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a month ago

Alternating Current Class 12th

An electrical power line, having a total resistance of 2Ω, delivers 1 kW at 220 V. The efficiency of the transmission line is approximately:

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alok kumar singh

Contributor-Level 10

Power delivered = 1000 W
Voltage = 220 V
Transmission 'I' = 1000/220
Power loss = I²R
Efficiency = (1000x100)/ (1000 + I²R)

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New answer posted

a month ago

Alternating Current Class 12th

In a series LR circuit, power of 400 W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8 . In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R. Taking the value of C as (n/3π) μF, then value of ' n ' is

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alok kumar singh

Contributor-Level 10

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New answer posted

2 months ago

Alternating Current Class 12th

AC voltage V(t) = 20sin $ω t$ of frequency 50Hz is applied to a parallel plate capacitor. The separation between the plates is 2mm and the area is 1m². The amplitude of the oscillating displacement current for the applied Ac voltage is__________.

[Take $ε_{0} = 8.85 * 1 0^{- 12} F / m]$

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alok kumar singh

Contributor-Level 10

$C = \frac{ε_{0} A}{d} \Rightarrow X_{C} = \frac{1}{C ω} = \frac{d}{ε_{0} A ω} \Rightarrow l_{0} = \frac{V_{0}}{X_{C}} = \frac{V_{0} ε_{0} A ω}{d} = \frac{2 π f V_{0} ε_{0} A}{d}$

$\Rightarrow l_{0} = \frac{2 * 3.14 * 50 * 20 * 8.85 * 1 0^{- 12} * 1}{2 * 1 0^{- 3}} = 27.79 * 1 0^{- 6} A = 27.79 μ A$

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New answer posted

2 months ago

Alternating Current Class 12th

A series LCR circuit driven by 300V at a frequency of 50 Hz contains a resistance R = $3 k Ω,$ an inductor of inductive reactances X_L = 250π $Ω$ and an unknown capacitor. The value of capacitance to maximize the average power should be:

(Take π² = 10)

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Vishal Baghel

Contributor-Level 10

Power is maximum at resonance

So, $X_{L} = X_{C} = \frac{1}{ω C}$ $_{}_{} \frac{}{}$

$\Rightarrow C = \frac{1}{ω * X_{L}}$

$= \frac{1}{250 * 100 * π^{2}}$

$= 4 μ F [π^{2} = 10]$

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New answer posted

2 months ago

Alternating Current Physics

What happens when a voltage is applied across an inductor?

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Vikash Kumar Vishwakarma

Contributor-Level 7

The inductor generates induced electromotive force (EMF) and opposes the changes in the current flow. Due to this the current lag the voltage by 90 degrees.

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New answer posted

2 months ago

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 $\frac{π}{2} .$ 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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Payal Gupta

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 $ω t$

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

When x and y both are connected in series :-

v = 100sin $ω t$

tan = 1 = 45°

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

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

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

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

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

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New answer posted

2 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

The effective current I in the given circuit at very high frequencies will be _________A.

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Vishal Baghel

Contributor-Level 10

At very high frequencies

$X_{C} = \frac{1}{ω C} \to 0$

$X_{L} = ω L \to \infty$

Thus equivalent circuit

$Z = 1 + 2 + 2 = 5 Ω$

$l = \frac{220}{5} = 44 A$

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New answer posted

2 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

In a series LR circuit X_L = R and power factor of the circuit is P₁. When capacitor with capacitance C such that X_L = X_C is put in series, the power factor becomes P₂. The ratio $\frac{P_{1}}{P_{2}}$ is:

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Vishal Baghel

Contributor-Level 10

$P_{1} = c o s ? = \frac{R}{z} = \frac{R}{\sqrt[]{X_{L}^{2} + R^{2}}}$

$P_{1} = c o s ? = \frac{R}{R \sqrt[]{2}} = \frac{1}{\sqrt[]{2}}$

So, $\frac{P_{1}}{P_{2}} = \frac{1}{2}$

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New answer posted

2 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

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

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Payal Gupta

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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