Consider a series combination of resistors R1 = 5 Ω, R2 = 5 Ω connected in parallel to a third resistor R3 = 20 Ω. This arrangement is connected across a cell of emf 12 V. The current through R3 is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.6 A
R1 and R2 in series (10 Ω) form one branch and R3 forms the other. Parallel branches have the same potential difference, and with no internal resistance given, the full 12 V appears across R3.
\( \displaystyle I_3 = \frac{V}{R_3} = \frac{12}{20} \)
\( \displaystyle I_3 = 0.6\,\mathrm{A} \)