Question

3.The phasor diagrams for a series RLC circuit are shown below. Rank the frequencies from lowest to highest. Which frequency is closest to the resonance? a. b. VI Vt Vc 4. Circuit / has a reonant frequency of wy, What is the resonant frequency wn of circuit // in terms of w? Circuit L Circuit IT 2L C/2 ε sin(40t) & sin(ot) 5. At time t -0 the capacitor in the circult below is fully charged with Qmax and the current through the circuit is 0. In terms of LC and Qmax what is the maximum current in the circuit? At what time t does this occur? a. b. L.

Answer 1

Solution 3

Step 1

a)The order of frequency from lowest to highest is given as:-

$\omega _{C}<\omega _{D}<\omega _{B}<\omega _{A}$

Step 2

As we know,

$X_{c}=\frac{1}{\omega c}$

Lower is the frequency, more is the capacitive reactance,$X_{c}$. Hence the circuit comes out to be capacitive($V_{c}>V_{L}$).

So, the third and the fourth circuits are capacitive. Since $V_{c}$ for the third circuit is more so it has the least frequency.

Step 3

Then comes the second circuit which is very close to resonance($V_{c}=V_{L}$).

And the first circuit has the maximum frequency because it is an inductive circuit. ( $V_{L}>V_{C}$)

The graph shown below can be considered regarding the same.

Capacitive Inductive Z0) Dynamic Impedance Frequency,f (fT) Series Resonance

b)The frequency which is very close to resonant frequency is $\omega _{B}$. The second circuit respective to $\omega _{B}$ is very close to resonance($V_{c}=V_{L}$).