Question

You have a 206 −Ω resistor, a 0.399 −H inductor, a 5.06 −μF capacitor, and a variable-frequency ac source with an amplitude of 3.00 V . You connect all four elements together to form a series circuit.

1) In units of rad/s, at what frequency will the current in the circuit be greatest?

2) What will be the current amplitude at this frequency?

3) What will be the current amplitude at an angular frequency of 391 rad/s ?

Answer 1

Given data:

R = 206 Ω

L = 0.399 H

C = 5.06 μF

A =  3.00 V

four elements together to form a series circuit.

current in the circuit be greatest when Z is smallest.Z is smallest when Z = R, X_L = X_R

X_L=ωL

X_R= 1/ωC

Z = R, X_L = X_R

ωL = 1/ωC

ω² = 1/LC

ω = (1/LC)^1/2

ω = ( 1 / 0.399 x 0.00000506 )^1/2

ω = (1/0.00000201894)^1/2

ω = ( 495309.419795)^1/2

ω = 703.78 rad/s

(2) What will be the current amplitude at this frequency?

When Z=R

Z =E/I = R

E/R = I

I= E/R

= 3/206

I = 0.0145 A

(3) What will be the current amplitude at an angular frequency of 391 rad/s?

I = E/Z

Z = { R² + (ωL-1/ωC) }^1/2

= { 203² + (391 X 0.399-1/391X 0.00000506) }^1/2

= { 41209 + (156.009 - 505.4436278) }^1/2

= { 41209 + (-349.4346278) }^1/2

= { 41209 + (-349.4346278) }^1/2

= { 41209 -349.4346278 }^1/2

= ( 40859.5653722 )^1/2

Z = 202.1374 Ω

I = E/Z

= 3 / 202.1374

I = 0.01484 A