<h2>
Answer:</h2>
786Ω and 20.32Ω respectively.
<h2>
Explanation:</h2>
(a) Given a number of resistors each with its own resistance, the largest resistance can be obtained when these resistors are connected in series.
From the question, the resistors have the following resistances;
36.0-Ω, 50.0-Ω , and 700-Ω
Now, when they are connected in series, the total resistance (R) obtainable is given by the sum of these individual resistances as follows;
R = 36.0-Ω + 50.0-Ω + 700-Ω
R = 786Ω
Therefore, the largest resistance that can be obtained by connecting a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together is 786Ω
(b) Similarly, given a number of resistors each with its own resistance, the smallest resistance can be obtained when these resistors are connected in parallel.
From the question, the resistors have the following resistances;
36.0-Ω, 50.0-Ω , and 700-Ω
Now, when they are connected in parallel, the total resistance (R) obtainable is given by using the relation as follows;
= 
+ 
+ 
= 
= 
= 
R = 
R = 20.32Ω
Therefore, the smallest resistance that can be obtained by connecting a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together is 20.32Ω
