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

Answer: Option A

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.

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