A 2 Ω resistor is connected in series with a parallel combination of a 6 Ω resistor and a 12 Ω resistor. The whole arrangement is connected to a 12 V battery. What current is drawn from the battery?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2 A
For resistors in parallel, \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\); resistances in series add. Then Ohm's law gives \(I = \frac{V}{R}\).
Parallel part: \(\frac{1}{R_p} = \frac{1}{6} + \frac{1}{12} = \frac{2+1}{12} = \frac{3}{12} = \frac{1}{4}\), so \(R_p = 4\,\Omega\)
Total resistance: \(R = 2 + 4 = 6\,\Omega\)
Current: \(I = \frac{V}{R} = \frac{12}{6}\)
\(I = 2\,\mathrm{A}\)