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Alternating Current

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6 days ago

Alternating Current Class 12th

A coil has resistance of 30 ohm and inductive reactance 20 ohm at 50 Hz frequency. If an ac source of 200 V, 100 Hz is connected across the coil, current in coil will be

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alok kumar singh

Contributor-Level 10

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New answer posted

6 days ago

Alternating Current Class 12th

Two bodies of mass 10 kg and 2 kg are moving with velocities m/s and m/s respectively. The velocity of their centre of mass is

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alok kumar singh

Contributor-Level 10

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New answer posted

a week ago

Alternating Current Class 12th

A capacitor having capacitance of 100 µF is charged with a potential difference of 12 V is connected to an inductor of inductance 10 mH. Find the maximum current through the inductor.

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alok kumar singh

Contributor-Level 10

$I = Q_{0 . ω}$

$= \frac{C V}{\sqrt[]{L C}} = V \sqrt[]{\frac{C}{L}}$

$= 12 \sqrt[]{\frac{100 * 1 0^{- 6}}{10 * 1 0^{- 3}}}$

= 1.2 A

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New answer posted

a week ago

Alternating Current Class 12th

In AC circuit with source voltage e = 20 sin 1000 t is connected to series L – R circuit whose power factor is $\frac{1}{\sqrt[]{2}}$ . If E = 25 sin 2000 t, the new power factor is

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alok kumar singh

Contributor-Level 10

Lω = 1000 L Lω = R.

New

R = 1000 L

2000 L

$c o s θ = \frac{R}{Z}$

$= \frac{1000 L}{\sqrt[]{{(1000 L)}^{2} + {(2000 ? L)}^{2}}}$

$= \frac{1}{\sqrt[]{1 + 4}} = \frac{1}{\sqrt[]{5}}$

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New answer posted

a week ago

Alternating Current Class 12th

Find average power in electric circuit if source voltage (V) = 20sin(100wt) and current in the circuit (I) = 2sin(100wt + $\frac{π}{3}$ )

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alok kumar singh

Contributor-Level 10

= IV cosf

$= \frac{20}{\sqrt[]{2}} * \frac{2}{\sqrt[]{2}} *$ cos 60°

= 10 W

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New answer posted

2 weeks ago

Alternating Current Class 12th

In a series $C - R$ circuit shown in figure, the applied voltage is $10 V$ and the voltage across $R$ the capacitor is found to be . The voltage across and the phase difference between current and the applied voltage will be respectively.

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alok kumar singh

Contributor-Level 10

$A s V^{2} = V_{R}^{2} + C_{C}^{2}$

$? V_{R} = \sqrt[]{V^{2} - V_{C}^{2}} = \sqrt[]{10^{2} - 8^{2}} = 6 V$

$\tan ? θ = \frac{X_{C}}{X_{R}} = \frac{V_{C}}{V_{R}} = \frac{8}{6} = \frac{4}{3}$

Now,

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New answer posted

2 weeks ago

Alternating Current Class 12th

In a series $C - R$ circuit shown in figure, the applied voltage is $10 V$ and the voltage across the capacitor is found to be . The voltage across $R$ and the phase difference between current and the applied voltage will be respectively.

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alok kumar singh

Contributor-Level 10

$A s V^{2} = V_{R}^{2} + C_{C}^{2}$

$? V_{R} = \sqrt[]{V^{2} - V_{C}^{2}} = \sqrt[]{10^{2} - 8^{2}} = 6 V$

$\tan ? θ = \frac{X_{C}}{X_{R}} = \frac{V_{C}}{V_{R}} = \frac{8}{6} = \frac{4}{3}$

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New answer posted

3 weeks ago

Alternating Current Class 12th

An inductor of 0.5 mH, a capacitor of 200 $μ F$ and a resistor of 2 $Ω$ are connected in series with a 220V ac source. If the current is in phase with the emf, the frequency of ac source will be__________ * 10² Hz.

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alok kumar singh

Contributor-Level 10

Since current is in phase with voltage, it means circuit is in resonance, so we can write

f = $\frac{1}{2 π \sqrt[]{L C}} = \frac{1}{2 π \sqrt[]{(0.5 * 1 0^{- 3}) * (200 * 1 0^{- 6})}}$

$\Rightarrow f = \frac{1 0^{4}}{2 π \sqrt[]{10}} \approx 5 * 1 0^{2} H z$ $[T a k i n g π = \sqrt[]{10}]$

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New answer posted

3 weeks ago

Alternating Current Class 12th

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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alok kumar singh

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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New answer posted

3 weeks ago

Alternating Current Class 12th

At t < 0, the capacitor is charged and the switch is opened. At t = 0 the switch is closed. The shortest time T at which the charge on the capacitor will be zero is given by:

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alok kumar singh

Contributor-Level 10

The time period of LC oscillations,

$T = 2 π \sqrt[]{L C}$

The time at which charge on the capacitor will be zero is $\frac{T}{4} .$

So t $= \frac{π}{2} \sqrt[]{L C}$

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