The two resistors are in series across an ideal 12 V source. What is the voltage across the 9 Ω resistor?
- A3 V
- B9 V
- C12 V
- D36 V
Show answer and method
Answer: B — 9 V
Series resistances add: 3 + 9 = 12 Ω. The current is V ÷ R = 12 ÷ 12 = 1 A through both resistors. The 9 Ω resistor therefore has a voltage drop of I × R = 1 × 9 = 9 V. The other resistor drops 3 V, so the drops add to the 12 V supply.