# Electrical Engineering - Ohm's Law - Discussion

### Discussion :: Ohm's Law - General Questions (Q.No.13)

13.

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 Explanation:

No answer description available for this question.

 Raghu said: (Mar 3, 2011) R=V/I 4.5/.040 =112.5

 Renugadevi said: (Sep 13, 2012) R=V/I =4.5/40*10^-3 =111.65

 Rajib said: (Apr 20, 2013) V = IR, R = V/I, R = 4.5/40(mA). R = 4.5/(40/1000)A. R = 4.5/0.04. R = 112.5 ohm.

 Arun said: (Jun 14, 2013) V = IR. R = V/I. R = 4.5/0.04. R = 112.5 Ohm.

 Sharanabasava said: (Aug 24, 2014) I = 40mA = 40*10^-3. V = 4.5v. Then R = V/I. = 4.5/(40*10^-3). = 4.5*10^3/40. = 112.5 ohms.

 Mamatha Paladugu said: (Dec 6, 2018) V = IR. R = V/I. R = 4.5/40*10^-3. =112.5ohms.

 Sham said: (Mar 3, 2019) I = V/R. R = V/A. R = 4.5/40x10^-3. R = 4.5x10^3/40. R = 4500/40. = 450/4. = 112.2ohm.

 Rahul Singh said: (Mar 8, 2019) R = V/I. R = 4.5/40 * 10^-3, R = 4.5 * 10^3/40. R = 4500/40. R = 450/4. R = 112.5ohm.

 Santhosh Elli said: (Mar 26, 2019) R = V/I. R = 4.5/40 * 10^-3, R = 4.5 * 10^3/40. R = 4500/40. R = 450/4. R = 112.5ohm.

 Vvk said: (Aug 10, 2019) 4.5/40 = .1125 unit is not changed.