Question

(2 points) Two driven inductors A R = 1 kΩ resistor, a L1 = 20 mH inductor and a L2 28 mH inductor are connected in series. A funtion generator drives the circuit with a 5 Vpp variable frequency sine wave. (a) What is the equivalent impedance Zeq of this circuit? O A. R +jwLIL2/(L1+ L2) E. None of these (b) For what angular frequency does |Zoq v2 R? 20833.3 radians/sec (c) What is the peak-to-peak value of the voltage across the rèsistor at this frequency? Volts (d) By how many degrees does the current lag the generator voltage at this frequency?

Answer 1

Problem-1

Let $\omega$ be the angular frequency of the supply voltage.

Given details are

$R=1K\Omega$

し 1-20111 H

L2 = 28m H

Now the inductive reactance will be

$X_{L1}=j\omega L_1$

$X_{L2}=j\omega L_2$

(a)

Since the three components are connected in series, the equivalent impedance of the circuit will be

$Z_{eq}=R+j\omega L_1+j\omega L_2$

$\Rightarrow Z_{eq}=R+j\omega (L_1+L_2)$

The magnitude of the equivalent impedance will be given as

$| Z_{eq}|=\sqrt{R^2+\omega^2(L_1+L_2)^2}$

(b)

Let $\omega_1$ be the angular frequency at which $|Z_{eq}|=\sqrt{2}R$

$\Rightarrow \sqrt{R^2+\omega_1^2(L_1+L_2)^2}=\sqrt{2}R$

$\Rightarrow R^2+\omega_1^2(L_1+L_2)^2=2R^2$

$\Rightarrow \omega_1^2(L_1+L_2)^2=R^2$

$\Rightarrow \omega_1(L_1+L_2)=R$

wi = L1 +L2

1K12 1 x 105 105 20833.33rad/sec 20mH + 28mH (20 + 28) x 10348

(c)

The equivalent impedance at the frequency of $\omega_1$ is

Zeg 103+)20833.33(20 +28) x 10-3-(1 +j) 103 Ω = V2 x 10345"D ()

Given the peak-to-peak voltage is $V_{p-p}=5V$

So, the RMS voltage is

VRAI s-24-252-1 7678V

So, the phasor representation of the source voltage is given as

V. = 1.7678 0°V

The current through the circuit will be

1.76782O° ls = _ = = 1.25 × 10-3L-45°ー1.25L-450mA Zeg V2x 10%450

The voltage drop across the resistor is

VR-Is × R = 1.25 × 10-3L-45° × 10°ー1.25L-45°

So, the peak-to-peak voltage across the resistor is

Vp-p-R = 2V/21 R = 2V2 × 1.25 = 3.53551

By observing the expression for the line current it is evident that the current lags the source voltage by 45 degrees.

Problem-2

Given details are

$E_m=5V$

$R=1K\Omega$

201nH

$C=0.01\mu F$

the resonant frequency is given by

fo = 0.1125 x 105-11.25kHz 2NLC 2π xy20x10-001-10 ー - 2n/LC 2π × V20 × 10-30.01 × 10-6

At the resonant frequency, we have $\omega L=\frac{1}{\omega C}$

The impedance of the circuit will be given by

$Z_{eq}=R+j(\omega L-\frac{1}{\omega C})=R=1K\Omega=1\times10^3\angle0^o\Omega$

The current through the circuit is

Em Zeg 1x 103200 -520%.A

So, the phase angle difference between the current and voltage at the resonance is 0 degrees

The voltage drop across the resistor is given by

$V_R=IR=5\angle0^omA\times1K\Omega=5\times10^{-3}\angle0^o\times10^3=5\angle0^oV$

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