Skip to content

Question

A \( 10 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of
12 resistors, each \( 1.5 \, \Omega \). What is the current through one edge from a corner?

Options

Choose one · Correct answer highlighted

Explanation

**Cells combination** series ε_eq = Σ ε_i, r_eq = Σ r_i, parallel for identical cells ε_eq = ε, r_eq = r/n, n number of cells. Maximum current when external R = r_eq, power transfer theorem, explaining why matching resistances maximizes power. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 1.5 = 1.25 Ω . Total current: Itₒtₐl = (V/Rₑq) = (10/1.25) = 8 A . Corner current: I = (Itₒtₐl/3) = (8/3) ≈ 2.67 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P =

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.