Consider a series RL circuit with R= 80.0 and an unknown inductor L, driven by a

Q: Consider a series RL circuit with R= 80.0 and an unknown inductor L, driven by a

Impedence of the ckt Z^2 = R^2 + X^2 Z^2 = 80^2 + (2*3.14 * 7300 * 4.1 e -3)^2 Z = 204.27 ohms Power across R = I R = 0.132 *80 power across L = i XL = 0.132* 187.9 ratio of this = 0.132 *80/(0.132* 187.9) = 0.425

ID: 1464131 • Letter: C

Question

Consider a series RL circuit with R= 80.0 and an unknown inductor L, driven by an AC emf with Erms= 27.0 Volts at frequency f= 7.30 kHz. If we measure the current in the circuit, we find Irms= 0.132 Amps.

What is the inductance L?

Answer: 4.10 x 10^-3 H

What is the magnitude of the phase angle between the applied voltage and the current?

Answer: 6.70 x 10^1 deg

What is the ratio of the power dissipated in the resistor to the power that would be dissipated if the inductor were not in the circuit? (Assume that Erms stays the same.)

Explanation / Answer

Impedence of the ckt Z^2 = R^2 + X^2

Z^2 = 80^2 + (2*3.14 * 7300 * 4.1 e -3)^2

Z = 204.27 ohms

Power across R = I R = 0.132 *80

power across L = i XL = 0.132* 187.9

ratio of this = 0.132 *80/(0.132* 187.9)

= 0.425

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Consider a series RL circuit with R= 80.0 and an unknown inductor L, driven by a

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Explanation / Answer

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