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← JEE Advanced 2015 · Paper 2
PhysicsInteger+4Question 7 of 60

JEE Advanced 2015 · Paper 2, Physics question 7

In the following circuit, the current through the resistor RR (=2 Ω)(=2\ \Omega) is II Amperes. The value of II is

A 6.5 V battery on the left in series with R (= 2 Ω) along the top. R feeds a network: from the top node, a 1 Ω resistor runs right along the top; a 6 Ω resistor runs down on the left to a middle node; inside the upper box a star of 2 Ω, 8 Ω and 4 Ω resistors joins the top-left node, the top-right node and the middle-right node; a 2 Ω resistor runs down on the right; a 10 Ω resistor joins the middle-left and middle-right nodes; below, a 12 Ω resistor (left) and a 4 Ω resistor (right) run down to the bottom wire returning to the battery.
Answer and worked solution

Official answer: 1

Label the nodes. A after R, B top-right, C middle-left, D middle-right, S the centre of the 2-8-4 Ω star, and E the bottom wire.

Node potentials. Inject 1 A at A with E at 0 V:
VA=4.5,VB=VS=4,VC=VD=3 ext(volts).V_A=4.5,\quad V_B=V_S=4,\quad V_C=V_D=3\ ext{(volts)}.
Check: from A, 0.5 A flows through the 1 Ω, 0.25 A through the 2 Ω and 0.25 A through the 6 Ω; into E, 0.25 A comes down the 12 Ω and 0.75 A down the 4 Ω.

The two bridges are balanced. The 8 Ω (S-B) and the 10 Ω (C-D) carry no current, and the network between A and E is 4.5 Ω.

Current.
I=6.52+4.5=1 A.I=\frac{6.5}{2+4.5}=1\text{ A}.

Answer: 1

Topics:Wheatstone bridge and metre bridgeSeries and parallel resistors
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