Question

A large electromagnet coil is connected to a 130 Hz ac source. The coil has a resistance 410 omega or Ohms, and at this source frequency the coil has an inductive reactance 230 omega or Ohms.

Part A.) What is the inductance of the coil? ( answer is L= ? H)
Part B.) What must the rms voltage of the source be if the coil is to consume an average electrical power of 830 W? (answer is V(rms)= ? V)

Answer 1

The inductance of the coil is 0.297 H and the rms voltage of the source should be 2024.39 V.

Step-by-step explanation:

Part A:

The inductance of the coil can be calculated using the formula:

L = XL / (2 * π * f)

Where:

• L is the inductance in henries (H)
• XL is the inductive reactance in ohms (Ω)
• f is the frequency in hertz (Hz)

Plugging in the given values, we get:

L = 230 Ω / (2 * π * 130 Hz) = 0.297 H

Part B:

The rms voltage of the source can be calculated using the formula:

Vrms = Pavg / Irms

Where:

• Vrms is the rms voltage in volts (V)
• Pavg is the average power in watts (W)
• Irms is the rms current in amperes (A)

Plugging in the given values, we get:

Vrms = 830 W / (0.410 Ω) = 2024.39 V

Answer 2

The inductance of the coil is approximately 0.278 H. The RMS voltage of the source should be approximately 265.2 V.

Step-by-step explanation:

Part A: To find the inductance of the coil, we can use the formula XL = 2πfL, where XL is the inductive reactance and f is the frequency. Given that the inductive reactance is 230 Ω and the frequency is 130 Hz, we can rearrange the formula to solve for L. Plugging in these values, we get:

230 Ω = 2π(130 Hz)L

L = 230 Ω / (2π(130 Hz)) ≈ 0.278 H

Therefore, the inductance of the coil is approximately 0.278 H.

Part B: To find the RMS voltage of the source, we can use the formula P = V(rms)^2 / R, where P is the average power, V(rms) is the RMS voltage, and R is the resistance. Given that the average power is 830 W and the resistance is 410 Ω, we can rearrange the formula to solve for V(rms). Plugging in these values, we get:

830 W = V(rms)^2 / 410 Ω

V(rms) = √(830 W * 410 Ω) ≈ 265.2 V

Therefore, the RMS voltage of the source should be approximately 265.2 V.