Question

Two rows of fluorescent lamps are installed in an office on the same branch circuit, with each row drawing 12.5 amperes. The source voltage is 277 volts, and the line or total circuit resistance is 0.5 Ω. The wire used has a constant (k) of 12.6. What is the voltage at the load? (Round the FINAL answer to one decimal place.)

Answer 1

The voltage at the load is calculated by subtracting the voltage drop due to the circuit's resistance from the source voltage, resulting in 264.5 volts.

Step-by-step explanation:

To calculate the voltage at the load for the fluorescent lamps, we need to account for the voltage drop due to the resistance in the circuit wires. Since both rows of lamps are on the same branch circuit and drawing 12.5 amperes each, the total current is 25 amperes. The voltage drop can be determined using Ohm's law, which states that V = IR, where V is the voltage drop, I is the current, and R is the resistance.

The voltage drop across the circuit wiring is therefore:

Vdrop = I × R = 25 A × 0.5 Ω = 12.5 V

To find the voltage at the load, we subtract the voltage drop from the source voltage:

Vload = Vsource - Vdrop = 277 V - 12.5 V = 264.5 V

Therefore, the voltage at the load is 264.5 volts when rounded to one decimal place.

Answer 2

`V_(load) = 264.5 Volts`

Step-by-step explanation:

As we know that the resistance of the wire is given as

`r = 0.5 ohm`

now we know that both the bulbs are getting 12.5 A current in their branches

so total current in the circuit is given as

`i = 12.5 + 12.5`

`i = 25 A`

so we have total voltage across the wire is given as

`V = ir`

`V = 25(0.5) = 12.5 volts`

now the load voltage is given as

`V_(load) = E - V`

`V_(load) = 277 - 12.5`

`V_(load) = 264.5 Volts`