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Series Circuits - Engineering

Q1:

A 12 V battery is connected across a series combination of 68 , 47 , 220 , and 33 . The amount of current is

A 326 mA

B 16.3 mA

C 32.6 mA

D 163 mA

ANS:C - 32.6 mA

R= R1+R2+R3+R4 = 68+47+220+33 = 368.
i = V/(R) = 12/368 = 0.0326086 In A.
So in milliampere 32.6 mA.