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← JEE Main 2022 · 26 Jun · Shift 1
PhysicsInteger+4 / -1Question 58 of 90

JEE Main 2022 · 26 Jun · Shift 1, Physics question 58

As per the given circuit, the value of current through the battery will be ______ A.

10 V cell on the left (positive terminal at the bottom). From the bottom-left node a 2 Ω resistor runs to the bottom-right node; from there a 10 Ω resistor rises to the top-right node. Between the top-left node and the top-right node are two branches: D1 (pointing towards the left node) in series with 6 Ω, and D2 (pointing towards the right) in series with 6 Ω. A diagonal branch from the top-left node to the bottom-right node holds 10 Ω, D3 (pointing towards the top-left node) and 6 Ω.
Answer and worked solution

Official answer: 1

Polarity. The positive terminal is at the bottom, so current goes through the 2 Ω to the bottom-right node, then back to the top-left node (the negative terminal).

Paths. Via 6 Ω, D3D_3 (forward) and 10 Ω: 16 Ω. Via 10 Ω up the right side, then the top 6 Ω and D1D_1 (forward): 16 Ω. D2D_2 is reverse biased and carries nothing.

Total. 2+(16∥16)=10 Ω2+(16\parallel16)=10\ \Omega, so I=1010=1I=\dfrac{10}{10}=1 A.

Answer: 1

Topics:Kirchhoff's lawsSemiconductors and p-n junctionSeries and parallel resistors
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