JEE PYQ: Current Electricity Question 23
Question 23 - 2020 (04 Sep 2020 Shift 2)
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
Solution
Current through $AB$: $i_1 = \frac{40}{40+60} = 0.4$ A. Current through $AD$: $i_2 = \frac{40}{90+110} = \frac{1}{5}$ A. $V_B - V_D = V_B + i_1(40) - i_2(90) = 18 - 16 = 2$ V.
BETA
