Curioustab

Home » Electrical Engineering » RL Circuits

The voltages in Problem 5 are measured at a certain frequency. To make the resistor voltage less than the inductor voltage, the frequency is

increased
decreased
doubled
not a factor

Correct Answer: increased

← Previous Question Next Question→

More Questions from RL Circuits

A 12 kΩ resistor is in series with a 90 mH coil across an 8 kHz ac source. Current flow in the circuit, expressed in polar form, is I = 0.3∠0° mA. The source voltage, expressed in polar form, is
Referring to Problem 18, the circuit impedance is
When the frequency of the voltage applied to a series RL circuit is decreased, the impedance
A 47 Ω resistor is in series with an inductive reactance of 120 Ω across an ac source. The impedance, expressed in polar form, is
A 3.3 kΩ resistor and a 120 mH coil are in parallel. Both components are across a 2 kHz, 12 V ac source. The total current in the circuit is
A 1.2 kΩ resistor is in series with a 15 mH coil across a 10 kHz ac source. The magnitude of the total impedance is
When the frequency of the voltage applied to a series RL circuit is increased, the phase angle
A 470 Ω resistor and a 200 mH coil are in parallel. Both components are across a 1.5 kHz ac source. The total admittance of the circuit, in polar form, is
A 470 Ω resistor and a coil with 125 Ω inductive reactance are in parallel. Both components are across a 15 V ac voltage source. Current through the inductor is
In a series RL circuit, 12 V rms is measured across the resistor, and 14 V rms is measured across the inductor. The peak value of the source voltage is

Discussion & Comments

No comments yet. Be the first to comment!

Join Discussion