Advertisements
Advertisements
Question
What is the ratio of inductive and capacitive reactance in an ac circuit?
Options
ω2LC
LC2
`(LC)/omega^2`
`omega^2L`
Advertisements
Solution
ω2LC
Explanation:
Inductive reactance, XL = ωL
Capacitive reactance, `X_C = 1/(omegaC)`
Therefore, `X_L/X_C = (omegaL)/((1"/"omegaC))`
= `omega^2LC`
APPEARS IN
RELATED QUESTIONS
Show that in an AC circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase.
Obtain if the circuit is connected to a high-frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?
Obtain if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.
An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L.
Alternating current is so called because _______.
An inductive circuit contains resistance of 10 ohms and an inductance of 2 henry. If an A.C. voltage of 120 Volts and frequency 60 Hz is applied to this circuit, the current would be nearly ______.
Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.
An electrical device draws 2kW power from AC mains (voltage 223V (rms) = `sqrt(50,000)` V). The current differs (lags) in phase by `phi(tan phi = (-3)/4)` as compared to voltage. Find (i) R, (ii) XC – XL, and (iii) IM. Another device has twice the values for R, XC and XL. How are the answers affected?
An ac voltage V = V0 sin ωt is applied across a pure inductor of inductance L. Find an expression for the current i, flowing in the circuit and show mathematically that the current flowing through it lags behind the applied voltage by a phase angle of `π/2`. Also draw graphs of V and i versus ωt for the circuit.
