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2 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A telegraph line of length 100km has a capacity of $0.01 μ F / k m$ and it carries an alternating current at 0.5 kilo cycle per second. If minimum impedance is required, then the value of the inductance that need to be introduced in series is _________mH. (if π = $\sqrt[]{10}$ )

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Payal Gupta

Contributor-Level 10

C = 10^-6 f

f = 0.5 * 10³ = 500

$z = \sqrt[]{x_{R}^{2} + {(x_{L} - x_{C})}^{2}}$

for minimum impedance :

x_L = x_C

$\Rightarrow ω^{2} = \frac{1}{L C}$

$\Rightarrow L = \frac{1}{π^{2}} = \frac{1}{10} * 1000 = 100 m H$

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2 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

An AC source is connected to an inductance of 100 mH, a capacitance of 100 $μ F$ and a resistance of 120 $Ω$ as shown in figure. The time in which the resistance having a thermal capacity 2 J/°C will get heated by 16°C is________ s.

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Payal Gupta

Contributor-Level 10

$x_{R} = 120 ?$

$x_{L} = ? L = 100 * 100 * 1 0^{? 3} 10 ?$

$x_{C} = \frac{1}{? c} = \frac{1}{100 * 100 * 1 0^{? 6}} = 100 ?$

$z = \sqrt[]{12 0^{2} + {(100 ? 10)}^{2}} = \sqrt[]{12 0^{2} + 9 0^{2}}$

= 150 $?$

$? 32 = \frac{160}{75} * t ? t = \frac{32 * 75}{160} = 15$

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3 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

If the potential barrier across a p-n junction is 0.6 V. Then the electric field intensity, in the depletion region having the width of 6 * 10^-⁶m, will be________ * 10⁵ N/C.

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Vishal Baghel

Contributor-Level 10

As, Ed = VB

$∴ E = \frac{V_{B}}{d} = \frac{0.6}{6 * 1 0^{- 6}} = 1 * 1 0^{5} N / C$

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3 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

An alternating emf E = 440 sin 100t is applied to a circuit containing an inductance of $\frac{\sqrt[]{2}}{π} H .$ If an a.c. ammeter is connected in the circuit, its reading will be:

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Vishal Baghel

Contributor-Level 10

E = 440 sin 100 t $ω = 100 π$

$L = \frac{\sqrt[]{2}}{π} H .$ $X_{L} = ω L = 100 π . \frac{\sqrt[]{2}}{π} = 100 \sqrt[]{2} Ω .$

$= \frac{220}{100} = 2.2 A$

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New answer posted

3 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

If L, C and R are the self inductance, capacitance and resistance respectively, which of the following does not have the dimension of time?

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Payal Gupta

Contributor-Level 10

Dimension

RC → [T]

$\frac{L}{R} \to [T]$

$\sqrt[]{L c} \to [T]$

$\frac{L}{C} \to$ dimensionless

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4 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

If an L-C circuit is considered analogous to a harmonically oscillating spring- block system, which energy of the L-C circuit would be analogous to potential energy and which one analogous to kinetic energy?

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, d) For house hold supplies, AC currents are used which are having zero average value over a cycle.

The line is having some resistance so power factor cos φ = R/Z≠0

so, φ not equal to π /2 ⇒ φ < /2

i.e., phase lies between 0 and π /2.

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4 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

When an AC voltage of 220 V is applied to the capacitor C

(a) The maximum voltage between plates is 220 V

(b) The current is in phase with the applied voltage

(c) The charge on the plates is in phase with the applied voltage

(d) The power delivered to the capacitor is zero

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c, d) When the AC voltage is applied to the capacitor, the plate connected to the positive terminal will be at higher potential and the plate connected to the negative terminal will be at lower potential.

The plate with positive charge will be at higher potential and the plate with negative charge will be at lower potential. So, we can say that the charge is in phase with the applied voltage.

P= E_rmsI_rmscos $\emptyset$

$\emptyset$ =90

So power, P = 0

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New answer posted

4 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

For an L-C-R circuit, the power transferred from the driving source to the driven oscillator is P = I²Z cos ?.

(a) Here, the power factor cos ? > 0, P > 0

(b) The driving force can give no energy to the oscillator (P = 0) in some cases

(c) The driving force cannot syphon out (P < 0) the energy out of oscillator

(d) The driving force can take away energy out of the oscillator

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

According to the question power transferred Is P = I² Z cos?

as we know cos? = R/Z

R>0 and Z>0

cos? >0 so P>0

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4 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Electrical energy is transmitted over large distances at high alternating voltages. Which of the following statements is (are) correct?

(a) For a given power level, there is a lower current

(b) Lower current implies less power loss

(c) Transmission lines can be made thinner

(d) It is easy to reduce the voltage at the receiving end using step-down transformers

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

We have to transmit energy (power) over large distances at high alternating voltages, so current flowing through the wires will below because for a given power (P).

P= E_rmsI_rms, I_rms is low when E_rms is high.

Power loss= I ²_rms R= low

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4 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?

(a) Only resistor

(b) Resistor and an inductor

(c) Resistor and a capacitor

(d) Only a capacitor

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c, d) According to the question, the current increases on increasing the frequency of supply. Hence, the reactance of the circuit must be decreases as increasing frequency.

For a capacitive circuit, capacitive reactance Xc= $\frac{1}{2 π ϑ C}$

For an CR circuit Z= $\sqrt[]{R^{2} + (X_{c}^{2})}$ when frequency increases X decreases.

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