Question

if the resistance across a 12V source is increased by 4 ohm the current drops by 0.5A the original resistance was? ​

Answer 1

Answer:

Say I1 is 1A. 12V/1A = 12 ohms. To cut current by 0.5A would mean 12V/(1A - 0.5A) = 24 ohms, meaning it would require 12 ohms to drop by 0.5A.

Say I1 = 2A. 12V/2A = 6 ohms. 12V/(2A - 0.5A) = 8 ohms. So that is 2 ohms to drop by 0.5A. That indicates to me that a solution is possible, somewhere between 6 and 12 ohms for the original resistance.

R1 = unknown original resistance

R2 = 4 ohms

I1 = original current

I1 = 12V/R1

I1 - 0.5A = 12V/(R1+R2)

I1 = 12V/(R1+R2) + 0.5A

12V/R1 = 12V/(R1+R2) + 0.5A

Hm. Kinda stuck here. I don’t find myself doing algebra every day, and I am a bit tired.

Let’s try something.

We know now the original resistance is between 6 and 12 ohms, the original current between 2A and 1A. Let’s split the difference on the current and see what happens.

12V/1.5A = 8 ohms

12V/(1.5A - 0.5A) = 12 ohms

Tada! That is a difference of 4 ohms.

Explanation:

Answer 2

Initially let I = 12/R amperes where, R is the original resistance.

When the R is increased by 4 ohms then:

12/(R + 4) = (I - 0.5) = (12/R) - 0.5

Rearranging:

12 = (R + 4)[ (12/R) - 0.5 ] = 12 - 0.5R + 48/R - 2 = 10 - 0.5R + 48/R

2 = 48/R - 0.5R

2R = 48 - 0.5R^2

0.5R^2 + 2R - 48 = 0

Solving the quadratic gives:

R = -12 or R = 8

Hence neglecting the negative value, the original resistance R = 8 ohms.

Check:

When R = 8 ohms, I = 12/8 = 1.5A

When R = 8 + 4 = 12 ohms, I = 12/12 = 1A