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. $4 \, \Omega$ and $6 \, \Omega$ are in parallel with $2 \, \Omega$ and $8 \, \Omega$ in series
• B. $6 \, \Omega$ and $8 \, \Omega$ are in parallel with $2 \, \Omega$ and $4 \, \Omega$ in series
• C. $2 \, \Omega$ and $6 \, \Omega$ are in parallel with $4 \, \Omega$ and $8 \, \Omega$ in series
• D. $2 \, \Omega$ and $4 \, \Omega$ are in parallel with $6 \, \Omega$ and $8 \, \Omega$ in series

Hint:

When combining resistances, remember that:
The equivalent resistance of resistors in series is the sum of their resistances.
The equivalent resistance of resistors in parallel is given by the reciprocal of the sum of the reciprocals of their resistances.

Answer 1

The Correct Option is D

Step 1: Understanding the Problem
We are given four resistances: $2 \, \Omega$, $4 \, \Omega$, $6 \, \Omega$, and $8 \, \Omega$. We need to connect them in such a way that the equivalent resistance is $\frac{46}{3} \, \Omega$.
Step 2: Analyzing the Options
We will evaluate each option to see which configuration gives the desired equivalent resistance.
Step 3: Evaluating Option (A)
In option (A), $4 \, \Omega$ and $6 \, \Omega$ are in parallel, and $2 \, \Omega$ and $8 \, \Omega$ are in series.
First, calculate the equivalent resistance of $4 \, \Omega$ and $6 \, \Omega$ in parallel: \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12}. \] \[ R_{\text{parallel}} = \frac{12}{5} \, \Omega. \]
Next, calculate the equivalent resistance of $2 \, \Omega$ and $8 \, \Omega$ in series: \[ R_{\text{series}} = 2 + 8 = 10 \, \Omega. \]
Now, combine the parallel and series resistances: \[ R_{\text{total}} = R_{\text{parallel}} + R_{\text{series}} = \frac{12}{5} + 10 = \frac{12}{5} + \frac{50}{5} = \frac{62}{5} \, \Omega. \]
This does not match the desired equivalent resistance of $\frac{46}{3} \, \Omega$. Therefore, option (A) is incorrect.
Step 4: Evaluating Option (B)
In option (B), $6 \, \Omega$ and $8 \, \Omega$ are in parallel, and $2 \, \Omega$ and $4 \, \Omega$ are in series.
First, calculate the equivalent resistance of $6 \, \Omega$ and $8 \, \Omega$ in parallel: \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{6} + \frac{1}{8} = \frac{4 + 3}{24} = \frac{7}{24}. \] \[ R_{\text{parallel}} = \frac{24}{7} \, \Omega. \]
Next, calculate the equivalent resistance of $2 \, \Omega$ and $4 \, \Omega$ in series: \[ R_{\text{series}} = 2 + 4 = 6 \, \Omega. \]
Now, combine the parallel and series resistances: \[ R_{\text{total}} = R_{\text{parallel}} + R_{\text{series}} = \frac{24}{7} + 6 = \frac{24}{7} + \frac{42}{7} = \frac{66}{7} \, \Omega. \]
This does not match the desired equivalent resistance of $\frac{46}{3} \, \Omega$. Therefore, option (B) is incorrect.
Step 5: Evaluating Option (C)
In option (C), $2 \, \Omega$ and $6 \, \Omega$ are in parallel, and $4 \, \Omega$ and $8 \, \Omega$ are in series.
First, calculate the equivalent resistance of $2 \, \Omega$ and $6 \, \Omega$ in parallel: \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{2} + \frac{1}{6} = \frac{3 + 1}{6} = \frac{4}{6} = \frac{2}{3}. \] \[ R_{\text{parallel}} = \frac{3}{2} \, \Omega. \]
Next, calculate the equivalent resistance of $4 \, \Omega$ and $8 \, \Omega$ in series: \[ R_{\text{series}} = 4 + 8 = 12 \, \Omega. \]
Now, combine the parallel and series resistances: \[ R_{\text{total}} = R_{\text{parallel}} + R_{\text{series}} = \frac{3}{2} + 12 = \frac{3}{2} + \frac{24}{2} = \frac{27}{2} \, \Omega. \]
This does not match the desired equivalent resistance of $\frac{46}{3} \, \Omega$. Therefore, option (C) is incorrect.
Step 6: Evaluating Option (D)
In option (D), $2 \, \Omega$ and $4 \, \Omega$ are in parallel, and $6 \, \Omega$ and $8 \, \Omega$ are in series.
First, calculate the equivalent resistance of $2 \, \Omega$ and $4 \, \Omega$ in parallel: \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{2} + \frac{1}{4} = \frac{2 + 1}{4} = \frac{3}{4}. \] \[ R_{\text{parallel}} = \frac{4}{3} \, \Omega. \]
Next, calculate the equivalent resistance of $6 \, \Omega$ and $8 \, \Omega$ in series: \[ R_{\text{series}} = 6 + 8 = 14 \, \Omega. \]
Now, combine the parallel and series resistances: \[ R_{\text{total}} = R_{\text{parallel}} + R_{\text{series}} = \frac{4}{3} + 14 = \frac{4}{3} + \frac{42}{3} = \frac{46}{3} \, \Omega. \]
This matches the desired equivalent resistance of $\frac{46}{3} \, \Omega$. Therefore, option (D) is correct. Final Answer: The correct configuration is (D) $2 \, \Omega$ and $4 \, \Omega$ are in parallel with $6 \, \Omega$ and $8 \, \Omega$ in series.