What is the potential difference between X and Y in the diagram above if the battery is of negligible internal resistance?
\(\frac{1}{R_{XY}}\) = \(\frac{1}{3} + \frac{1}{2} + \frac{1}{6}\) = \(\frac{5}{6}\)
\(\therefore\) R\(_{XY}\) = \(\frac{5}{6}\)\(\Omega\)
\(\therefore\) R\(_T\) = \(\frac{5}{6}\) + 0.8 = \(\frac{5}{6}\) + \(\frac{4}{5}\) = \(\frac{10}{5}\) = 2\(\Omega\)
But, V = IR\(_T\) where V = 5v and R\(_T\) = 2\(\Omega\)
I = \(\frac{\text{V}}{R_T}\) = \(\frac{5}{2}\)A
\(\therefore\) V\(_{across XY}\) = I R\(_{XY}\) = \(\frac{5}{2}\) x \(\frac{6}{5}\) = 3.0V.
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