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

Capacitance, C = 60 $\mu F$ = 60 $\times {10}^{-6}$ 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 $\pi \times \nu$ = 2 $\pi \times 50$ = 100 $\pi$ rad/s

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

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

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

= $\sqrt[]{225+({25.133-53.052)}^2}$

= 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 = ${\left(7.26\right)}^2\times 15$ = 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 $\times {10}^{-3}$ H

Capacitance, C = 60 $\mu F$ = 60 $\times {10}^{-6}$ 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 $\pi \times \nu$ = 2 $\pi \times 50$ = 100 $\pi$ rad/s

Maximum current is given as:

$I_0$ = $\frac{V_0}{(\omega L-\frac{1}{\omega C})}$ = $\frac{325.27\mathrm{}}{(100\pi \times 80\mathrm{}\times {10}^{-3}-\frac{1}{100\pi \times 60\mathrm{}\times {10}^{-6}})}$ = $-$ 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 $\times \omega L$ = 8.24 $\times 100\pi \times$ 80 $\times {10}^{-3}$ V = 207.09 V

(ii) Potential drop across capacitor, $V_C$ = I $\times \frac{1}{\omega C}$ = 8.24 $\times \frac{1}{100\pi \times 60\mathrm{}\times {10}^{-6}}$ = 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 $\times {10}^{-6}$ 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 ${\omega }^2$ = $\frac{1}{LC}$

$\omega =\sqrt[]{\frac{1}{LC}}$ = $\frac{1}{\sqrt[]{5\times 80\mathrm{}\times {10}^{-6}}}$ = 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,

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

rms current flowing through the Capacitor C is given as,

$I_L$ = $\frac{V}{\frac{1}{\omega C}}$ = 230 $\times 50\times$ 80 $\times {10}^{-6}$ = 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 $\times {10}^{-6}$ F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

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

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

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

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

Peak voltage, $V_0$ = $\surd 2\times V$ = $\surd 2\times 110$ = 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

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

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

$\mathrm{T}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{a}\mathrm{g},t=\frac{\varnothing }{\omega }$ = $\frac{\frac{0.19°\times \pi }{180}}{24\pi \times {10}^3}$ = $\frac{0.19\times \pi }{24\pi \times {10}^3\times 180}$ = 4.4 $\times {10}^{-8}$ 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 $\times {10}^{-6}$ 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 $\pi \times 60$ rad/s

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

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

Peak voltage, $V_0$ = $\surd 2\times V$ = $\surd 2\times 110$ = 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

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

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

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

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

7.14 Inductance of the Inductor, L = 0.50 H

Resistance of the resistor, R = 100 Ω

Potential of supply voltage, V = 240 V

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

Peak voltage is given as $V_0$ = $\sqrt[]{2V}=\sqrt[]{2}\times 240$ = 339.41 V

Angular frequency of the supply, $\omega =2\pi \nu$ = 2 $\pi \times {10}^{4}rad/s$

Maximum current in the supply is given as

$I_0$ = $\frac{V_0}{\sqrt[]{R^2+{\left(\omega L\right)}^2}}$ = $\frac{339.41}{\sqrt[]{{100}^2+{(2\pi \times {10}^4\times 0.50)}^2}}$ = 10.80 $\times {10}^{-3}$ A