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Chapter 12: Problem 1779

Two resistances \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) have effective resistance \(\mathrm{R}_{\mathrm{s}}\) when connected in sires combination and \(R_{p}\) when connected in parallel combination if $\mathrm{R}_{8} \mathrm{R}_{\mathrm{p}}=16\( and \)\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)=4\( the values of \)\mathrm{R}_{1}\( and \)\mathrm{R}_{2}$ are (A) \(2 \Omega\) and \(0.5 \Omega\) (B) \(1 \Omega\) and \(0.25 \Omega\) (C) \(8 \Omega\) and \(2 \Omega\) (D) \(4 \Omega\) and \(1 \Omega\)

Short Answer

Expert verified
The values of \(R_1\) and \(R_2\) can be found by utilizing the formulas for series and parallel resistance combinations and the given relationships. By expressing \(R_s\) in terms of \(R_p\) and substituting into the series formula, you can replace \(R_1\) with the given ratio, simplify the equation and substitute it into the parallel formula. By solving for \(R_p\) and substituting the value back into previous equations, the values for \(R_2\) and \(R_1\) can be found. Comparing these with the given answer options, we find that \(R_1 = 8 \Omega\) and \(R_2 = 2 \Omega\), which aligns with option (C).

Step by step solution

01

Write formulas for series and parallel combinations

For two resistances connected in series, the effective resistance \(R_s\) is given by: \[R_s = R_1 + R_2\] For two resistances connected in parallel, the effective resistance \(R_p\) is given by: \[\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\]
02

Write given relationships

We are given the following relationships: \[R_s R_p = 16\] \[\frac{R_1}{R_2} = 4\]
03

Substitute first relationship into the series formula

Since \(R_s R_p = 16\), we can rewrite this as: \[R_s = \frac{16}{R_p}\] Now substitute this into the series formula: \[\frac{16}{R_p} = R_1 + R_2\]
04

Replace R1 with the given ratio

We are given that \(R_1 = 4R_2\). Substitute this into the current equation: \[\frac{16}{R_p} = 4R_2 + R_2\]
05

Simplify the equation and substitute into the parallel formula

Combine the terms and divide both sides by 5: \[\frac{16}{5R_p} = R_2\] Substitute this into the parallel formula: \[\frac{1}{R_p} = \frac{1}{4(\frac{16}{5R_p})} + \frac{1}{\frac{16}{5R_p}}\]
06

Solve for Rp and substitute back into our equations

Simplify the equation and solve for \(R_p\): \[\frac{1}{R_p} = \frac{5R_p}{16(4R_p) + 16R_p}\] Upon solving the equation, we get \( R_p = \frac{16}{5} \). Now, we can substitute the value of \(R_p\) back into earlier equations to get the value of \( R_2\): \[R_2 = \frac{16}{5 R_p}\] Substitute the value of \( R_p \) into the equation: \[R_2 = \frac{16}{5 (\frac{16}{5})}\] Upon solving, we get \( R_2 = 2 \Omega \). Finally, substitute the value of \( R_2 \) into the given ratio \(R_1/R_2 = 4 \): \[\frac{R_1}{2 \Omega} = 4 \] And thus, we get \( R_1 = 8 \Omega \).
07

Check the answer choices

Comparing the calculated values of \(R_1\) and \(R_2\) to the given answer choices: (A) \(2 \Omega\) and \(0.5 \Omega\) (B) \(1 \Omega\) and \(0.25 \Omega\) (C) \(8 \Omega\) and \(2 \Omega\) (D) \(4 \Omega\) and \(1 \Omega\) We find that the correct answer is (C) \(8 \Omega\) and \(2 \Omega\).

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