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Find the total resistance |
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Answer» YZP given resistance are serial then, total resistance R = 10Ω + 2Ω R = 12Ω this resistance and YP resistance are parallel \(\frac{1}{R} = \frac{1}{6}+\frac{1}{12}\) \(\frac{1}{R}=\frac{2+1}{12}\) \(\frac{1}{R}=\frac{3}{12}\) R = 4Ω In YPQ will be YP is 4Ω and PQ 8Ω are serial then R = 4Ω + 8Ω R = 12Ω then YPQ and YQ are parallel \(\frac{1}{R}=\frac{1}{6}+\frac{1}{12}\) R = 4 YQ and YQR are in serial then R = 4Ω + 4Ω R = 8Ω YR and YQR in parallel then \(\frac{1}{R}=\frac{1}{8}+\frac{1}{8}\) R = 4Ω XY and XYR in serial then R = 4Ω + 4Ω R = 8Ω XYR and XR are in parallel \(\frac{1}{R}=\frac{1}{8}+\frac{1}{8}\) R = 4Ω AX and XR will be serial then R = 4Ω + 6Ω R = 10Ω Then total equivalent resistance is 10Ω |
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