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Ohm''s Law - Engineering

Q1:

The current through a flashlight bulb is 40 mA and the total battery voltage is 4.5 V. The resistance of the bulb is approximately

A 112 

B 11.2 

C 1.2 

D 18 

ANS:A - 112 

R = V/I.
R = 4.5/40 * 10^-3,
R = 4.5 * 10^3/40.
R = 4500/40.
R = 450/4.
R = 112.5ohm.