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2 months ago

Physics Alternating Current Class 12th

A direct current of 4 A and an alternating current of peak value 4 A flow through resistance of $3 Ω a n d 2 Ω$ respectively. The ratio of heat produced in the two resistances in same interval of time will be:

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alok kumar singh

Contributor-Level 10

In DC :- In AC:-

i₁ = 4A i₂ = 4 sin $ω t$

R₁ = 3 $Ω$ R₂ = $2 Ω$

In same time internal of 't' - $\frac{H_{1}}{H_{2}} = \frac{i_{1}^{2} R_{1} t}{{(l_{2^{- r m s}})}^{2} R_{2} t} = \frac{16 * 3 t}{{(\frac{4}{\sqrt[]{2}})}^{2} * 2 * t} = \frac{48}{(\frac{32}{2})} = 3 : 1$

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2 months ago

Physics Alternating Current Class 12th

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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alok kumar singh

Contributor-Level 10

v = 100 sin $ω t$

$i_{0} = \frac{100}{R_{x}} = 5 \Rightarrow R_{x} = 20 Ω$

i = $5 s i n (ω t - \frac{π}{2})$

i = 100 5sin $ω t$ &nb

...more

v = 100 sin $ω t$

$i_{0} = \frac{100}{R_{x}} = 5 \Rightarrow R_{x} = 20 Ω$

i = $5 s i n (ω t - \frac{π}{2})$

i = 100 5sin $ω t$

$5 = \frac{100}{R_{y}} \Rightarrow R_{y} = \frac{100}{5} = 20 Ω$

When x and y both are connected in series :-

v = 100sin $ω t$

tanθ = 1 = q = 45°

R₀ = $\sqrt[]{R_{x}^{2} + R_{y}^{2}} = 20 \sqrt[]{2} Ω$

$∴ \overset{o}{l_{0}} = \frac{v_{0}}{R_{0}} = \frac{100}{20 \sqrt[]{2}} = \frac{5}{\sqrt[]{2}} A .$

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

$= \frac{5}{\sqrt[]{2} \cdot \sqrt[]{2}}$

$= \frac{5}{2} A = \frac{5}{2} A$

less

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2 months ago

Physics Alternating Current Class 12th

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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Vishal Baghel

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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2 months ago

Physics Alternating Current Class 12th

To light, a 50 W, 100 V lamp is connected, in series with a capacitor of a capacitance $\frac{50}{π \sqrt[]{x}} μ F,$ with 200V, 50Hz AC source. The value of x will be________.

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Payal Gupta

Contributor-Level 10

$R = \frac{v_{2}}{P} = \frac{10000}{50} \Rightarrow R = 200 Ω$

$V = \sqrt[]{V_{C}^{2} + V_{R}^{2}}$

$\Rightarrow 200 = \sqrt[]{V_{C}^{2} + {(100)}^{2}}$

$x_{c} = \frac{1}{ω_{c}} = \frac{π \sqrt[]{x}}{100 π * 50} = \frac{\sqrt[]{x}}{5000 * 1 0^{- 6}} = \frac{\sqrt[]{x} * 1 0^{3}}{5}$

$= 200 \sqrt[]{x}$

$V_{C} = i x_{C}$

$\Rightarrow \sqrt[]{3} . \sqrt[]{1 + x} = 2 \sqrt[]{x}$

$\Rightarrow 3 (1 + x) = 4 x \Rightarrow x = 3$

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2 months ago

Physics Alternating Current Class 12th

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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Vishal Baghel

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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2 months ago

Physics Alternating Current Class 12th

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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Payal Gupta

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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2 months ago

Physics 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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Vishal Baghel

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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2 months ago

Physics Alternating Current Class 12th

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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Vishal Baghel

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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New answer posted

2 months ago

Physics 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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Payal Gupta

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

2 months ago

Physics Alternating Current Current Electricity Analysis

A zener of breakdown voltage V_Z = 8V and maximum zener current, I_ZM = 10mA is subjected to an input voltage V_i = 10V with series resistance R = 100 $Ω$ In the given circuit R_L represents the variable load resistance. The ratio maximum and minimum value of R_L is__________.

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alok kumar singh

Contributor-Level 10

Across zener diode & RL

8 = I_LR_L

8 = $(20 - 10) R_{L (m a x)}$

$R_{L m a x} = \frac{8}{10} k Ω$

At max current in loop (1)

10 = i_max R + 8

i_max= $\frac{2}{R} = \frac{2}{100} = 0.02 A$

= 20 mA

R_L (max) = $\frac{8}{10} k Ω$

At minimum curre3nt through zener

i_max R_L (minimum) = 8

R_L (minimum) = $\frac{8}{20} k Ω$

$\frac{R_{L m a x}}{R_{L m i n}} = 2$

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