# MCQ related to electric current and ohms law

Joined Jan 14, 2016
11
hi,
a carbon resistor has to meet the following requirements
IR drop : 5V
Current : 100mA
Safety factor for power dissipation : 2
Which one of the following resistors will be suitable
(A)5 ohm, 10 W
(B)0.5ohm, 100W
(C)50ohm, 1W

#### nerdegutta

Joined Dec 15, 2009
2,684
This smells like homework. Moved to Homework Forum.

#### WBahn

Joined Mar 31, 2012
30,295
hi,
a carbon resistor has to meet the following requirements
IR drop : 5V
Current : 100mA
Safety factor for power dissipation : 2
Which one of the following resistors will be suitable
(A)5 ohm, 10 W
(B)0.5ohm, 100W
(C)50ohm, 1W

Joined Jan 14, 2016
11
This smells like homework. Moved to Homework Forum.

i thought the correct option is B because
I=(100/0.5)^0.5
I=14.14Amp
so it satisfies the current and power rating thats why i think the correct option is B but it is not..
i dont know how C can be the correct option as it does not satisfies the power rating.

Last edited:

#### crutschow

Joined Mar 14, 2008
34,841
How did you get 100/0.5 and I=14.4Amp?
The problem stated the voltage was 5V and the current was 100mA.

#### shteii01

Joined Feb 19, 2010
4,644
Power dissipated by the resistor == Voltage across resistor * Current through resistor
P=5*0.1=0.5 Watt
Safety factor of 2 means: 2*P=2*0.5=1 Watt
Therefore the answer is C because all you need is the resistor that will dissipate 1 Watt, you don't need 10 Watt, you don't need 100 Watt.

And lol for trying to use the answers to figure out the question.

Joined Jan 14, 2016
11
Power dissipated by the resistor == Voltage across resistor * Current through resistor
P=5*0.1=0.5 Watt
Safety factor of 2 means: 2*P=2*0.5=1 Watt
Therefore the answer is C because all you need is the resistor that will dissipate 1 Watt, you don't need 10 Watt, you don't need 100 Watt.

And lol for trying to use the answers to figure out the question.
thanks a lot.but i still cant figure out what is meant by safety factor??

Joined Jan 14, 2016
11
How did you get 100/0.5 and I=14.4Amp?
The problem stated the voltage was 5V and the current was 100mA.
again go through the calculations and question bro..

#### WBahn

Joined Mar 31, 2012
30,295
i thought the correct option is B because
I=(100/0.5)^0.5
I=14.14Amp
so it satisfies the current and power rating thats why i think the correct option is B but it is not..
i dont know how C can be the correct option as it does not satisfies the power rating.
If you had bothered to track your units you would have immediately seen that your answer is wrong. Instead, you threw pure numbers at some formula and then tacked on the units that you wanted the answer to have.

What you SHOULD have done:

I = (V/R)^0.5 = (100 V / 0.5 Ω)^0.5 = (200 A)^0.5 = 14.14 √A

At this point you could have noticed that √A is not amperes and therefore known that your answer had to be wrong.

Joined Jan 14, 2016
11
I=(V/R)^0.5

#### WBahn

Joined Mar 31, 2012
30,295
Looking (guessing, really) again I suspect that you were trying to use

P = I²R => I = √(P/R)

and then using the power rating given in the answer.

Okay, so what you found is that a 0.5 Ω resistor dissipates 100 W at a current of 14.14 A.

So what?

The question is asking for a resistor that drops 5V at a current of 100 mA. What voltage will a 0.5 Ω resistor drop at 100 mA?

#### WBahn

Joined Mar 31, 2012
30,295
I=(V/R)^0.5
You didn't GIVE a formula! I'm having to GUESS what your starting point was because you couldn't be bothered to furnish it nor could you be bothered to properly supply the units of the quantities in your expression. I am NOT a mind reader and engineering is NOT about guessing!

#### WBahn

Joined Mar 31, 2012
30,295
Power dissipated by the resistor == Voltage across resistor * Current through resistor
P=5*0.1=0.5 Watt
Safety factor of 2 means: 2*P=2*0.5=1 Watt
Therefore the answer is C because all you need is the resistor that will dissipate 1 Watt, you don't need 10 Watt, you don't need 100 Watt.

And lol for trying to use the answers to figure out the question.
So you are picking the resistor based ONLY on the power rating matching exactly what you need? The resistance doesn't matter at all?

(A) 5 ohm, 10 W
(B) 50 ohm, 100W
(C) 0.5 ohm, 1W

#### WBahn

Joined Mar 31, 2012
30,295
thanks a lot.but i still cant figure out what is meant by safety factor??
Safety factor means that you choose the component so that it can handle more than you expect it will actually have to.

If you weigh 200 lb and you buy a chair that is rated for 400 lb, you have a safety factor of 2.

Joined Jan 14, 2016
11
You didn't GIVE a formula! I'm having to GUESS what your starting point was because you couldn't be bothered to furnish it nor could you be bothered to properly supply the units of the quantities in your expression. I am NOT a mind reader and engineering is NOT about guessing!

Joined Jan 14, 2016
11
Safety factor means that you choose the component so that it can handle more than you expect it will actually have to.

If you weigh 200 lb and you buy a chair that is rated for 400 lb, you have a safety factor of 2.
thanks a lot... now i can grasp it.

Joined Jan 14, 2016
11
You didn't GIVE a formula! I'm having to GUESS what your starting point was because you couldn't be bothered to furnish it nor could you be bothered to properly supply the units of the quantities in your expression. I am NOT a mind reader and engineering is NOT about guessing!
Sir, i used the formula P=(V^2)/R but i have now figured out that what i was doing wrong. Thanks a lot again for your help.

#### WBahn

Joined Mar 31, 2012
30,295
You're welcome. Good luck.

#### EM Fields

Joined Jun 8, 2016
583
Since the drop across the resistor is 5 volts when there's 100 milliamperes through it then, from Ohm's law, its resistance must be R = E/I =5V/0.1A = 50 ohms.

The power it must dissipate is then P = E I = 5V X 0.1A = 0.5 watts.

Safety factor refers to giving the load a break and not running it balls - to - the - wall, (its maximum rating) so the only answer that fits is (C).

Looking at it another way, if you trusted the test, (which you never should) you could blow off everything but the 50 ohm choice and buy some time time for the rest of the test.