Solving circuit design problem with two approaches

A voltage of 200V is applied to a tapped resistor of 500 ohm. Find the resistance between two tapping points connected to a circuit needing 0.1A at 25V. I have solved this problem with two approach. But the correct answer is coming only by one approach.

Total resistance is Ra+Rb = 500 ohm.
Va+Vb = 200V ; Vb = 25V ;

Hence Va = 175V.
Applying KCL at the tapping point we get:
-I + 0.1 + 25/Rb = 0 ; I = 0.1 + 25/Rb;
Va = I.Ra;
175 = (0.1 + 25/Rb)(500-Rb)
If we solve we will get Ra = 420.85 ohm and Rb = 79.15 ohm.

So far so good.

Consider second image in which Ra and Rb variable are taken such that Ra + Rb = 500 ohm
Va = 175V
Applying KCL at the tapping point we get:
-I + 0.1 + 25/(500-Ra) = 0 ; I = 0.1 + 25/500-Ra;
Va = I.Ra;
175 = (0.1 + 25/500-Ra)(Ra)
If we solve we will get Ra = 2079.15 ohm.
which is wrong, but I am unable to understand the flaw in the second case. It is perfectly fine. The answer is not correct. Just by changing the way variable are taken the answer is changing.

• Shouldn't it be Va = IRa in both cases? – Transistor Aug 26 '18 at 8:35
• Yes, Va = I.Ra ; I have changed. – TapasX Aug 26 '18 at 8:52
• And did you recalculate? Did it solve the problem? – Transistor Aug 26 '18 at 9:04
• There is nothing to recalculate. I had forgotten to put Ra in place of R. Calculations remains same. – TapasX Aug 26 '18 at 11:10

1 Answer

The flaw is that this equation has some restrictions when applied to a real circuit. From a mathematical perspective the second solution is correct. But as still applied that

$Ra=500\Omega-Rb$

you would end up with a negative resistance for

$Ra = -1579\Omega$

In the real world such a thing doesn't exist and therefore the solution is not applicable for your problem.

• If Rb is less than 500, then Ra will be always positive. – TapasX Aug 26 '18 at 11:14
• I agree yes... but that doesn't change the fact that one of your solutions is resulting in a negative Ra – Humpawumpa Aug 26 '18 at 15:33