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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 $\times {10}^3$ 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 $\times 15\times 0.5\Omega$ = 15 Ω

rms current in the wire lines is given as

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

Line power loss = $I^{2}R$ = ${20}^2\times$ 15 = 6 $\times {10}^3$ 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 $\times 15$ = 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 = $\frac{6}{806}$ $\times 100$ = 0.744 %

In the previous exercise, power loss during transmission = $\frac{600}{1400}$ $\times 100$ = 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 $\times {10}^3$ 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 $\times 15\times 0.5\Omega$ = 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 = $\frac{P}{V_1}$ = $\frac{800\mathrm{}\times {10}^3}{4000}$ = 200 A

Line power loss = $I^{2}R$ = ${200}^2\times$ 15 = 600 $\times {10}^3$ 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 $\times 15$ = 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

7.22 (a) It is true that in any AC circuit, the applied voltage is equal to the average sum of instantaneous voltages across the series elements of the circuit. However, this is not true for rms voltages because voltage across different elements may not be in phase.

(b) A capacitor is used in the primary circuit of an induction coil. This is because when the circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparks.

(c) The dc signal will appear across capacitor C because for dc signals, the impedance of an inductor (L) is negligible while the impedance of a capacitor © is very high (almost infinite). Hence, a dc signal appears across C

For an AC signal of high frequency, the impedance of L is high and that of C is very low. Hence, an AC signal of high frequency appears across L.

(d) If an iron core is inserted in the choke coil (which is in series with a lamp connected to AC line), then the lamp will glow dimly. This because the choke coil and the iron core increase the impedance of the circuit.

(e) A choke coil is needed in the use of fluorescent tubes with AC Mains because it reduces the voltage across the tube without wasting much power. An ordinary resistor cannot be used instead of choke coil for this purpose because it wastes power in the form of heat.

7.20 Inductance, L = 0.12 H

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

Resistance, R = 23 Ω

Supply voltage, V = 230 V

Peak voltage is given as, $V_0$ = $\sqrt[]{2}\times 230$ = 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, ${\omega }_R$ = $\frac{1}{\sqrt[]{LC}}$ = $\frac{1}{\sqrt[]{0.12\times 480\mathrm{}\times {10}^{-9}}}$ = 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 = $\frac{1}{2}\times {14.14\mathrm{}}^2\times 23$ = 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 $\omega =\frac{R}{2L}$ = $\frac{23}{2\times 0.12}$ = 95.83 rad/s

$\Delta \nu$ = $\frac{\mathrm{\Delta }\omega }{2\pi }$ = $\frac{95.83}{2\times \pi }$ = 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’ = $\frac{1}{\surd 2}\times {I_0}_{max}$ = $\frac{14.14\mathrm{}}{\surd 2}$ = 10 A

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

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