SSC JE Electrical Engineering 2015 Paper

Question : 118 of 199

Marks: +1, -0

The voltage across the 1kΩ resistor of the network shown in the given figure is

Solution:

{V _A- 0}/{1k}+{V _A- 10}/{2k}+{V _A+2}/{2k}=0

4V_A=8

V_A = 2

volt

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