Four resistors of 8 Ω each are connected in parallel, five such combinations are then connected in series. What is the total resistance?

10 Ω

The correct answer is Option (1) → 10 Ω

Resistance, $R=8Ω$

$\frac{1}{R_{eq}}$, four R in parallel = $\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}$

$\frac{1}{R_{eq}}=\frac{4}{8}$

$R_{eq}=2Ω$

$R_T$, fine $R_{eq}$ are connected in series = $R_{eq}×5$

$=2Ω×5=10Ω$