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: Find the equivalent resistance of four \(8\,\Omega\) resistors in parallel. For \(n\) identical resistors \(R\) connected in parallel: \[ R_p=\frac{R}{n} \] Here, \[ R=8\,\Omega, \qquad n=4 \] Therefore, \[ R_p=\frac{8}{4} \] \[ R_p=2\,\Omega \]

Step 2: Connect five such combinations in series. Five equivalent resistances of \(2\,\Omega\) each are connected in series. Hence, \[ R_{\text{total}} = 2+2+2+2+2 \] \[ R_{\text{total}} = 10\,\Omega \]

Step 3: Identify the correct option. \[ \boxed{R_{\text{total}}=10\,\Omega} \]