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: Problem Definition
Given resistances: $2 \, \Omega$, $4 \, \Omega$, $6 \, \Omega$, $8 \, \Omega$. Objective: achieve an equivalent resistance of $\frac{46}{3} \, \Omega$ through a specific connection.
Step 2: Option Analysis
Each provided configuration will be assessed for its resulting equivalent resistance.
Step 3: Option (A) Evaluation
Configuration: $4 \, \Omega$ and $6 \, \Omega$ in parallel; $2 \, \Omega$ and $8 \, \Omega$ in series.
Parallel resistance ($4 \, \Omega$, $6 \, \Omega$):\[\frac{1}{R_{\text{parallel}}} = \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12}.\]\[R_{\text{parallel}} = \frac{12}{5} \, \Omega.\]
Series resistance ($2 \, \Omega$, $8 \, \Omega$):\[R_{\text{series}} = 2 + 8 = 10 \, \Omega.\]
Total resistance:\[R_{\text{total}} = R_{\text{parallel}} + R_{\text{series}} = \frac{12}{5} + 10 = \frac{12}{5} + \frac{50}{5} = \frac{62}{5} \, \Omega.\]
Result: $\frac{62}{5} \, \Omega eq \frac{46}{3} \, \Omega$. Option (A) is incorrect.
Step 4: Option (B) Evaluation
Configuration: $6 \, \Omega$ and $8 \, \Omega$ in parallel; $2 \, \Omega$ and $4 \, \Omega$ in series.
Parallel resistance ($6 \, \Omega$, $8 \, \Omega$):\[\frac{1}{R_{\text{parallel}}} = \frac{1}{6} + \frac{1}{8} = \frac{4 + 3}{24} = \frac{7}{24}.\]\[R_{\text{parallel}} = \frac{24}{7} \, \Omega.\]
Series resistance ($2 \, \Omega$, $4 \, \Omega$):\[R_{\text{series}} = 2 + 4 = 6 \, \Omega.\]
Total resistance:\[R_{\text{total}} = R_{\text{parallel}} + R_{\text{series}} = \frac{24}{7} + 6 = \frac{24}{7} + \frac{42}{7} = \frac{66}{7} \, \Omega.\]
Result: $\frac{66}{7} \, \Omega eq \frac{46}{3} \, \Omega$. Option (B) is incorrect.
Step 5: Option (C) Evaluation
Configuration: $2 \, \Omega$ and $6 \, \Omega$ in parallel; $4 \, \Omega$ and $8 \, \Omega$ in series.
Parallel resistance ($2 \, \Omega$, $6 \, \Omega$):\[\frac{1}{R_{\text{parallel}}} = \frac{1}{2} + \frac{1}{6} = \frac{3 + 1}{6} = \frac{4}{6} = \frac{2}{3}.\]\[R_{\text{parallel}} = \frac{3}{2} \, \Omega.\]
Series resistance ($4 \, \Omega$, $8 \, \Omega$):\[R_{\text{series}} = 4 + 8 = 12 \, \Omega.\]
Total resistance:\[R_{\text{total}} = R_{\text{parallel}} + R_{\text{series}} = \frac{3}{2} + 12 = \frac{3}{2} + \frac{24}{2} = \frac{27}{2} \, \Omega.\]
Result: $\frac{27}{2} \, \Omega eq \frac{46}{3} \, \Omega$. Option (C) is incorrect.
Step 6: Option (D) Evaluation
Configuration: $2 \, \Omega$ and $4 \, \Omega$ in parallel; $6 \, \Omega$ and $8 \, \Omega$ in series.
Parallel resistance ($2 \, \Omega$, $4 \, \Omega$):\[\frac{1}{R_{\text{parallel}}} = \frac{1}{2} + \frac{1}{4} = \frac{2 + 1}{4} = \frac{3}{4}.\]\[R_{\text{parallel}} = \frac{4}{3} \, \Omega.\]
Series resistance ($6 \, \Omega$, $8 \, \Omega$):\[R_{\text{series}} = 6 + 8 = 14 \, \Omega.\]
Total resistance:\[R_{\text{total}} = R_{\text{parallel}} + R_{\text{series}} = \frac{4}{3} + 14 = \frac{4}{3} + \frac{42}{3} = \frac{46}{3} \, \Omega.\]
Result: $\frac{46}{3} \, \Omega$ matches the target. Option (D) is correct.Final Answer: The correct configuration is (D) $2 \, \Omega$ and $4 \, \Omega$ in parallel with $6 \, \Omega$ and $8 \, \Omega$ in series.