Exam question- Current realtion to voltage and resistance changes

Thread Starter

copacabana81

Joined Oct 29, 2017
12
Hi everyone.
My exam is coming soon (grhhh).
Could you please check for me my answers, if they correct?. Questions are from previous yaers exams. I will appreciate it. Please see attached file.

My answers:

a) 72mA
b) 42 mA
c) 52 mA
d) current will stay same 60 mA
 

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Thread Starter

copacabana81

Joined Oct 29, 2017
12
Also, in another question is : The current in linear resistive circuit is 50 mA.
Determinate current if voltage is halfed and resistance i tripled.

Will current be 8.3 mA?
 

crutschow

Joined Mar 14, 2008
23,136
Show us how you calculated answers a) and c).
Those are in error.

Below is the LTspice simulation of a circuit that gives 60mA current at a nominal resistance of 1kΩ.
See if you can determine why the currents for 800Ω (20% reduced) and 1.13kΩ (13% increase) don't quite match your answers.
Remember that current equals V/R.

upload_2017-12-28_7-53-44.png
 

crutschow

Joined Mar 14, 2008
23,136
Also, in another question is : The current in linear resistive circuit is 50 mA.
Determinate current if voltage is halfed and resistance i tripled.

Will current be 8.3 mA?
Yes.
So now you should be able to determine why your other two answers are not correct.
 

Thread Starter

copacabana81

Joined Oct 29, 2017
12
Oh yes!

Massive Thank you!

So,
Current is 60mA

If you reduce resistance by 20%, current will increase by 20%

20% of 60 is 12 therfore current will be 72mA

point c) should be then 52.2mA
 

Thread Starter

copacabana81

Joined Oct 29, 2017
12
Ok, so

20% is 1/5 therefore 0.2

resistance reduced by 20% so its reduded by fifth part, is that make sense?

Because resistance is reduced by 20% the current will increase so...

60mA * 0.2= 12mA
 

MrAl

Joined Jun 17, 2014
6,402
Oh yes!

Massive Thank you!

So,
Current is 60mA

If you reduce resistance by 20%, current will increase by 20%

20% of 60 is 12 therfore current will be 72mA

point c) should be then 52.2mA
Hi,

72ma is close but not right. If you start with any resistance that draws 60ma at some voltage then reduce the resistance by 20 percent, the current does not go up by 20 percent.

Try it and see what you get. You can use any resistance and any voltage as long as the current is 60ma to start, then reduce the resistance by 20 percent, then keep the voltage constant and calcualte the current again.
You should do it this way at least once so you can see how it works.

If you have trouble picking a resistance to start and a voltage, try 50 ohms and 3 volts to start, then go from there.
 

Thread Starter

copacabana81

Joined Oct 29, 2017
12
Thank you guys for your support. Its a week today since I try to figure it out lol.

Ok, so if the V=3 and R=50 ohms I=V/R so I=0.06 A
reduced resistance by 20%

V=3 R=40ohms I=0.075 A, change in current 0.015A whitch is 25%
And were is the logic?
how come current going up by 25% when resistance is reduced by 20%???
 

LesJones

Joined Jan 8, 2017
2,304
If you add 20% to 100 you get 120. If you then take 20% off 120 you don't get back to 100. (The result is 96) I think this is the error in your thinking.

Les
 

MrAl

Joined Jun 17, 2014
6,402
Thank you guys for your support. Its a week today since I try to figure it out lol.

Ok, so if the V=3 and R=50 ohms I=V/R so I=0.06 A
reduced resistance by 20%

V=3 R=40ohms I=0.075 A, change in current 0.015A whitch is 25%
And were is the logic?
how come current going up by 25% when resistance is reduced by 20%???

Hi,

Because R is in the denominator and I (the current) is not.

In the expression:
I=E/R

if we reduce R by 20 percent that means we multiply it by 0.20 and subtract it from the original R, so we end up with:
I=E/(R-0.20*R)

and now we see we have two R's in the denominator so let's move them to the numerator:
I=(E/R)/(1-0.2)

and of course that is equal to:
I=(E/R)/0.8

and now let's get that 0.8 into the numerator also so we can see this more clearly:
I=(E/R)*1.25

and since we started with:
I=E/R

and now we are multiplying that by 1.25, we can easily see that now the current goes up by a factor of 1.25 which is 25 percent higher.

So the resistance is in the denominator, and when we think of the change as being in the numerator by thinking of what factor the current changes by, we have to find the reciprocal of that.
 
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