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JEE MainPhysics
In the circuit shown V_d - V_a is ( in V )

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[IMAGE] Along agbfcd
V_a + 1 - 2 - 4 + 8 - 1 = V_d
⇒ V_a + 2 = V_d ⇒ V_d - V_a = 2 V

V_a + 1 - 2 - 4 + 8 - 1 = V_d
⇒ V_a + 2 = V_d ⇒ V_d - V_a = 2 V

Source: Grass-G / P-02 Physics Paper-02_Update.docx
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