Final answer:
The voltage of a circuit with three resistors of 2.4 kΩ, 680 Ω, and 1.5 kΩ arranged in series with a current of 54.6 mA is approximately 250.07 V.
Step-by-step explanation:
To calculate the voltage of the circuit with three resistors arranged in series, one can apply Ohm's Law, which states V = I × R, where V is the voltage, I is the current, and R is the total resistance. The total resistance in this case is the sum of the individual resistances: 2.4 kΩ + 680 Ω + 1.5 kΩ. Given the current is 54.6 mA (which is 0.0546 A),
the total resistance is 2.4k + 0.68k + 1.5k = 4.58 kΩ or 4580 Ω. Therefore, the voltage across the circuit is V = 0.0546 A × 4580 Ω.
After calculating, we get V = 250.068 V.
Hence, the voltage of the circuit is approximately 250.07 V.