Question

If you are provided a set of resistances 2 \( \Omega \), 4 \( \Omega \), 6 \( \Omega \) and 8 \( \Omega \). Connect these resistances so as to obtain an equivalent resistance of \( \frac{46}{3} \Omega \).

• A. 6 \( \Omega \) and 8 \( \Omega \) are in parallel with 2 \( \Omega \) and 4 \( \Omega \) in series
• B. 2 \( \Omega \) and 6 \( \Omega \) are in parallel with 4 \( \Omega \) and 8 \( \Omega \) in series
• C. 2 \( \Omega \) and 4 \( \Omega \) are in parallel with 6 \( \Omega \) and 8 \( \Omega \) in series
• D. 4 \( \Omega \) and 6 \( \Omega \) are in parallel with 2 \( \Omega \) and 8 \( \Omega \) in series

Hint:

When testing options for equivalent resistance, it's helpful to first convert the target fraction to a decimal to quickly eliminate combinations that are clearly too large or too small. Here, \(46/3 \approx 15.33\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find a combination of the four given resistors that results in a specific equivalent resistance, \( \frac{46}{3} \Omega \). We can test the combinations described in the options.
Step 2: Key Formula or Approach:
- For resistors in series: \(R_{eq} = R_1 + R_2 + \dots\)
- For two resistors in parallel: \(R_{eq} = \frac{R_1 R_2}{R_1 + R_2}\)
The target resistance is \( \frac{46}{3} \Omega \approx 15.33 \Omega \).
Step 3: Detailed Explanation:
Let's evaluate the equivalent resistance for each option. The phrasing "A and B are in parallel with C and D in series" means that A and B form a parallel block, which is then connected in series with C and D.
(A) 6 \( \Omega \) and 8 \( \Omega \) in parallel, in series with 2 \( \Omega \) and 4 \( \Omega \)
Parallel part: \( R_p = \frac{6 \times 8}{6 + 8} = \frac{48}{14} = \frac{24}{7} \, \Omega \).
Total resistance: \( R_{eq} = R_p + 2 + 4 = \frac{24}{7} + 6 = \frac{24 + 42}{7} = \frac{66}{7} \, \Omega \). This is not \( \frac{46}{3} \Omega \).
(B) 2 \( \Omega \) and 6 \( \Omega \) in parallel, in series with 4 \( \Omega \) and 8 \( \Omega \)
Parallel part: \( R_p = \frac{2 \times 6}{2 + 6} = \frac{12}{8} = \frac{3}{2} \, \Omega \).
Total resistance: \( R_{eq} = R_p + 4 + 8 = \frac{3}{2} + 12 = 1.5 + 12 = 13.5 \, \Omega \). This is not \( \frac{46}{3} \Omega \).
(C) 2 \( \Omega \) and 4 \( \Omega \) in parallel, in series with 6 \( \Omega \) and 8 \( \Omega \)
Parallel part: \( R_p = \frac{2 \times 4}{2 + 4} = \frac{8}{6} = \frac{4}{3} \, \Omega \).
Total resistance: \( R_{eq} = R_p + 6 + 8 = \frac{4}{3} + 14 \).
To add these, find a common denominator: \( R_{eq} = \frac{4}{3} + \frac{14 \times 3}{3} = \frac{4 + 42}{3} = \frac{46}{3} \, \Omega \). This matches the target resistance.
(D) 4 \( \Omega \) and 6 \( \Omega \) in parallel, in series with 2 \( \Omega \) and 8 \( \Omega \)
Parallel part: \( R_p = \frac{4 \times 6}{4 + 6} = \frac{24}{10} = 2.4 \, \Omega \).
Total resistance: \( R_{eq} = R_p + 2 + 8 = 2.4 + 10 = 12.4 \, \Omega \). This is not \( \frac{46}{3} \Omega \).
Step 4: Final Answer:
The combination described in option (C) gives the required equivalent resistance of \( \frac{46}{3} \Omega \).