Alternating Current Overview

7.26 The rating of step-down transformer = 40000 V – 220 V

Hence, the input voltage,  $V_1$ = 40000 V

Output voltage,  $V_2$ = 220 V

Total electric power required, P = 800 kW = 800 $*{10}^3$ W

Source potential, V = 220 V

Voltage at which electric plant generates power, V' = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wires lines carrying power = 0.5 Ω /km

Total resistance of the wire, R = 2 $*15*0.5\Omega$ = 15 Ω

rms current in the wire lines is given as

I = $\frac{P}{V_1}$ = $\frac{800\mathrm{}*{10}^3}{40000}$ = 20 A

Line power loss = $I^{2}R$ = ${20}^2*$ 15 = 6 $*{10}^3$ W = 6 kW

Since the leakage power loss is negligible, the total power to be supplied by plant = 800 + 6 = 806 kW

Voltage drop in the transmission line = IR = 20 $*15$ = 300 V. Hence, the total voltage transmitted from plant = 300 + 40000 = 40300 V

Since the power generated is 440V, the transformer rating at the plant should be 440 V – 40300 V

Hence, power loss during transmission = $\frac{6}{806}$ $*100$ = 0.744 %

In the previous exercise, power loss during transmission = $\frac{600}{1400}$ $*100$ = 42.86 %

Since the power loss is less for a high voltage transmission, high voltage transmission is preferred.

7.25 Total electric power required, P = 800 kW = 800 $*{10}^3$ W

Supply voltage, V = 220 V

Electric plant generating voltage, V' = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wires lines carrying power = 0.5 Ω /km

Total resistance of the wire, R = 2 $*15*0.5\Omega$ = 15 Ω

Step-down transformer rating 4000 – 220 V, hence

Input voltage to the transformer,  $V_1$ = 4000 V

Output voltage from the transformer,  $V_2$ = 220 V

rms current in the wire lines is given as

I = $\frac{P}{V_1}$ = $\frac{800\mathrm{}*{10}^3}{4000}$ = 200 A

Line power loss = $I^{2}R$ = ${200}^2*$ 15 = 600 $*{10}^3$ W = 600 kW

Since the leakage power loss is negligible, the total power to be supplied by plat = 800 + 600 = 1400 kW

Voltage drop in the transmission line = IR = 200 $*15$ = 3000 V. Hence, the total voltage transmitted from plant = 3000 + 4000 = 7000 V

Since the power generated is 440V, the transformer rating at the plant should be 440 V – 7000 V

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 = $P_R+P_C+P_L=789.97+0+0=789.97W$