Tags Alternating Current

Alternating Current

Get insights from 94 questions on Alternating Current, answered by students, alumni, and experts. You may also ask and answer any question you like about Alternating Current

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

7 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Study the circuits (a) and (b) shown in figure and answer the following questions.

(a) Under which conditions would the rms currents in the two circuits be the same?

(b) Can the rms current in circuit (b) be larger than that in (a)?

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

(I_rms)a= $\frac{V_{r m s}}{R}$ = $\frac{V}{R}$

(I_rms)b= $\frac{V_{r m s}}{Z}$ = $\frac{V}{\sqrt[]{R^{2} + (X_{L} - X_{C}) 2}}$

When $(I r m s) a$ = (I_rms)b

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

XL=Xc resonance condition

As Z>R $\frac{(I r m s) a}{(I r m s) b}$ = $\frac{\sqrt[]{R^{2} + (X_{L} - X_{C}) 2}}{R}$ = $\frac{Z}{R}$ >1

(I_rms)a > (I_rms)b

0 0

New answer posted

7 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Draw the effective equivalent circuit of the circuit shown in figure, at very high frequencies and find the effective impedance.

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

Inductive reactance XL= 2 $π ϑ$ L

And capacitive reactance Xc= $\frac{1}{2 π ϑ C}$

For very high frequencies XL= infinity and Xc=0

When reactance of the circuit is infinite it will be considered as open circuit . when reactance of a circuit is zero it will be considered as short circuited.

So C1, C2 = shorted and L1, L2=opened

So R_eff= R₁+R₂

0 0

New answer posted

7 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Draw the effective equivalent circuit of the circuit shown in figure, at very high frequencies and find the effective impedance

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

If we consider a L-C circuit analogous to a harmonically oscillating spring block system. The electrostatic energy $\frac{1}{2}$ CV2 is analogous to potential energy and energy associated with moving charges (current) that is magnetic energy $\frac{1}{2}$ LI2 is analogous to kinetic energy.

0 0

New answer posted

7 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

In the LCR circuit shown in figure the AC driving voltage isV=V_msinwt

Write the equation of motion for q(t) describe subsequent motion of charges.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

0 0

New answer posted

7 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

For a LCR circuit at frequency w the equation reads

Multiply the equation by i and simplify where is possible

0 Follower 1 View

Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

0 0

New answer posted

7 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Consider the LCR circuit shown in figure find the net current I and the phase of i. show that i=V/Z. find the impedance Z for this circuit.

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Total current i=i1+i2

Vmsinwt=Ri1

I1= $\frac{V_{m} s i n w t}{R}$

If q2 is charge on the capacitor at any time t then for series combination of C and L

Applying KVL in below circuit

So q2=qmsin(wt $\frac{q_{2}}{C} + \frac{L d i_{2}}{d t} - V_{m} s i n w t = 0$ )……….1

qm[ $+ \emptyset$ sin(wt+ $\frac{q_{2}}{C} + \frac{L d_{2} q_{2}}{d t^{2}} = V_{m} s i n w t$ )]= Vmsinwt

if $\frac{d q_{2}}{d t} = - q_{m} w^{2} s i n ? (w t + \emptyset)$

qm= $u s e t h i s v a l u e i n e q n 1$

from above equation i2= $\frac{1}{C} + L (- w^{2})$

i2= $\emptyset$ when $\emptyset = 0$

so total current I = i1+i2

I= $\frac{V_{m}}{(\frac{1}{c} - L w^{2})}$

I= i1+i2= Ccos $\frac{d q_{2}}{d t} = w q_{m} \cos ? (w t + \emptyset)$ +Csin $\frac{w V_{m} c o s ? (w t + \emptyset)}{\frac{1}{c} - L w^{2}}$

I= Csin(wt+ $\emptyset = 0, i_{2} = \frac{V_{m} c o s w t}{\frac{1}{w C} - L w}$ )

C= $\frac{V_{m} s i n w t}{R} + \frac{V_{m} c o s w t}{\frac{1}{w C} - L w}$

$\emptyset s i n w t$ 1/2

$\emptyset c o s w t$ = tan-1 $\emptyset$

$\sqrt[]{A^{2} + B^{2}}$ 1/2

This is the expression for impedance.2

0 0

New answer posted

7 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if

(i) Power is transmitted at 220V. Comment on the feasibility of doing this.

(ii) A step-up transformer is used to boost the voltage to 11000V, power transmitted, then a step-down transformer is used to bring voltage to 220 V. (ρ_cu= 1.7*10^-8 SI unit)

0 Follower 1 View

Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

0 0

New answer posted

7 months ago

Alternating Current Physics NCERT Exemplar Solutions Class 12th Chapter Seven

An electrical device draws 2KW power from AC mains (voltage 223V(rms)=) $\sqrt[]{50000} V$ . The current differs (lags) in phase by $\emptyset$ (tan $\emptyset = - 3 / 4$ ) as compared to voltage . find

(a) R

(b) X_L-X_C(c) I_m. Another device has twice the values for R,X_candX_L. how the answer affected?

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Power drawn, p= 2KW= 2000W

tan $\emptyset$ =-3/4

P=V2/Z

Z=V2/P= $\frac{223 * 223}{2 * 10^{3}}$ =25

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

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

625 = $R^{2} + ({X_{L} - X_{C})}^{2}$ ………… (1)

tan $\emptyset$ = $\frac{X_{L} - X_{C}}{R}$ = ¾

XL-XC= 3R/4

Use this in eqn 1

625= R2+ (3R/4)2 = R2+9R2/16

625 = 25R2/16, R= 20ohm

XL-XC=15ohm

Im= $\sqrt[]{2}$ I= $\sqrt[]{2} V / Z$ = $\frac{223 * \sqrt[]{2}}{25}$ = 12.6A

If R, XL and Xc are all doubled, tan $\emptyset$ does not change . Z is doubled, current is halved. So power is also halved.

0 0

New answer posted

7 months ago

Alternating Current Alternating Current Overview

7.26 Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

7.26 The rating of step-down transformer = 40000 V – 220 V

Hence, the input voltage, $V_{1}$ = 40000 V

Output voltage, $V_{2}$ = 220 V

Total electric power required, P = 800 kW = 800 $* 10^{3}$ W

Source potential, V = 220 V

Voltage at which electric plant generates power, V' = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wires lines carrying power = 0.5 Ω /km

Total resistance of the wire, R = 2 $* 15 * 0.5 Ω$ = 15 Ω

rms current in the wire lines is given as

I = $\frac{P}{V_{1}}$ = $\frac{800 * 10^{3}}{40000}$ = 20 A

Line power loss = $I^{2} R$ = $20^{2} *$ 15 = 6 $* 10^{3}$ W = 6 kW

Since the leakage power loss is

...more

7.26 The rating of step-down transformer = 40000 V – 220 V

Hence, the input voltage, $V_{1}$ = 40000 V

Output voltage, $V_{2}$ = 220 V

Total electric power required, P = 800 kW = 800 $* 10^{3}$ W

Source potential, V = 220 V

Voltage at which electric plant generates power, V' = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wires lines carrying power = 0.5 Ω /km

Total resistance of the wire, R = 2 $* 15 * 0.5 Ω$ = 15 Ω

rms current in the wire lines is given as

I = $\frac{P}{V_{1}}$ = $\frac{800 * 10^{3}}{40000}$ = 20 A

Line power loss = $I^{2} R$ = $20^{2} *$ 15 = 6 $* 10^{3}$ W = 6 kW

Since the leakage power loss is negligible, the total power to be supplied by plant = 800 + 6 = 806 kW

Voltage drop in the transmission line = IR = 20 $* 15$ = 300 V. Hence, the total voltage transmitted from plant = 300 + 40000 = 40300 V

Since the power generated is 440V, the transformer rating at the plant should be 440 V – 40300 V

Hence, power loss during transmission = $\frac{6}{806}$ $* 100$ = 0.744 %

In the previous exercise, power loss during transmission = $\frac{600}{1400}$ $* 100$ = 42.86 %

Since the power loss is less for a high voltage transmission, high voltage transmission is preferred.

less

0 0

New answer posted

7 months ago

Alternating Current Alternating Current Overview

7.25 A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.

(a) Estimate the line power loss in the form of heat.

(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?

(c) Characterize the step up transformer at the plant.

0 Follower 12 Views

Payal Gupta

Contributor-Level 10

7.25 Total electric power required, P = 800 kW = 800 $* 10^{3}$ W

Supply voltage, V = 220 V

Electric plant generating voltage, V' = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wires lines carrying power = 0.5 Ω /km

Total resistance of the wire, R = 2 $* 15 * 0.5 Ω$ = 15 Ω

Step-down transformer rating 4000 – 220 V, hence

Input voltage to the transformer, $V_{1}$ = 4000 V

Output voltage from the transformer, $V_{2}$ = 220 V

rms current in the wire lines is given as

I = $\frac{P}{V_{1}}$ = $\frac{800 * 10^{3}}{4000}$ = 200 A

Line power loss = $I^{2} R$ = $200^{2} *$ 15 = 600 $* 10^{3}$ W = 600 kW

Since the leakage po

...more

7.25 Total electric power required, P = 800 kW = 800 $* 10^{3}$ W

Supply voltage, V = 220 V

Electric plant generating voltage, V' = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wires lines carrying power = 0.5 Ω /km

Total resistance of the wire, R = 2 $* 15 * 0.5 Ω$ = 15 Ω

Step-down transformer rating 4000 – 220 V, hence

Input voltage to the transformer, $V_{1}$ = 4000 V

Output voltage from the transformer, $V_{2}$ = 220 V

rms current in the wire lines is given as

I = $\frac{P}{V_{1}}$ = $\frac{800 * 10^{3}}{4000}$ = 200 A

Line power loss = $I^{2} R$ = $200^{2} *$ 15 = 600 $* 10^{3}$ W = 600 kW

Since the leakage power loss is negligible, the total power to be supplied by plat = 800 + 600 = 1400 kW

Voltage drop in the transmission line = IR = 200 $* 15$ = 3000 V. Hence, the total voltage transmitted from plant = 3000 + 4000 = 7000 V

Since the power generated is 440V, the transformer rating at the plant should be 440 V – 7000 V

less

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
681k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Alternating Current

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience