3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Conductor X is a solid truncated right circular cone of length L whose end faces have radii a and 2a. Conductor Y is a solid right circular cylinder of the same length L and radius 23a. Both are made of the same ohmic material of resistivity ρ. They are joined in series (end face to end face, with negligible contact resistance) and carry a steady current I. Take the current density to be uniform over every cross-section perpendicular to the axis. What is the ratio (rate of Joule heating in X)/(rate of Joule heating in Y)? Exactly one option is correct.
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Answer: C
Same current in series, so the heating ratio equals the resistance ratio.
Cone: radius at distance x from the narrow end is r(x)=a(1+x/L). RX=∫0Lπa2(1+x/L)2ρdx=πa2ρL[1−21]=2πa2ρL (in general ρL/(πr1r2)).
Cylinder: RY=π(9a2/4)ρL=9πa24ρL.
Ratio =4/91/2=89.
Traps: using the mean radius 1.5a for the cone gives 1; using the mean of the end areas (2.5πa2) gives 9/10.
Difficulty: HardQuestion 2Level 2 · ExamSingle correct
Twenty-four identical cells, each of emf 2.0 V and internal resistance 1.0 Ω, are all used to drive current through an external resistor of 1.5 Ω. The cells are arranged as m identical rows connected in parallel, each row containing n cells in series with like polarities aiding (mn=24). Which arrangement makes the power dissipated in the external resistor as large as possible, and what is that power? Exactly one option is correct.
An=6, m=4; 24 W
Bn=8, m=3; 22.12 W
Cn=4, m=6; 20.45 W
Dn=12, m=2; 15.36 W
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Answer: A
The battery is equivalent to emf nE with internal resistance nr/m, so I=R+nr/mnE and P=I2R.
With m=24/n: I=1.5+n2/242n. Maximising n/(1.5+n2/24) gives n2/24=1.5, i.e. internal resistance nr/m=R: n=6, m=4 (an allowed integer pair).
Then I=12/(1.5+1.5)=4 A and P=16×1.5=24 W.
Check of the others: (8,3): I=3.84 A, P=22.12 W; (4,6): I=48/13 A, P=20.45 W; (12,2): I=3.2 A, P=15.36 W.
Difficulty: HardQuestion 3Level 3 · AdvancedSingle correct
Twelve identical wires, each of resistance R, form the edges of a cube, joined at the eight corners. One edge wire, joining corners P and Q, is now removed completely; the other eleven wires are unchanged. What is the equivalent resistance between P and Q? Exactly one option is correct.
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Answer: B
Insight: in the complete cube, the edge PQ is in parallel with "the rest of the network" between P and Q, so first find the full-cube edge resistance.
Full cube, current in at P, out at Q. Reflection in the plane that perpendicularly bisects PQ swaps the face through P perpendicular to PQ with the parallel face through Q, so potentials are antisymmetric: VP=u, VQ=−u; P's two other neighbours on its face are at w, the corner diagonally opposite P on that face at z, and their mirror images at −w, −z.
Node at w (neighbours P, the z corner, its mirror at −w): (w−u)+(w−z)+2w=0, so 4w=u+z.
Node at z (neighbours the two w corners and its mirror at −z): 2(z−w)+2z=0, so z=w/2. Hence w=2u/7.
Current out of P: [2u+2(u−w)]/R=24u/(7R), so Rfull=2u/7R24u=127R.
Then Rfull1=R1+Rrest1: 7R12−R1=7R5, so Rrest=57R.
Removing PQ leaves exactly "the rest", so the answer is 7R/5.
Traps: 7R/12 is the full-cube value; 5R/7 is the reciprocal slip.
Difficulty: Very hard