Alternating Current Overview

If wattles current flows in the AC circuit, then the circuit is:

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Purely inductive circuit

$θ = \frac{π}{2}$

$c o s \frac{π}{2} = 0$

Average power = 0

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New question posted

2 months ago

Match List – I with List – II.

(A) AC generator (I) Detects the presence of current in the circuit

(B) Galvanometer (II) Converts mechanical energy into electrical energy

(C) Transformer (III) Works on the principle of resonance in AC circuit

(D) Metal detector (IV) Changes an alte

...more

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New answer posted

3 months ago

In a series LCR circuit, the inductance, capacitance and resistance are L = 100mH, C = $100 μ F$ and R = $10 Ω$ respectively. They are connected to an AC source of voltage 220V and frequency of 50 Hz. The approximate value of current in the circuit will be……….A.

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$ω = 2 π f = 2 π * 50 = 100 π$

$X_{L} = w L = 100 π * 100 * 1 0^{- 3}$

$X_{C} = \frac{1}{w C} = \frac{1}{100 π} * 100 * 1 0^{- 6} = \frac{100}{π}$

$z = 10.0086 Ω$

$l = \frac{v}{z} = \frac{220}{10.008} = 22 A$

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New answer posted

3 months ago

A sinusoidal voltage V(t) = 210 sin3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 $μ F$ and R = $100 Ω .$ The phase difference $(?)$ between the applied voltage and resultant current will be:

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V = 210 sin 3000 t

$ω$ = 3000

$X_{L} = ω L = 30$

$X_{C} = \frac{1}{ω c} = \frac{40}{3}$

$t a n ? = \frac{V_{L} - V_{C}}{V_{R}} = \frac{X_{L} - X_{C}}{R}$

$\begin{array}{l} t a n ? = 0.17 \\ ? = t a n^{- 1} (0.17) \end{array}$

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New answer posted

3 months ago

A 220 V, 50 Hz AC source is connected to a 25 V, 5 W lamp and an additional resistance R in series (as shown in figure)

to run the lamp at its peak brightness, then the value of R (in ohm) will be__________.

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for lamp, $P = \frac{v^{2}}{R}$

$\Rightarrow R = \frac{25 * 25}{5} = 125 Ω$

$l_{m a x} = \frac{v}{R} = \frac{25}{125} = \frac{1}{5} A$

$\Rightarrow I_{m a x} = \frac{220}{125 + R} = \frac{1}{5}$

$\Rightarrow 1100 = 125 + R$

R = 975 $Ω$

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3 months ago

The current flowing through and ac circuit is given by I = 5sin(12πt) A

How long will the current take to reach the peak value staring from zero?

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Given, I = 5 sin (120πt), $ω$ = 120

$T = \frac{2 π}{ω}$

$T = \frac{2 π}{120 π} = \frac{1}{60} s e c, \frac{T}{4} = \frac{1}{60 * 4} = \frac{1}{240} s e c$

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New answer posted

3 months ago

The rms value of conduction current in a parallel plate capacitor is 6.9 $μ A$ . The capacity of this capacitor, if it is connected to 203 V ac supply with an angular frequency of 600 rad/s, will be:

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Current in capacitor l = $\frac{V}{X_{C}}$ $\frac{}{_{}}$

I = v * $(ω c)$

$C = \frac{1}{v ω} = \frac{6.9 * 1 0^{- 6}}{230 * 600} = 50 p F$

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New answer posted

3 months ago

To increase the resonant frequency in series LCR circuit,

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$f = \frac{1}{2 π \sqrt[]{L C}}$

$f α \frac{1}{\sqrt[]{C}}$

Another capacitor should be added in series with the first capacitor.

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New answer posted

3 months ago

An emf of $20 V$ is applied at time $t = 0$ to a circuit containing in series $10 m H$ inductor and $5 Ω$ resistor. The ratio of the currents at time $t = \infty$ and at $t = 40 s$ is close to: (take $e^{2} = 7.389)$

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$\begin{array}{r} i & = \frac{V}{r} (1 - e^{- R t / L}) \\ = 4 (1 - e^{- 500 t}) \end{array}$

At $t = \infty$

$i_{1} = 4 A$

at $t = 40 s$

$i_{2} = 4 (1 - e^{- 20000})$

= 4 [1 - \frac{1}{{(e^{2})}^{10000}}]

= 4 [1 - \frac{1}{(7.389)^{10000}}]

\frac{i_{1}}{i_{2}}

is slightly greater than 1 .

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New answer posted

3 months ago

A circuit element X when connected to an a.c. supply of peak voltage 100V given a peak current of 5A which is in phase with the voltage. A second element Y when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

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