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**MULTIPLE CHOICE QUESTIONS I**

**Q.1. If the rms current in a 50 Hz ac circuit is 5 A, the value of the current 1/300 seconds after its value becomes zero is(a) 5√2 A(b) (c) 5/6 A(d) 5/√2 AAns. **(b)

I

I = I

at t = 1/300 sec., I = 5√2 sin 2πvt

Hence option (b) verifies.

(a) Zero

(b) X

(c) – X

(d) R

Ans.

For maximum power to be delivered from the generator (or internal reactance X

⇒ X

⇒X

**Q.3. When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220V. This means(a) Input voltage cannot be AC voltage, but a DC voltage.(b) Maximum input voltage is 220V(c) The meter reads not v but <v**(c)

(a) The generator frequency should be reduced

(b) Another capacitor should be added in parallel to the first

(c) The iron core of the inductor should be removed

(d) Dielectric in the capacitor should be removed

Ans.

So to reduce resonant frequency, we have either to increase L or to increase C.

To increase capacitance another capacitor must be connected in parallel with the first.

**Q.5. Which of the following combinations should be selected for better tuning of an LCR circuit used for communication?(a) R = 20 Ω, L = 1.5 H, C = 35µF(b) R = 25 Ω, L = 2.5 H, C = 45µF(c) R = 15 Ω, L = 3.5 H, C = 30µF(d) R = 25 Ω, L = 1.5 H, C = 45µFAns. **(c)

So for higher Q, L must be large, and R and C must be of smaller value.

This condition is satisfied in (c) part.

**Q.6. An inductor of reactance 1 Ω and a resistor of 2 Ω are connected in series to the terminals of a 6V (rms) a.c. source. The power dissipated in the circuit is(a) 8 W(b) 12 W(c) 14.4 W(d) 18 WAns. **(c)

E

Average power dissipated in the circuit

P

(a) 1/√2 A

(b) √2 A

(c) 2 A

(d) 2√2 A

Ans.

V

P

I

I

I

I

**MULTIPLE CHOICE QUESTIONS II**

**Q.8. As the frequency of an ac circuit increases, the current first increases and then decreases. What combination of circuit elements is most likely to comprise the circuit?(a) Inductor and capacitor(b) Resistor and inductor(c) Resistor and capacitor(d) Resistor, inductor and capacitorAns. **(a,d)

Compare the given circuit by predicting the variation in their reactances with frequency. So, that then we can decide the elements.

Reactance of an inductor of inductance L is X

X

With an increase in frequency (

For an L-C-R circuit,

Z = Impedance of the circuit

As frequency (v) increases, Z decreases and at certain value of the frequency known as resonant frequency (v

(a) Only resistor

(b) Resistor and an inductor

(c) Resistor and a capacitor

(d) Only a capacitor

Ans.

Here in question on increasing the frequency(v) the current also increase.

So capacitive reactance of circuit decreases and resistance does not depend on frequency. So on increasing frequency, X

(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

Ans.

As we know that Power = I

So to decrease power loss I

Output power=Input power

V

∵ V

(a) Here, the power factor cos φ ≥ 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

Ans.

In the given problem power transferred,

P = I

Where I is the current, Z = Impedance and cos φ is power factor

(a) As power factor, cos φ = R/Z

where R > 0 and Z > 0

⇒ cos φ > 0 ⇒ P > 0

(b) when φ = π/2 ( in case of L or C), P = 0.

(c) from (a), it is clear that P < 0 is not possible.

(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) Power delivered to the capacitor is zero

Ans.

When capacitor is connected to ac supply the plate of capacitor will be at higher potential which is connected to the positive terminal than the plate connected to the negative

terminal.

Power applied to circuit is

P

φ = 90º for pure capacitor circuit

∴ P

**Q.13. The line that draws power supply to your house from street has(a) Zero average current(b) 220 V average voltage(c) Voltage and current out of phase by 90°(d) Voltage and current possibly differing in phase ϕ such that |ϕ| < π/2Ans. **(a, d)

AC supply are used in houses.

So average current over a cycle of AC is zero. In household circuit L and C are connected, so R and Z cannot be equal.

So, Power factor

as ϕ ≠ π/2 ⇒ ϕ < π/2

i.e. phase angle between voltage and current lies between 0 and π/2.

**VERY SHORT ANSWER TYPE QUESTIONS**

**Q.14. If a 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?Ans. **An L-C circuit is analogous to a harmonically oscillating spring block system. Therefore energy due to motion of charge particle i.e., magnetic energy 1/2LI

Ans.

In similar way,

X

At high frequency, X

So reactance of capacitance can be considered negligible and capacitor can be considered as short circuited.

Here the above figure is equivalent circuit to given circuit.

Total impedance, Z=R

(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)?

Ans.

(a) (I

Squaring both sides

R

or (X

X

So I

As V

Squaring both sides

R

(X

Square of any number can never be negative. Reactance of X

Ans.

I = I

Instantaneous power output of ac source

P = EI

=E

= E

= E

Taken phase angle ϕ±ve.

Instantaneous Power as cos ϕ = R/Z.

R and Z can never be negative and value of cos θ(θ =2ωt ± ϕ ) can vary from (1 to 0 to -1) in any case P can never be negative.

We know that average power of LCR series ac circuit is

Again as cos ϕ = R/Z is always positive, because R and Z, the reactances are always positive.

So Pav can never be negative.

Ans.

Where ω

From graph ω

So Bandwidth ω

If v is small X

For v is large, X

So X

For X

X

So phase angle in series LCR ac circuit will change from a negative to zero and then zero to positive value.

**SHORT ANSWER TYPE QUESTIONS**

**Q.21. A device ‘X’ is connected to an a.c source. The variation of voltage, current and power in one complete cycle is shown in Figure. ****(a) Which curve shows power consumption over a full cycle?****(b) What is the average power consumption over a cycle?****(c) Identify the device ‘X’.****Ans. **(a) Power is the product of voltage and current (Power = P = VI).

So, the curve of power will be having maximum amplitude, equals to the product of amplitudes of voltage (V) and current (I) curve. Frequencies , of B and C are-equal, therefore they represent V and I curves. So, the curve A represents power.

(b) The full cycle of the graph (as shown by shaded area in the diagram) consists of one positive and one negative symmetrical area.

Hence, average power consumption over a cycle is zero.

(c) Here phase difference between V and I is π/2 therefore, the device ‘X’ may be an inductor (L) or capacitor (C) or the series combination of L and C.**Q.22. Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current?Ans. **For a Direct Current (DC),

1 ampere = 1 coulomb/sec

Direction of AC changes with the frequency of source with the source frequency and the attractive force would average to zero. Thus, the AC ampere must be defined in terms of some property that is independent of the direction of current. Joule’s heating effect is such property and hence it is used to define rms value of AC.

So, r.m.s. value of AC is equal to that value of DC, which when passed through a resistance for a given time will produce the same amount of heat as produced by the alternating current when passed through the same resistance for same time.

Ans. R=1 Ω, L=0.01H, V=200 V, v=50 Hz.

Impedance of the circuit

∴ For phase angle, tan ϕ =Z/R

tan ϕ =3.14

ϕ = tan

Phase difference

ϕ = 1.20 radian

Time lag between alternating voltage and current

**Q.24. A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the type of transformer being used.Ans. P _{S}=60W, I_{S}=0.54A, I_{P}=?**

P

60=V

60/0.54 = V

V

In multiple of 11

V

Ratio factor of transformer 'r ' = output voltage/input voltage

Or r<1, so transformer is step down transformer.

In transformer, output power = input power

I

I

Ans.

A capacitor does not allow a direct current (having zero frequency) to pass through it. But as frequency of current increases capacitor will pass more current through it as on increasing frequency, the charging and discharging happens at fast rate. It implies that the reactance offered by capacitor decreases on increasing frequency.

So the reactance of capacitor can be written as X

**Q.26. Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.Ans.** According to Lenz’s law, when current in an inductor change, the direction of induced e.m.f. will oppose the change in current in inductor. The magnetic flux will be in opposite direction with the magnetic flux produced by changing e.m.f. or current in coil and vice-versa.

Since the induced e.m.f is directly proportional to the rate of change of current, so an inductor produces greater reactance to flow of current through it.

If direct current passes through an inductor the reactance produced by inductor is zero.

So on applying an AC current its reactance increases with increasing the frequency of AC.

So the reactance is directly proportional to frequency or reactance of inductor is

X

**LONG ANSWER TYPE QUESTIONS**

**Q.27. An electrical device draws 2kW power from AC mains (voltage 223V The current differs (lags) in phase by as compared to voltage. Find (i) R, (ii) X _{C} – X_{L}, and (iii) I_{M}. Another device has twice the values for R, X_{C} and X_{L}. How are the answers affected?**P= 2000W Current lags the voltage so

Ans.

V^{2 }= 50,000 V X_{C}>X_{L}_{}

I_{m}_{ = }I_{0}

P = V^{2}/Z

2000 = 50000/Z

Z = 50000/2000 = 25 Ω

R^{2 }+ (X_{C }- X_{L})^{2}= 25^{2 }...(I)

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

tan ϕ = -3/4

Put the value of (X_{C}-X_{L})^{2 }in (I)

R^{2 }= 16×25

R=4×5 = 20Ω

X_{C}-X_{L }= -15Ω

I_{0} = 12.6 A

As (i) R, X_{C}, X_{L} all are double tan will remain same. (ii) Z will become double then I = V/Z become half as value of V does not change. (iii) As I become half P=VI will become again half as voltage remains same.**Q.28. 1 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 11000 V, power transmitted, then a step-down transformer is used to bring voltage to 220 V.(ρ**When power is transmitted at 220 V.

Power lost in transmission P = I

P = VI

P=1 MW=10

V=220 Volt

⇒

l=10 km × 2 = 20,000 m

∴ A = πr

r = 0.5 cm = 0.5 cm×10

ρ

∴ Power loss = I

Power loss in heating = 82.6 MW

As 82.6 MW> 1 MW

So this method cannot be used to transmit the power.

(ii) When power is transmitted at 11000 V

P = 10

VI = 1000000

11000 I = 1000000

∴ R

∴ Power loss = P = I

P = 3.3×10

Fractional power loss

Power loss in % = 3.3%

Total current i from the source V

capacitor and inductor and part i

Potential across R= Potential of source

P.D. across R=V

i

q_{1} is charge on the capacitor at any time t, then for series combination of C, L. applying Kirchhoff’s voltage law in loop ABEFA.

V_{C}+V_{L}= V_{M} sin ωt

Let q_{1}=q_{m} sin (ωt +ϕ ) ...III

Substitute the values of equations (III) and (V) in equation (II)

at ϕ = 0

Applying Kirchhoff’s junction rule at junction B, i=i_{2}+i_{1} using relation I, IV

Now using relation VI for q_{m} and at ϕ=0

Let

i= C cos ϕ sin ωt + C sinϕ . cos ωt

= C [cos ϕ sin ωt + sin ϕ cos ωt]

i= C sin (ωt + ϕ)

Squaring and adding (VII), (VIII)

A^{2}+B^{2} = C^{2} cos^{2}ϕ + C^{2} sin^{2}ϕ

=C^{2}[cos^{2 }ϕ + sin^{2 }ϕ)

A^{2} + B^{2 }= C^{2}

And

Comparing (IX) and (X)

This is the impedance Z for the circuit.**Q.30. For an LCR circuit driven at frequency ω, the equation reads(i) Multiply the equation by i and simplify where possible(ii) Interpret each term physically(iii) Cast the equation in the form of a conservation of energy statement(iv) Integrate the equation over one cycle to find that the phase difference between v and i must be acuteAns. **Consider L-C-R series circuit with AC supply

V=V

Applying voltage Kirchhoff’s law over the circuit

∴ V

(a) Multiply the above equation by i on both the sides.

Multiply above equation by 1/2 on both sides

(b)

(i) represents the rate of change of potential energy in inductance L.

(ii) Represents energy stored in dt time in the capacitor.

(iii) i^{2}R represents Joules heating loss.

(iv) is the rate at which driving force pours in energy. It goes into ohmic loss and increase of stored energy in capacitor and inductor.

(c) Here equation (I) is in the form of conservation of energy statement.

(d) Integrating eqn. (I) both sides with respect to dt over a cycle we get

[because V=V_{m} sin ωt]

As i^{2}RT is +ve [ ∵ i^{2}, R and T can never be negative]

So, is positive which is only possible if phase difference ϕ is constant and the angle is acute.**Q.31. In the LCR circuit shown in Figure, the ac driving voltage is v = v _{m} sin ωt. (i) Write down the equation of motion for q (t). (ii) At t = t0, the voltage source stops and R is short circuited. Now write down how much energy is stored in each of L and C. (iii) Describe subsequent motion of charges.**

V

⇒ As charge q(t) changes in circuit with time in AC,

Then,

This is the equation for variation of motion of charge with respect to time.

(ii) Let time dependent charge in circuit is at phase angle with voltage then q=q

At t = t_{0}, R is short circuited, then energy stored in L and C, when K is closed will be,

At t= t_{0},

i=i_{m} sin (ωt_{0} + ϕ) ...(IV)

From (II),

Comparing (IV) and (I) I_{m}=q_{m}ω

Using equation (II)

(c) When R is short circuited, the circuit becomes L-C oscillator. The capacitor will go on discharging and all energy will transfer to L, and back and forth. Hence there is oscillation of energy from electrostatic to magnetic and vice versa.

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