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← JEE Advanced 2022 · Paper 1
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JEE Advanced 2022 · Paper 1, Physics question 10

The figure shows a circuit having eight resistances of 1 Ω1\ \Omega each, labelled R1R_1 to R8R_8, and two ideal batteries with voltages E1=12\mathcal E_1 = 12 V and E2=6\mathcal E_2 = 6 V.

Which of the following statement(s) is(are) correct?

A diamond of four outer nodes: left L, top T, right Rt, bottom Bt, with centre node O. Outer edges: R6 (L-T), R7 (T-Rt), R5 (L-Bt), R8 (Bt-Rt). Spokes: R2 (T-O) and R4 (O-Bt). Horizontal branch from L to O: battery E2 (longer, positive plate on the L side) in series with R3. Horizontal branch from O to Rt: battery E1 (longer, positive plate on the O side) in series with R1. All resistances are 1 Ω; E1 = 12 V, E2 = 6 V.
Choose one or more options
Answer and worked solution

Official answer: A, B, C, D

Take the centre node O at 0 V.

Symmetry. The network is symmetric top to bottom, so the top node T and bottom node B are at the same potential tt. Call the left and right nodes ll and rr.

Branch equations. Both batteries have their positive plate on the left.

  • In the left branch (E2\mathcal E_2, R3R_3), the current from O to L is I3=6−lI_3=6-l.
  • In the right branch (E1\mathcal E_1, R1R_1), the current from R to O is I1=12+rI_1=12+r.

Kirchhoff's current law.

  • At L: (6−l)=(l−t)+(l−t)(6-l)=(l-t)+(l-t).
  • At R: (12+r)=(t−r)+(t−r)(12+r)=(t-r)+(t-r).
  • At T: (l−t)=(t−r)+t(l-t)=(t-r)+t.

Solution. l=1.2l=1.2, r=−4.8r=-4.8, t=−1.2t=-1.2 V. So:

  • ∣I1∣=12−4.8=7.2|I_1|=12-4.8=7.2 A ✓
  • ∣I2∣=∣t∣=1.2|I_2|=|t|=1.2 A ✓
  • ∣I3∣=6−1.2=4.8|I_3|=6-1.2=4.8 A ✓
  • ∣I5∣=l−t=2.4|I_5|=l-t=2.4 A ✓

All four are correct. A full nodal solve confirms these values.

Topics:Kirchhoff's lawsSeries and parallel resistors
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