JEE Main 2015 · 11 Apr · Shift 1 · Physics Q17

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PhysicsSingle correct+4 / -1Question 17 of 90

In the electric network shown, when no current flows through the 4 Ω resistor in the arm EB, the potential difference between the points A and D will be :

Rectangle F-E-D on top and A-B-C at the bottom. FE has a 2 V cell and 2 Ω; AF has 2 Ω; AB has a 9 V cell; BC has a 3 V cell; CD has a resistor R; the middle arm EB has 4 Ω and a 4 V cell.

3 V

4 V

5 V

6 V

Answer and worked solution

Official answer: C

With no current in EB, the potential difference between E and B is just the 4 V cell's emf. E and D are joined by a plain wire, so $V_A-V_D=V_A-V_E$ .

Going from A to E along A→B→E: the 9 V cell in AB and the 4 V cell in BE are oriented in opposite senses in the figure, so $|V_A-V_E|=9-4=5$ V.

Topics:Kirchhoff's laws

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JEE Main 2015 · 11 Apr · Shift 1, Physics question 17