Alternating Current Overview

7.20 Inductance, L = 0.12 H

Capacitance, C = 480 nF = 480 $*{10}^{-9}$ F

Resistance, R = 23 Ω

Supply voltage, V = 230 V

Peak voltage is given as, $V_0$ = $\sqrt[]{2}*230$ = 325.27 V

Current flowing in the circuit is given by the relation,

$I_0=\frac{V_0}{\sqrt[]{R^2+({\omega L-\frac{1}{\omega C})}^2}}$ where $I_0$ = maximum at resonance

$\mathrm{A}\mathrm{t}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{n}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{w}\mathrm{e}\mathrm{}\mathrm{h}\mathrm{a}\mathrm{v}\mathrm{e}\mathrm{}{\omega }_R\mathrm{L}-\mathrm{}\frac{1}{{\mathrm{\omega }}_{\mathrm{R}}\mathrm{C}}=0$ , where ${\omega }_R$ = Resonance angular frequency.

Hence, ${\omega }_R$ = $\frac{1}{\sqrt[]{LC}}$ = $\frac{1}{\sqrt[]{0.12*480\mathrm{}*{10}^{-9}}}$ = 4166.67 rad/s

Therefore, resonating frequency ${\nu }_R$ = $\frac{{\omega }_R}{2\pi }$ = 663.15 Hz

Maximum current ${I_0}_{max}$ = $\frac{V_0}{R}$ = $\frac{325.27\mathrm{}}{23}$ = 14.14 A

Maximum average power absorbed by the circuit is given as

${P_{avg}}_{max}$ = $\frac{1}{2}{{I_0}_{max}}^2$ R = $\frac{1}{2}*{14.14\mathrm{}}^2*23$ = 2300 W

The power transferred to the circuit is half the power at resonant frequency

Frequencies at which power transferred is half = ${\omega }_R\pm$ Δ $\omega$ = 2 $\pi ({\nu }_R\pm \Delta \nu )$ where,

Δ $\omega =\frac{R}{2L}$ = $\frac{23}{2*0.12}$ = 95.83 rad/s

$\Delta \nu$ = $\frac{\mathrm{\Delta }\omega }{2\pi }$ = $\frac{95.83}{2*\pi }$ = 15.25 Hz

Therefore ${\nu }_R\pm \Delta \nu$ = 663.15 $\pm$ 15.25 = 647.9 Hz and 678.4 Hz

Hence, at 647.9 Hz and 678.4 Hz frequencies, the power transferred is half.

At these frequencies, current amplitude can be given as:

I' = $\frac{1}{\surd 2}*{I_0}_{max}$ = $\frac{14.14\mathrm{}}{\surd 2}$ = 10 A

Q factor of the given circuit can be obtained using the relation,

Q = $\frac{{\omega }_{R}L}{R}$ = $\frac{4166.67*0.12}{23}$ = 21.74

7.19 Inductance, L = 80 mH = 80 $*{10}^{-3}$ H

Capacitance, C = 60 $\mu F$ = 60 $*{10}^{-6}$ F

Resistance of the resistor, R = 15 Ω

Supply voltage, V = 230 V

Supply frequency, $\nu$ = 50 Hz

Peak voltage, $V_0$ = V $\surd 2$ = 325.27 V

Angular frequency, $\omega$ = 2 $\pi \nu$ = 2 $\pi *\nu$ = 2 $\pi *50$ = 100 $\pi$ rad/s

Since the elements are connected in series, the impedance is given by

Z = $\sqrt[]{R^2+({\omega L-\frac{1}{\omega C})}^2}$

= $\sqrt[]{{15}^2+({100\pi *80\mathrm{}*{10}^{-3}-\frac{1}{100\pi *60\mathrm{}*{10}^{-6}})}^2}$

= $\sqrt[]{225+({25.133-53.052)}^2}$

= 31.693 Ω

Current flowing in the circuit, I = $\frac{V}{Z}$ = $\frac{230}{31.693}$ = 7.26 A

Average power transferred to resistance is given as $P_R$ = $I^2$ R = ${\left(7.26\right)}^2*15$ = 789.97 W

Average power transferred to capacitor, $P_C$ = 0

Average power transferred to Inductor, $P_L$ = 0

Total power absorbed by the circuit = $P_R+P_C+P_L=789.97+0+0=789.97W$

7.18 Inductance, L = 80 mH = 80 $*{10}^{-3}$ H

Capacitance, C = 60 $\mu F$ = 60 $*{10}^{-6}$ F

Supply voltage, V = 230 V

Supply frequency, $\nu$ = 50 Hz

Peak voltage, $V_0$ = V $\surd 2$ = 325.27 V

Angular frequency, $\omega$ = 2 $\pi \nu$ = 2 $\pi *\nu$ = 2 $\pi *50$ = 100 $\pi$ rad/s

Maximum current is given as:

$I_0$ = $\frac{V_0}{(\omega L-\frac{1}{\omega C})}$ = $\frac{325.27\mathrm{}}{(100\pi *80\mathrm{}*{10}^{-3}-\frac{1}{100\pi *60\mathrm{}*{10}^{-6}})}$ = $-$ 11.65 A

The negative sign is due to $\omega L$ $<\frac{1}{\omega C}$

Amplitude of maximum current $\left|I_0\right|$ = 11.65 A

rms value of the current, I = $\frac{I_0}{\surd 2}$ = $\frac{-11.65}{\surd 2}$ = -8.24 A

(i) Potential difference across inductor, $V_L$ = I $*\omega L$ = 8.24 $*100\pi *$ 80 $*{10}^{-3}$ V = 207.09 V

(ii) Potential drop across capacitor, $V_C$ = I $*\frac{1}{\omega C}$ = 8.24 $*\frac{1}{100\pi *60\mathrm{}*{10}^{-6}}$ = 437.14 V

Average power consumed by the inductor is zero as actual voltage leads the current by $\frac{\pi }{2}$

Average power consumed by the capacitor is zero as the voltage lags current by $\frac{\pi }{2}$

The total power absorbed ( average over one cycle) is zero.

7.17 Inductance of the Inductor, L = 5.0 H

Resistance of the resistor, R = 40 Ω

Capacitance of the capacitor, C = 80 $\mu F=$ 80 $*{10}^{-6}$ F

Potential of the voltage source, V = 230 V

Impedance Z of the given parallel LCR circuit is given as

$\frac{1}{Z}$ = $\sqrt[]{\frac{1}{R^2}+{(\frac{1}{\omega L}-\omega C)}^2}$ where $\omega$ = angular frequency

At resonance $\frac{1}{\omega L}-\omega C$ = 0 or ${\omega }^2$ = $\frac{1}{LC}$

$\omega =\sqrt[]{\frac{1}{LC}}$ = $\frac{1}{\sqrt[]{5*80\mathrm{}*{10}^{-6}}}$ = 50 rad/s

The magnitude of Z is the maximum at $\omega$ = 50 rad/s. As a result, total current is minimum.

rms current flowing through the Inductor L is given as,

$I_L$ = $\frac{V}{\omega L}$ = $\frac{230}{50*5}$ = 0.92 A

rms current flowing through the Capacitor C is given as,

$I_L$ = $\frac{V}{\frac{1}{\omega C}}$ = 230 $*50*$ 80 $*{10}^{-6}$ = 0.92 A

rms current flowing through resistor R is given as, $I_R$ = $\frac{V}{R}$ = $\frac{230}{40}$ = 5.75 A

7.16 Capacitance of the capacitor, C = 100 $\mu F$ = 100 $*{10}^{-6}$ F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

Frequency of the supply voltage, $\nu$ = 12 kHz = 12 $*{10}^3$ Hz

Angular frequency, $\omega$ = 2 $\pi \nu$ = 2 $\pi *12\mathrm{}*{10}^3$ rad/s = 24 $\pi *{10}^3$ rad/s

For a RC circuit, the impedance Z, is given by Z = $\sqrt[]{R^2+\frac{1}{{\omega }^2{C}^2}}$

Z = $\sqrt[]{{40}^2+\frac{1}{{(24\pi *{10}^3)}^2{*(100*{10}^{-6})}^2}}$ = 40 Ω

Peak voltage, $V_0$ = $\surd 2*V$ = $\surd 2*110$ = 155.56 V

Peak current, $I_0$ = $\frac{V_0}{Z}$ = $\frac{155.56\mathrm{}}{40}$ = 3.8

9 A

In a capacitor circuit, the voltage lags behind the current by a phase angle of $\varnothing$ . This angle is given by the relation

$\tan ?\varnothing$ = $\frac{\frac{1}{\omega C}}{R}$ = $\frac{1}{\omega CR}$ = $\frac{1}{24\pi *{10}^3*100*{10}^{-6}*40}$ = 3.315 $*$ ${10}^{-3}$

$\varnothing =0.19°=\frac{0.19°*\pi }{180}$ rad

$\mathrm{T}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{a}\mathrm{g},t=\frac{\varnothing }{\omega }$ = $\frac{\frac{0.19°*\pi }{180}}{24\pi *{10}^3}$ = $\frac{0.19*\pi }{24\pi *{10}^3*180}$ = 4.4 $*{10}^{-8}$ s = 0.044 $\mu$ s

Hence, $\varnothing$ tends to become zero at high frequencies. At a high frequency, capacitor C acts as a conductor.

In DC circuit, after the steady state is achieved, $\omega$ = 0

Hence, capacitor C amounts to an open circuit.

7.15 Capacitance of the capacitor, C = 100 $\mu F$ = 100 $*{10}^{-6}$ F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

Frequency of the supply voltage, $\nu$ = 60 Hz

Angular frequency, $\omega$ = 2 $\pi \nu$ = 2 $\pi *60$ rad/s

For a RC circuit, the impedance Z, is given by Z = $\sqrt[]{R^2+\frac{1}{{\omega }^2{C}^2}}$

Z = $\sqrt[]{{40}^2+\frac{1}{{(2\pi *60)}^2{*(100*{10}^{-6})}^2}}$ = 47.996 Ω

Peak voltage, $V_0$ = $\surd 2*V$ = $\surd 2*110$ = 155.56 V

Peak current, $I_0$ = $\frac{V_0}{Z}$ = $\frac{155.56\mathrm{}}{47.996}$ = 3.24 A

In a capacitor circuit, the voltage lags behind the current by a phase angle of $\varnothing$ . This angle is given by the relation

$\tan ?\varnothing$ = $\frac{\frac{1}{\omega C}}{R}$ = $\frac{1}{\omega CR}$ = $\frac{1}{2\pi *60\mathrm{}*100*{10}^{-6}*40}$ = 0.663

$\varnothing =33.55°=\frac{33.55°*\pi }{180}$ rad

$\mathrm{T}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{a}\mathrm{g},t=\frac{\varnothing }{\omega }$ = $\frac{\frac{33.55°*\pi }{180}}{2\pi *60}$ = $\frac{33.55*\pi }{2\pi *60*180}$ = 1.55 $*{10}^{-3}$ s = 1.55 ms

Hence the time lag between maximum current and maximum voltage is 1.55 ms.