Current Electricity

JEE Main level, numerical answer, medium

108 of 380

Question 108 of 380

A network of four resistances is connected to 9V9V battery, as shown in figure. The magnitude of voltage difference between the points AA and BB is __________ VV.

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Answer: 3

The resistance in each branch is 6Ω6\Omega. They are in parallel. So, net resistance is R′=62=3ΩR'=\dfrac{6}{2}=3\Omega.
The total current flowing through the loop,
I=VR′=9V3Ω=3AI=\dfrac{V}{R'}=\dfrac{9V}{3\Omega }=3A
Thus, through each branch the current is 1.5A1.5A.
Hence, the voltage difference across the points is
VA−VB=(4×1.5)−(2×1.5)=3V{V}_{A}-{V}_{B}=\left(4\times 1.5\right)-\left(2\times 1.5\right)=3V

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