1.

A battery of 9V is connected in series with resistors of 0.2Omega,3Omega,0.4Omega and 12Omega respectively. How much current would flow through the 12Omega resistors ?

Answer»

Solution :`R= 0.2+0.3 +0.4 +0.5 +12 = 13.4 Omega`
`I =(V)/(R) = (9)/(13.4) =0.67 A `
SINCE, all resistors are in SERIES hence the current in 12` Omega` RESISTOR will also be 0.67 A.


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