Question

You have a 192 −Ω−Ω resistor, a 0.409 −H−H inductor, a 4.95 −μF−μF capacitor, and a variable-frequency ac source with an amplitude of 3.02 VV . You connect all four elements together to form a series circuit.
(a) At what frequency will the current in the circuit be greatest? What will be the current amplitude at this frequency?
(b) What will be the current amplitude at an angular frequency of 400 rad/s? At this frequency, will the source voltage lead or lag the current?

Answer 1

Answer:

Explanation:

Given that,

The resistance of the resistor is

R = 192 Ω

The inductance of the inductor is

L = 0.409 H

The capacitance of the capacitor is

C = 4.95 μF

a. At what frequency is current max?

The current will be maximum at resonance, and resonance frequency is given as

f = 1/2π√L•C

f = 1/2π√(0.409×4.95×10^-6)

f = 111.86Hz

b. Currency at frequency 400rad/s.

w = 400rad/s

So, we need to find the inductive reactance

XL = wL

XL = 400 ×0.409

XL = 163.6 ohms

We also need the capacitive reacttance

XC = 1/wC

XC = 1/400×4.95×10^-6

XC = 505.05 ohms

Then, the impedance of the circuit is given as

Z = √(R²+(XL-XC)²)

Z = √(192²+(505.05—163.6)²)

Z = √(192²+341.9²)

Z = 392.12 ohms

Then, the current that flows can be calculated using

V= IZ

I = V/Z

I = 3.02/392.12

I = 7.7 × 10^-3 A

I = 7.7 mA

Since Xc>XL, then, the source voltage lags the current