Question
Download Solution PDFWhat is the value of voltage at node VA shown in the network below?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation:
Applying KCL on the node NA
\(\frac{V_A -50}{5}-10 +\frac{V_A}{2}+ \frac{V_A-10}{3} = 0\)
\(\frac{V_A}{5}+\frac{V_A}{2}+\frac{V_A}{3} = 20 + \frac{10}{3}\)
\(6 V_A + 15 V_A + 10 V_A = \frac{70}{3}\times 30\)
\(V_A = \frac{70\times30}{3\times31} = 22.58~ V\)
Last updated on May 28, 2025
-> UPSC ESE admit card 2025 for the prelims exam has been released.
-> The UPSC IES Prelims 2025 will be held on 8th June 2025.
-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.
-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.