Alternating Current Overview

The equation of current in a purely inductive circuit is 5sin(49pt – 30°). If the inductance is 30 mH then the equation for the voltage across the inductor, will be: ${L e t π = \frac{22}{7}}$

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Contributor-Level 10

$i = 5 s i n (49 π t - 30 °)$

$L = 30 m H, π = \frac{22}{7}$

$X_{L} = ω L = 49 π * 30 * 1 0^{- 3}$

$= 49 \frac{22}{7} * 30 * 1 0^{- 3}$

$V_{L} = 23.1 s i n (49 π t + 60 °)$

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3 months ago

The frequencies at which the current amplitude in an LCR series circuit becomes $\frac{1}{\sqrt[]{2}}$ times its maximum value, are 212 rad s^-¹ and 232 rad^-¹. The value of resistance in the circuit is R = $5 Ω .$ The self inductance in the circuit is_________mH.

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Contributor-Level 10

$\frac{_{}}{}$ $i = \frac{l_{0}}{\sqrt[]{2}}$ at $ω_{1}$ $_{}$ = 212 rad/s $R = 5 Ω$

$_{}$ $ω_{2}$ = 232 rad/s $Δ ω = ω_{2} - ω_{1} = 20 r a d / s$ $_{}_{}$

$P = \frac{P_{m a x}}{2} a t ω_{1} a n d ω_{2}$

So, $Δ ω = ω_{2} - ω_{1} = \frac{R}{L}$ $_{}_{} \frac{}{}$

$\Rightarrow L = \frac{R}{Δ ω} = \frac{5}{20} = . 25 H$

= 250 mH

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Statement – I : The reactance of an ac circuit is zero. It is possible that the circuit contains a capacitor and an inductor.

Statement – II : In ac circuit, the average power delivered by the source never becomes zero.

In the light of the above statements, choose the correct answer from the options given below

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Contributor-Level 10

Z = $\sqrt[]{R^{2} + {(x_{C} - x_{L})}^{2}},$ if only L & C are present then R = 0 then p = 0

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As shown in the figure an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is $\frac{\sqrt[]{a}}{π} A .$ The value of a is __________.

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Contributor-Level 10

Let l = l₀ cos wt

Then v = v₀ sinwt

at t = 0, v = 0

but l = l₀

$l_{r m s} = \frac{l_{0}}{\sqrt[]{2}}$

$V_{r m s} = l_{r m s} z$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (X_{L})$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (2 π * 50 * 200 * 1 0^{- 3})$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (20 π)$

$l_{0} = \frac{220 \sqrt[]{2}}{20 π}$

$l_{0} = \frac{11 \sqrt[]{2}}{π} = \frac{\sqrt[]{a}}{π}$

a = 121 * 2

a = 242

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3 months ago

A resistance of 40Ω is connected to a source of alternating current rated 220 V, 50 Hz. Find the time taken by the current to charge from its maximum value to the rms value :

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Alternating Current Overview Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Contributor-Level 10

Let l = l₀ cos wt

l = l₀, at t₁ = 0

& $l = \frac{l_{0}}{\sqrt[]{2}}, a t ω t = \frac{π}{4} \Rightarrow t_{2} = \frac{π}{4 ω}$

$t_{2} = \frac{π}{4 ω} = \frac{π}{4} * \frac{T}{2 π} = \frac{T}{8}$

$t_{2} = \frac{1}{8} * \frac{1}{υ} (T = \frac{1}{υ})$

$t_{2} = \frac{1}{8 * 50}$

^{$t_{2} = \frac{1}{4 * 1 0^{+ 2}}$}

= 0.25 * 10^-2

t₂ = 2.5 ms

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3 months ago

For a series LCR circuit, I vs curve is shown:

Choose the most appropriate answer from the options given below:

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alok kumar singh

Contributor-Level 10

At w_r impedance of the circuit is equal to the resistance of the circuit.

Left of w_r, circuit is mainly capacitive

Right of w_r, the circuit is mainly Inductive.

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5 months ago

7.26 Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?

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Contributor-Level 10

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 $* 15 * 0.5 Ω$ = 15 Ω

rms current in the wire lines is given as

I = $\frac{P}{V_{1}}$ = $\frac{800 * 10^{3}}{40000}$ = 20 A

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

Since the leakage power loss is

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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 $* 10^{3}$ 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 $* 15 * 0.5 Ω$ = 15 Ω

rms current in the wire lines is given as

I = $\frac{P}{V_{1}}$ = $\frac{800 * 10^{3}}{40000}$ = 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.

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7.25 A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.

(a) Estimate the line power loss in the form of heat.

(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?

(c) Characterize the step up transformer at the plant.

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Contributor-Level 10

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 $* 15 * 0.5 Ω$ = 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 = $\frac{P}{V_{1}}$ = $\frac{800 * 10^{3}}{4000}$ = 200 A

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

Since the leakage po

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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 $* 15 * 0.5 Ω$ = 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 = $\frac{P}{V_{1}}$ = $\frac{800 * 10^{3}}{4000}$ = 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

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7.24 At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m³s^–1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms^–2 ).

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Contributor-Level 10

7.24 Height of the water pressure head, h = 300 m

Volume of water flow rate, V = 100 $m^{3}$ /s

Efficiency of turbine generator, $η$ = 60 % = 0.6

Acceleration due to gravity, g = 9.8 m/ $s^{2}$

Density of water, $ρ$ = $10^{3}$ kg/ $m^{3}$

Therefore, electric power available from the plant = $η$ $* h ρ g V$

= 0.6 $* 300 * 10^{3} * 9.8 * 100$

= 176.4 $* 10^{6}$ W

= 176.4 MW

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5 months ago

7.23 A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?

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