Math Problem Statement

In the first network that we analyzed, with the three resistors in series, we just added to get that they acted together like a single resistor of 10 ohms. We can do a similar thing for parallel circuits. In the second circuit analyzed, the electric current through the battery is 25/6 amperes. Thus, the parallel portion is equivalent to a single resistor of 20/(25/6) = 4.8 ohms. (a) What is the equivalent resistance if we change the 12 ohm resistor to 5 ohms? Page 4 (b) What is the equivalent resistance if the two are each 8 ohms? (c) Find the formula for the equivalent resistance if the two resistors in parallel are r1 ohms and r2 ohms. In the first network that we analyzed, with the three resistors in series, we just added to get that they acted together like a single resistor of 10 ohms. We can do a similar thing for parallel circuits. In the second circuit analyzed, the electric current through the battery is 25/6 amperes. Thus, the parallel portion is equivalent to a single resistor of 20/(25/6) = 4.8 ohms. (a) What is the equivalent resistance if we change the 12 ohm resistor to 5 ohms? Page 4 (b) What is the equivalent resistance if the two are each 8 ohms? (c) Find the formula for the equivalent resistance if the two resistors in parallel are r1 ohms and r2 ohms.

Solution

Let's break down the problem step by step and solve it:

Part (a): Changing the 12-ohm resistor to 5 ohms

We are asked to find the equivalent resistance of two resistors in parallel when one resistor changes from 12 ohms to 5 ohms. In this case, the two resistors in parallel are:

  1. r1=5r_1 = 5 ohms (instead of 12 ohms)
  2. r2=4.8r_2 = 4.8 ohms (as determined from the previous analysis)

For resistors in parallel, the formula for the equivalent resistance ReqR_{\text{eq}} is:

1Req=1r1+1r2\frac{1}{R_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2}

Substituting the values r1=5r_1 = 5 ohms and r2=4.8r_2 = 4.8 ohms:

1Req=15+14.8\frac{1}{R_{\text{eq}}} = \frac{1}{5} + \frac{1}{4.8}

We can now calculate this:

1Req=0.2+0.20830.4083\frac{1}{R_{\text{eq}}} = 0.2 + 0.2083 \approx 0.4083

So,

Req10.40832.448ohmsR_{\text{eq}} \approx \frac{1}{0.4083} \approx 2.448 \, \text{ohms}

Thus, the equivalent resistance when the 12-ohm resistor is replaced by a 5-ohm resistor is approximately 2.448 ohms.

Part (b): Both resistors are 8 ohms

Now, we calculate the equivalent resistance when both resistors are 8 ohms.

Again, using the formula for parallel resistors:

1Req=1r1+1r2\frac{1}{R_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2}

Substituting r1=8r_1 = 8 ohms and r2=8r_2 = 8 ohms:

1Req=18+18=14\frac{1}{R_{\text{eq}}} = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}

Thus,

Req=4ohmsR_{\text{eq}} = 4 \, \text{ohms}

So, when both resistors are 8 ohms, the equivalent resistance is 4 ohms.

Part (c): Formula for the equivalent resistance of two resistors r1r_1 and r2r_2 in parallel

The general formula for the equivalent resistance ReqR_{\text{eq}} of two resistors r1r_1 and r2r_2 in parallel is:

1Req=1r1+1r2\frac{1}{R_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2}

Solving for ReqR_{\text{eq}}, we get:

Req=r1r2r1+r2R_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2}

Thus, the formula for the equivalent resistance of two resistors r1r_1 and r2r_2 in parallel is:

Req=r1r2r1+r2R_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2}


Would you like further clarification or details on any of these steps? Also, here are some related questions to expand the understanding of parallel circuits:

  1. What happens to the equivalent resistance if more resistors are added in parallel?
  2. How do you calculate the total current if you know the total resistance in the circuit?
  3. How can the formula for equivalent resistance be adapted for resistors in series?
  4. What are the differences in behavior between resistors in series and resistors in parallel?
  5. How can you measure the equivalent resistance in a physical circuit using a multimeter?

Tip: In parallel circuits, the equivalent resistance is always less than the smallest resistor in the parallel group.

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Math Problem Analysis

Mathematical Concepts

Electric Circuits
Resistors in Parallel
Equivalent Resistance

Formulas

Equivalent resistance for parallel resistors: 1/R_eq = 1/r1 + 1/r2
Simplified formula for two resistors in parallel: R_eq = (r1 * r2) / (r1 + r2)

Theorems

Principle of Parallel Resistor Combination

Suitable Grade Level

Grades 10-12