Question
Download Solution PDFIn the following DC circuit, current in R3
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFBy re-drawing the dc circuit, we get
Now applying nodal analysis at node ‘A’, we get
\(\frac{{{V_A} - 7.5}}{5} + \frac{{{V_A}}}{5} + \frac{{{V_A}}}{5} = 0\;\)
VA - 7.5 + VA + VA = 0
3 VA = 7.5
∴ VA = 2.5 V
Hence current flowing through ‘R3’ branch is
\(I = \frac{{{V_A}}}{5} = \frac{{2.5}}{5}\)
∴ I = 0.5 A
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