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Parallel Circuits - Engineering

Q1:

Four resistors are connected in parallel. Fifteen mA flows through resistor R. If the second resistor is 2R, the third resistor 3R, and the fourth resistor 4R, the total current in the circuit is

A 60 mA

B 15 mA

C 135 mA

D 31.25 mA

ANS:D - 31.25 mA

Since voltage remains constant.
branch 1->V=IR.
branch 2->V=(I/2)2R.
So on
thus Itotal=15+(15/2)+(15/3)+(15/4)=31.25mA.