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JEE Main 2024 (4 Apr, Shift 1) · Combination of Resistors

Combination of Resistors question from JEE Main 2024 (4 Apr, Shift 1)

Q27JEE Main 2024 · 4 APR · SHIFT 1Difficult

Combination of Resistors

Twelve wires each having resistance 2 Ω2\ \Omega are joined to form a cube. A battery of 6 V6\text{ V} emf is joined across point aa and cc. The voltage difference between ee and ff is ________ V.

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Answer: 1

  1. Battery across face diagonal a–c: take Va=6V_a = 6, Vc=0V_c = 0. By symmetry b, d, e, g are all at 33 V; let Vh=yV_h = y and Vf=6−yV_f = 6 - y.
  2. KCL at h (neighbours a, e, g; equal resistances): (y−6)+(y−3)+(y−3)=0⇒y=4(y - 6) + (y - 3) + (y - 3) = 0 \Rightarrow y = 4 V, so Vf=2V_f = 2 V.
  3. Ve−Vf=3−2=1V_e - V_f = 3 - 2 = 1 V
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