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Physics Alternating Current

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Physics Alternating Current 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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Physics Alternating Current Class 12th

A 10 Ω resistance is connected across 220 V - 50Hz AC supply. The time taken by the current to change from its maximum value to the rms value is :

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alok kumar singh

Contributor-Level 10

I = I? cosωt
Current is changing from its maximum value to rms value (I? /√2)
I? /√2 = I? cosωt
cosωt = 1/√2
ωt = π/4
2π * 50t = π/4
t = 1/400 s = 2.5 ms

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Physics Alternating Current Class 12th

A common transistor ratio set requires 12V (D.C) for its operation. The D.C source is constructed by using a transformer and a rectifier circuit, which are operated at 220V (A.C) on standard domestic A.C, supply. The number of turns of secondary coil are 24, then the number of turns of primary are___________.

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alok kumar singh

Contributor-Level 10

$\frac{N_{p}}{N_{s}} = \frac{V_{p}}{V_{s}} \Rightarrow N_{p} = \frac{V_{p}}{V_{s}} * N_{s} = \frac{220}{12} * 24 = 440$

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Physics Alternating Current Class 12th

A resonance circuit having inductance and resistance 2 * 10^-4 H and 6.28 $Ω$ respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is__________.

[p = 3.14]

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alok kumar singh

Contributor-Level 10

Quality factor = $\frac{ω L}{R} = \frac{2 * 3.14 * 10 * 1 0^{6} * 2 * 1 0^{- 4}}{6.28} = 2000$

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Physics Alternating Current Class 12th

In a series LCR resonant circuit, the quality factor is measured as 100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be--------.

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Payal Gupta

Contributor-Level 10

As we know that Q = $\frac{ω L}{R}$

$\Rightarrow \frac{Q'}{Q} = \frac{L'}{R'} * \frac{R}{L} = (\frac{L'}{L}) * (\frac{R}{R'}) = 2 * 2 \Rightarrow Q' = 4 Q = 4 * 100 = 400$

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Physics Alternating Current 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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Physics Alternating Current Class 12th

An alternating is given by the equation i = i₁ $s i n ω t + i_{2} c o s ω t .$ Then rms current will be

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Payal Gupta

Contributor-Level 10

$l = l_{1} s i n ω t + l_{2} c o s ω t = \sqrt[]{l_{1}^{2} + l_{2}^{2}} [(\frac{l_{1}}{\sqrt[]{l_{1}^{2} + l_{2}^{2}}}) s i n ω t + \frac{l_{2}}{\sqrt[]{l_{1}^{2} + l_{2}^{2}}} c o s ω t]$

$= \sqrt[]{l_{1}^{2} + l_{2}^{2}} s i n (ω t + α) = l_{0} s i n (ω t + α$

$l_{r m s} = \frac{l_{0}}{\sqrt[]{2}} = \frac{\sqrt[]{l_{1}^{2} + l_{2}^{2}}}{\sqrt[]{2}}$

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Physics Alternating Current 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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Physics Alternating Current 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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Physics Alternating Current 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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