Question

Consider a series LRC-circuit in which C-120.0 uF. When driven at a frequency w = 200.0 rad s-1 the com ples impedance is given by Z-(100.0-J10.0) Ω. (a) Calculate (i) the reactances X1 and Xc of the circuit at w = 200.0 rad s-1, ii) the inductance of the circuit. (b) Calculate the resonant angular frequency of the circuit. Hence, calculate the driving frequencies for which the average power in the circuit is half the reso- nant power [Hint: the solutions to the quadratic formula: az? + bx + c 0 are given by: -4ac) /2a.

Answer 1

(a). (i). The reactants are
`X_(L) =31.66\ \Omega` .

(II). The inductance of the circuit is 0.1583 Henry.

(b). The resonant angular frequency is 229.4 rad/s.

Step-by-step explanation:

Given that,

Capacitor = 120.0 μC

Frequency = 200.0 rad/s

Impedance = 100.0 -10j

(I). We need to calculate the
`X_(C)`

`X_(C)=(1)/(C*\omega)`

Put the value into the formula

`X_(C)=(1)/(120*10^(-6)*200)`

`X_(C)=41.66\ \Omega`

(II). We know that,

Formula of impedance is

`Z=\sqrt{R^2+X_(L)^2+X_(C)^2}`...(I)

Given equation of impedance is

`Z=(100-10j)`...(II)

On Comparing of equation (I) and (II)

`R = 100`

`X_(L)-X_(C)=-10`

Now, put the value of
`X_(C)`

`X_{L=41.66-10`

`X_(L)=31.66\ \Omega`

We need to calculate the inductance

Using formula of inductance

`X_(L)=\omega* L`

Put the value into the formula

`L=(X_(L))/(\omega)`

`L=(31.66)/(200)`

`L=0.1583\ Henry`

(b). We need to calculate the resonant angular frequency

Using formula of the resonant angular frequency

`angular\ frequency =(1)/(√(L* C))`

`angular\ frequency =\frac{1}{\sqrt{0.1583*120*10^(-6)}}`

`angular\ frequency =229.4\ rad/s`

Hence, (a). (i). The reactants are
`X_(L) =31.66\ \Omega` .

(II). The inductance of the circuit is 0.1583 Henry.

(b). The resonant angular frequency is 229.4 rad/s.