If a maximum resistance produced by two resistors is #5 Omega# and minimum resistance produced by them is #1.2 Omega#, then what is the value of the resistances?

Question

Leo Abeyta · Accepted Answer

The maximum resistance is when the two resistances are in series, and the minimum when they are in parallel

#R_1+R_2 = 5 Omega# #(R_1R_2)/(R_1+R_2) = 1.2Omega# Resolve the system to discover: #R_1 = 2Omega# #R_1 = 3Omega#

Lily Adams · Answer

#r_1=2# and #r_2=3#

With both in series, the maximum resistance is found.

#r_1+r_2= 5#

and when in parallel, the minimum

#1/r_1+1/r_2=1/1.2# so

#{(r_1+r_2= 5),((r_1+r_2)/(r_1r_2)=1/1.2):}#

or

#{(r_1+r_2= 5),(r_1r_2=1.2 xx 5):}# giving

#r_1=2# and #r_2=3#

Charles Aeschliman · Answer

undefined To find the values of the resistances, we need to consider the maximum and minimum resistance combinations produced by the two resistors. Let's denote the resistances of the two resistors as $ R_1 $ and $ R_2 $.

The maximum resistance occurs when the resistors are in series, so we have:

$$ R_{\text{max}} = R_1 + R_2 $$

Given that $ R_{\text{max}} = 5\, \Omega $, we have:

$$ R_1 + R_2 = 5 $$

The minimum resistance occurs when the resistors are in parallel, so we have:

$$ \frac{1}{R_{\text{min}}} = \frac{1}{R_1} + \frac{1}{R_2} $$

Given that $ R_{\text{min}} = 1.2\, \Omega $, we have:

$$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{1.2} $$

Now, we can solve these equations simultaneously to find the values of $ R_1 $ and $ R_2 $.