JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 6)
Four resistances 40 $$\Omega $$, 60 $$\Omega $$, 90 $$\Omega $$ and 110 $$\Omega $$ make the arms of a quadrilateral ABCD. Across AC is
a battery of emf 40 V and internal resistance negligible.The potential difference across BD in V is
_______.
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Answer
2
Explanation
_4th_September_Evening_Slot_en_6_2.png)
$${i_1} = {{40} \over {40 + 60}} = 0.4$$
$${i_2} = {{40} \over {90 + 110}} = {1 \over 5}$$
$${V_B} + {i_1}(40) - {i_2}(90) = {V_D}$$
$${V_B} - {V_D} = {1 \over 5}(90) - {4 \over {10}} \times 40$$
$${V_B} - {V_D} = 18 - 16 = 2$$
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