Correct Answer - Option 2 : 4 Ω
Concept:
When the resistances are connected in series, the equivalent resistance is given by
\({R_{eq}} = {R_1} + {R_2} + \ldots + {R_n}\)
The equivalent resistance is greater than the largest resistance in the series circuit
When the resistances are connected in parallel, the equivalent resistance is given by
\(\frac{1}{{{R_{eq}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \ldots + \frac{1}{{{R_n}}}\)
The equivalent resistance is lesser than the smallest resistance in the series circuit
Calculation:
R1 = 12 ohm, R2 = 6 ohm
\(\frac{1}{{{R_{eq}}}} = \frac{1}{{12}} + \frac{1}{{6}} \)
⇒ Req = 4 ohm
