CFAA Fire Alarm Technician · Question
In a series circuit with a 24V source, R1 is 4 Ohms and R2 is 8 Ohms. What is the voltage drop across R2?
Total R = R1 + R2 = 4 + 8 = 12 Ohms. Total I = V / R = 24 / 12 = 2A. Voltage across R2 = I x R2 = 2A x 8 Ohms = 16 Volts.
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Question: In a series circuit with a 24V source, R1 is 4 Ohms and R2 is 8 Ohms. What is the voltage drop across R2?
Answer options: ✅ 16 V
- 8 V
- 24 V
- 12 V
Correct answer: 16 V
Explanation: Total R = R1 + R2 = 4 + 8 = 12 Ohms. Total I = V / R = 24 / 12 = 2A. Voltage across R2 = I x R2 = 2A x 8 Ohms = 16 Volts.
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Question explanations
- The primary power for a fire alarm system must be:
- If a 120VAC primary power circuit for a fire alarm control unit is consuming 60 Watts of power, what is the cu
- In a steady-state DC circuit, if the conductor length is doubled while maintaining voltage and wire gauge, wha
- A resistor is rated at 10 Watts at 24VDC. If the voltage drops to 20VDC, what is the new power dissipation?
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