Question

Four resistors of \(8\,\Omega\) each are connected in parallel. Then five such equivalent combinations are connected in series. What will be the total resistance?

• A. \(10\,\Omega\)
• B. \(2\,\Omega\)
• C. \(12\,\Omega\)
• D. \(2.5\,\Omega\)

Hint:

For \(n\) identical resistors: Parallel: \[ R_{\text{eq}}=\frac{R}{n} \] Series: \[ R_{\text{eq}}=nR \] Always simplify the parallel block first and then combine the resulting equivalent resistances.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem represents a compound or mixed resistor circuit network containing both parallel and series blocks. To solve complex mixed-combination circuits easily, break the entire network down into distinct groups. First, solve for the single equivalent resistance of one parallel block, and then treat each block as an individual component linked together in a series chain.
Step 2: Key Formula or Approach:
1. Parallel rule for identical resistors: The equivalent resistance ($R_p$) of $n$ identical resistors connected in parallel is: $$ R_p = \frac{R_{\text{individual}}}{n} $$ 2. Series rule for identical blocks: The total resistance ($R_{\text{total}}$) of $m$ identical resistance blocks wired end-to-end in series is: $$ R_{\text{total}} = m \times R_p $$
Step 3: Detailed Explanation:
Let's calculate the answer step-by-step using our formulas: 1. Step 1: Find the resistance of one parallel combination block: - Each individual resistor value ($R_{\text{individual}}$) = $8\,\Omega$ - Number of resistors connected in parallel inside one block ($n$) = $4$ $$ R_p = \frac{8\,\Omega}{4} = 2\,\Omega $$ Each parallel cluster can be replaced conceptually by a single $2\,\Omega$ resistor. 2. Step 2: Connect five identical parallel combination blocks in series: - Now we have a circuit with 5 identical blocks connected in a line, where each block has an effective resistance of $R_p = 2\,\Omega$. - Number of blocks in series ($m$) = $5$ $$ R_{\text{total}} = R_{p1} + R_{p2} + R_{p3} + R_{p4} + R_{p5} $$ $$ R_{\text{total}} = 2\,\Omega + 2\,\Omega + 2\,\Omega + 2\,\Omega + 2\,\Omega $$ $$ R_{\text{total}} = 5 \times 2\,\Omega = 10\,\Omega $$ The total accumulated resistance of this mixed network configuration is $10\,\Omega$. This matches option (A).
Step 4: Final Answer:
The total resistance of the network configuration is 10$\Omega$.