The resistance of two conductors in parallel is 12 ohm and in series is 50 ohm. Find the resistance of each conductor.
Answers
This is a very straightforward problem, and not too difficult to do in your head. If we have two nearly equal resistances, the series resistance (Rs) is twice the average value (midpoint between the two) and the parallel resistance Rp) is approximately equal to half the midpoint value. So a rough initial guess might be 12 and 13 ohms, where Rs is 25 ohms but Rp is approximately 6.25 ohms. Since Rp needs to be lower, we can play around with resistor values. Here’s an easy shortcut:
Let’s start with the equation for two parallel resistors:
Rp = (R1 * R2)/(R1 + R2)
Since Rs = R1 + R2,
Rp = (R1 * R2)/Rs.
We can rewrite this as
Rp * Rs = R1 * R2
Plugging in our numbers (Rp = 6, Rs = 25), we see that R1 + R2 = 150. What two numbers in the vicinity of 12 and 13 have the product of 150 and a sum of 25? Immediately 10 and 15 come to mind! Easy enough
Solution:- It is given that the resistance of two conductors in parallel is 12 ohm and in series is 50 Ohm.
Let r1 and r2 be the unknown resistance.
(i).
When the resistances are in series combination.
r1 + r2 = 50 Ohm.
(ii).
When the resistances are in parallel combination.
=>1/r1 + 1/r2
=> r1r2/r1 + r2 = 1/R(say)
where R is given as 12 Ohm.
Therefore, r1 + r2 = 50 Ohm
and r1r2/r1 + r2 = 12 Ohm.
=> r1r2 = 12 × 50
=> r1r2 = 600
Also,
=> ( r1 - r2 )^2 = ( r1 + r2 )^2 - 4r1r2
=> ( r1 - r2 )^2 = 2500 - 4 × 600
=> ( r1 - r2 )^2 = 100
=> ( r1 - r2 )^2 = (10)^2
=> r1 - r2 = 10
From (i) and (iii), 2r1 = 60
Therefore, r1 = 30 Ohm and r2 = 20 Ohm.