Resistors When two resistors of resistances R1 and R2 are...

Key Concepts

Parallel Resistor Networks

In electrical circuits, resistors connected in parallel provide multiple paths for current to flow, causing the overall or equivalent resistance to be lower than any of the individual resistances. This network configuration is fundamental in circuit analysis and design because it influences how current divides among the various branches.

Equivalent Resistance Calculation

The equivalent resistance in a parallel network is determined by the reciprocal sum of the individual resistances. This means that the overall resistance is found by taking the inverse of the sum of the inverses of each resistor's resistance, a concept that is key in predicting and controlling circuit behavior.

Reciprocal Relationships in Circuit Analysis

The process of calculating equivalent resistance in parallel circuits illustrates the concept of reciprocal relationships, where increasing the number of parallel branches results in a lower equivalent resistance. This fundamental understanding is critical for both theoretical analysis and practical applications in electronics.

Circuit Design Under Constraints

Designing circuits often involves ensuring that component values meet specific criteria for safe and efficient operation. This includes applying constraints to the circuit's equivalent resistance, requiring careful selection of resistor values so that the overall performance matches the desired parameters.

Transcript

00:01 We're asked to apply the methods of nonlinear inequalities to a question about resistors.

00:08 So, we're told that when two resistors of resistances r1 and r2 are connected in parallel, as shown in the figure following exercise 81 in this book, we're told that the total resistance r satisfies the equation.

00:26 1 over r is equal to 1 over r1 plus 1.

00:33 Over r2.

00:37 We're asked to find r1 for a parallel circuit in which r2 is equal to 2 oms and r we're told must be at least one ome.

00:53 So in other words r is greater than or equal to 1.

00:57 We want to find r1.

01:04 So plugging these values in, well first of all, let's solve our equation here for r.

01:16 To do so, recognize i can simply take the reciprocal of both sides and i get r is equal to 1 over 1 over r1 plus 1 over r2.

01:33 And now we know that since r is going to be greater than or equal to 1, it follows that 1 over 1 over r1 plus 1 over r2 is greater than or equal to of course, we also know that r2 is equal to 2.

01:49 So this is the same as 1 over 1 over r1 plus 1 half is greater than or equal to 1.

01:57 And in this way, we obtain a rational inequality.

02:05 Now, to solve this rational inequality for r1, first you want to write it so that one side's a rational expression and the other side is 0.

02:14 Then we want to find the critical values of this rational expression.

02:17 Then we'll want to test values lying in the intervals between the critical values, and finally we'll want to determine whether or not these intervals line the solution set.

02:30 So first i'm going to isolate the rational expression by subtracting one from both sides.

02:46 Now i'm going to simplify the first term here.

02:49 I'm going to add the terms in the denominator together.

02:53 To do this, i'm going to use the least common multiples of their denominators.

02:57 So we see that this term is the denominator of r1, and this term has the denominator of 2.

03:02 So the least common denominator, sorry, least common multiple of the denominators is going to be 2r1.

03:09 So i'm going to multiply the first term in the denominator by 2 over 2, and the second term by r1 over r1.

03:22 In both cases, i'm just multiplying by 1, which is allowed.

03:31 So this simplifies to 1 over, and then we have 2 r1 plus r1 over the denominator 2r1 minus 1 is greater than or equal to 0.

03:51 Of course, this simplifies to, well, we have flipping 2r1 over 2r1 plus r1 is 3r1 minus 1 is greater than or equal to 0.

04:01 Next, i made a mistake.

04:10 I'm sorry.

04:11 This should actually be 2 plus r1 over 2r1, and therefore this is going to be 3, sorry, 2r1 over 2 plus r1, minus 1 is greater than or equal to 0.

04:34 Once again, i'm going to combine terms by using the least common multiple denominators, 2 plus r1, and 1.

04:42 The least common multiple is just 2 plus r1.

04:47 So i'm going to multiply the second term by 2 plus r1 over 2 plus r1.

04:58 This is the same as multiplying by 1.

05:07 So combining i get 2r1 minus 2 minus r1 over 2 plus r1 is greater than or equal to 0...

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