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Q. Three resistance of 10$\Omega$ , 5$\Omega$ , 2$\Omega$ respectively are connected in parallel. New resistance will be

Current Electricity

Solution:

$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{10} + \frac{1}{5} + \frac{1}{2} = \frac{8}{10}$
Hence $R_p = 1.25 \, \Omega$