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Alternating Current Overview

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10 months ago

Alternating Current Overview Class 12th

A 750 Hz, 20 V (rms) source is connected to a resistance of 100Ω, an inductance of 0.1803H and a capacitance of 10µF all in series. The time in which the resistance (heat capacity 2 J/°C) will get heated by 10°C. (assume no loss of heat to the surroundings) is close to:

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alok kumar singh

Contributor-Level 10

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10 months ago

Alternating Current Overview Class 12th

For the given input voltage waveform Vin(t), the output voltage waveform Vo(t), across the capacitor is correctly depicted by:

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Vishal Baghel

Contributor-Level 10

τ = RC = 10µS
For 0

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10 months ago

Alternating Current Overview Class 12th

An AC circuit has R = 100Ω, C = 2µF and L = 80mH, connected in series. The quality factor of the circuit is:

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Vishal Baghel

Contributor-Level 10

Q = (1/R)√ (L/C) = (1/100)√ (80e-3/2e-6) = 2

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10 months ago

Alternating Current Overview Class 12th

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant ' $b$ ', the connect equivalence would be:

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Vishal Baghel

Contributor-Level 10

In damped oscillation

$∴ v_{2} = \frac{1}{8} * \frac{1}{T} = \frac{1}{8} * \frac{10^{16}}{1.6} = 7.8 * 10^{14}$

$m a + b v + k x = 0$

In the circuit

$m \frac{d^{2} x}{{d t^{2}}^{2}} + b \frac{d x}{d t} + k x = 0$

$- i R - L \frac{d i}{d t} - \frac{q}{C} = 0$

Comparing equation (i) and (ii)

$L \frac{d^{2} q}{d t^{2}} + R \frac{d q}{d t} + \frac{1}{c} q = 0$

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10 months ago

Alternating Current Overview Class 12th

A 100Ω resistance, a 0.1 µF capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz.

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Vishal Baghel

Contributor-Level 10

ω = 2πf = 1/√ (LC)
L = 1/ (4π²f²C) = 1/ (4π² * 60² * 0.1*10? ) = 70.3 mH

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10 months ago

Alternating Current Overview Class 12th

A 0.07 H inductor and a 12Ω resistor are connected in series to a 220 V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take π as 22/7]

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Vishal Baghel

Contributor-Level 10

L = 0.07H and R = 12Ω
Vs = 220V, 50Hz
XL = 2πfL = 2 (22/7) (50) (0.07) = 22Ω
Z = √ (R² + XL²) = √ (12² + 22²) = √ (144+484) = √628 ≈ 25Ω
i = V/Z = 220/25 = 8.8A
tan φ = XL/R = 22/12 = 11/6
φ = tan? ¹ (11/6)

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11 months ago

Alternating Current Overview Class 12th

Find the peak current and resonant frequency of the following circuit (as shown in figure).

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Vishal Baghel

Contributor-Level 10

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

$Z = \sqrt[]{{(120)}^{2} + {(10 - 100)}^{2}} = 150 Ω$

$ω = \frac{1}{\sqrt[]{L C}} = \frac{1}{\sqrt[]{1 0^{- 1} * 1 0^{- 4}}} = \sqrt[]{1 0^{5}}$

$? ω = 2 π f$

$\Rightarrow f = \frac{1 0^{3}}{2 π \sqrt[]{10}} = 50 H z$

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11 months ago

Alternating Current Overview Class 12th

A series LCR circuit is designed to resonate at an angular frequency $ω_{0} = 1 0^{5} r a d / s .$ The circuit draws 16W power from 120 V source at resonance. The value of resistance 'R' in the circuit is…………………………. $Ω .$

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Vishal Baghel

Contributor-Level 10

$ω_{0} = 1 0^{5} r a d / s e c$

P = 16w, 120v at resonance

$P = \frac{v^{2}}{R} \Rightarrow 16 = \frac{{(120)}^{2}}{R} \Rightarrow R = \frac{14400}{16} = 900 Ω$

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11 months ago

Alternating Current Overview Class 12th

A transmitting station releases waves of wavelength 960m. A capacitor of 2.56 $μ F$ is used in the resonant circuit. The self inductance of coil necessary for resonance is……………. * 10^-8 H.

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Vishal Baghel

Contributor-Level 10

2 π f $\frac{1}{\sqrt[]{L C}} \Rightarrow L = \frac{λ^{2}}{4 π^{2} C c^{2}} = \frac{96 0^{2}}{4 * 3.1 4^{2} * 2.56 * 1 0^{- 6} * 9 * 1 0^{16}}$ =10^-7 = 10 * 10^-8 H

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11 months ago

Alternating Current Overview Class 12th

The current (i) at time t = 0 and t = $\infty$ respectively for the given circuit is:

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Vishal Baghel

Contributor-Level 10

At t = 0, no current flows through inductor.

$i = \frac{E}{\frac{6 * 9}{6 + 9}} = \frac{5 E}{18}$

At t = $\infty$ , current flows through circuit as if inductor is shorted.

$i = \frac{E}{\frac{5}{2} + \frac{4}{5}} = \frac{10 E}{33}$

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