IIT-JEE 2012 · Paper 1 · Physics Q12

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PhysicsMultiple correct+4Question 12 of 60

For the resistance network shown in the figure, choose the correct option(s).

A symmetric network fed by a 12 V battery (current I₁). From the left node, 2 Ω goes up to P and 4 Ω down to Q; P-S is 2 Ω (current I₂ from P to S) and Q-T is 4 Ω; P-Q and S-T are 1 Ω each (vertical); S connects to the right node through 2 Ω and T through 4 Ω; the right node returns to the battery.

The current through

PQ

is zero.

I_1=3

The potential at

S

is less than that at

Q

I_2=2

Answer and worked solution

Official answer: A, B, C, D

Balanced bridge. The upper path (2 Ω + 2 Ω + 2 Ω) and the lower path (4 Ω + 4 Ω + 4 Ω) have the same ratio at every node: P and Q are each one third of the way along their paths, and so are S and T two thirds of the way. So $V_P=V_Q$ and $V_S=V_T$ , and the 1 Ω links carry no current.

(A) No current in PQ. True.
(B) The network is 6 Ω in parallel with 12 Ω, i.e. 4 Ω: $I_1=12/4=3$ A. True.
(C) Current flows from the left node to the right node; S is further along (two thirds of the drop) than Q (one third): $V_S<V_Q$ . True.
(D) The upper path carries $I_2=12/6=2$ A. True.

(A), (B), (C), (D)

Topics:Wheatstone bridge and metre bridge Series and parallel resistors Kirchhoff's laws

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IIT-JEE 2012 · Paper 1, Physics question 12