{ A}. P=VI=12×3=36 WP = VI = 12\times3 = 36\text{ W}; check P=I2R=9×4=36 WP = I^2R = 9\times4 = 36\text{ W}. ✓ W=Pt=36×10=360 JW = Pt = 36\times10 = 360\text{ J}.

Q5 — Subtopic 1.2.3 (KCL). Why: "at a node, in vs out" = current law. ∑Iin=∑Iout\sum I_{in} = \sum I_{out}: 3+1.5+Ix=4+2⇒4.5+Ix=6⇒Ix=1.5 A3 + 1.5 + I_x = 4 + 2 \Rightarrow 4.5 + I_x = 6 \Rightarrow I_x = 1.5\text{ A}.

Q6 — Subtopic 1.2.7 (RC time constant). Why: R and C together + "charging" = transient, not steady Ohm's law. τ=RC=10 000×47×10−6=0.47 s\tau = RC = 10\,000 \times 47\times10^{-6} = 0.47\text{ s}. At t=τt=\tau: V=Vf(1−e−1)=5×0.6321=3.16 VV = V_f(1-e^{-1}) = 5\times0.6321 = 3.16\text{ V}.

Q7 — Subtopic 1.2.2 (mixed network) + 1.2.1 (series/parallel). Why: "series with a parallel combination" = compute parallel first, then add. Parallel: 4×124+12=4816=3 Ω\dfrac{4\times12}{4+12} = \dfrac{48}{16} = 3\ \Omega. Total: 6+3=9 Ω6 + 3 = 9\ \Omega.

Q8 — Subtopic 1.1.9 (conventional vs electron flow). Why: Pure concept-recall placed after a calculation to break rhythm. Conventional current flows opposite to electron flow (electrons go from − to +; conventional current is defined from + to −). The difference is historical: Franklin fixed the positive direction before the electron was discovered.

Q9 — Subtopic 1.2.4 (KVL). Why: "single loop, sum of drops" = voltage law. ∑Vsource=∑Vdrops\sum V_{source} = \sum V_{drops}: 10=3+4+V3⇒V3=3 V10 = 3 + 4 + V_3 \Rightarrow V_3 = 3\text{ V}.

Q10 — Subtopic 1.2.9 (Thevenin) + 1.2.10 (Norton). Why: Given Thevenin form; must apply it and convert to Norton (source transformation). IL=VthRth+RL=205+15=1 AI_L = \dfrac{V_{th}}{R_{th}+R_L} = \dfrac{20}{5+15} = 1\text{ A}. VL=ILRL=1×15=15 VV_L = I_L R_L = 1\times15 = 15\text{ V}. Norton: IN=Vth/Rth=20/5=4 AI_N = V_{th}/R_{th} = 20/5 = 4\text{ A}, RN=Rth=5 ΩR_N = R_{th} = 5\ \Omega (in parallel).

[
  {"claim":"Q2 divider gives V2 = 6 V","code":"V=9; R1=3000; R2=6000; result = (V*R2/(R1+R2))==6"},
  {"claim":"Q7 equivalent resistance = 9 ohm","code":"par=(4*12)/(4+12); result = (6+par)==9"},
  {"claim":"Q10 load current 1A and Norton current 4A","code":"Vth=20; Rth=5; RL=15; IL=Vth/(Rth+RL); IN=Vth/Rth; result = (IL==1) and (IN==4)"}
]