NCERT Exemplar: Alternating Current | Physics for JEE Main & Advanced PDF Download

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 A
Ans. (b)
Solution.
I₀ = 5A,  V=50Hz, t = 1/300s
I = I₀ sin ωt
at t = 1/300 sec., I = 5√2 sin 2πvt



Hence option (b) verifies.

Q.2. An alternating current generator has an internal resistance R_g and an internal reactance X_g. It is used to supply power to a passive load consisting of a resistance R_g and a reactance X_L. For maximum power to be delivered from the generator to the load, the value of X_L is equal to
(a) Zero
(b) X_g
(c) – X_g
(d) R_g
Ans. (c)
Solution.
For maximum power to be delivered from the generator (or internal reactance X_g) to the load (of reactance, X_L),
⇒ X_L + X_g = 0 (the total reactance must vanish)
⇒X_L = -X_g

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 <v²> and is calibrated to read  
(d) The pointer of the meter is stuck by some mechanical defect
Ans. (c)
Solution.
The voltmeter connected to AC mains calibrated to read rms value  

Q.4. To reduce the resonant frequency in an LCR series circuit with a generator
(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. (b)
Solution. The resonant frequency of LCR series circuit is

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 = 45µF
Ans. (c)
Solution. We know that for communication, quality factor must be higher and quality factor is Q.

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 W
Ans. (c)
Solution. According to the problem, XL = 1 Ω , R = 2 Ω,
E_rms = 6 V, P_av = ?
Average power dissipated in the circuit
P_av = E_rmsI_rms cos φ    ..(i)




Q.7 The output of a step-down transformer is measured to be 24 V when connected to a 12 watt light bulb. The value of the peak current is
(a) 1/√2 A
(b) √2 A
(c) 2 A
(d) 2√2 A
Ans. (a)
Solution.
V_s=24 V
P_S=12 W
I_SV_S= 12
I_s = 12/V_s  = 12/24 = 0.5 ampere
I₀ = I_s √2 = 0.5√2
I₀ =  5√2/10 = 1/√2 ampere

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 capacitor
Ans. (a,d)
Solution.
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_L = 2πvL, where v is the frequency of the AC circuit.
X_c = Reactance of the capacitive circuit

With an increase in frequency (f) of an AC circuit, R remains constant, inductive reactance (X_L) increases and capacitive reactance (X_c) decreases.
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 (v₀), impedance Z is minimum that is Z_min = R current varies inversely with impedance and at Z_min current is maximum.

Q.9. 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
Ans. (c, d)
Solution.
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_C decreases to increase current in circuit.

Q.10. 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
Ans. (a, b, d)
Solution.
As we know that Power = I²_rmsR
So to decrease power loss I_rms, and R must be lower for a constant power supply. To decrease I_rms, V_rms must be increased by step up transformer to get same power in step up transformer
Output power=Input power
V_sI_s = V_pI_p
∵ V_S>V_P so I_P ≫ I_S, so loss of power during transmission become lower. We can again reduce voltage by using step down transformer.

Q.11. For an LCR circuit, the power transferred from the driving source to the driven oscillator is P = I²Z cos φ.
(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. (a, b, c)
Solution.
In the given problem power transferred,
P = I²Zcos φ
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.

Q.12. 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) Power delivered to the capacitor is zero
Ans. (c, d)
Solution.
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_av = V_rms I_rms cos φ
φ = 90º for pure capacitor circuit
∴ P_av = V_rmsI_rms cos 90º = 0

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/2
Ans. (a, d)
Solution.
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 1/2LI² is analogous to kinetic energy. The electrostatic energy due to charging of capacitor 1/2CV² is analogous to potential energy.

Q.15. Draw the effective equivalent circuit of the circuit shown in Figure, at very high frequencies and find the effective impedance.
Ans. ∵ We know that reactance due to inductance X_L= 2πvL. As frequency is high so X_L will be high or L can be considered as open circuit for high frequency of ac circuit.
In similar way,
X_c= 1/2πvC
At high frequency, X_C will be low.
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=R₁+R_3.

Q.16. 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)?
Ans. 


(a) (I_rms)_a=(I_rms)_b

Squaring both sides
R² = R² + (X_L - X_C)²
or (X_L - X_C)² = 0
X_L = X_C
So I_rms in circuits (a) and (b) will be equal if X_L=X_C.


As V_rms  =V so,

Squaring both sides
R² + (X_L - X_C)² < R²
(X_L - X_C)² < 0
Square of any number can never be negative. Reactance of X_L and X_C cannot be negative. So the rms current in circuit (b) cannot be larger than that in (a).

Q.17. Can the instantaneous power output of an ac source ever be negative? Can the average power output be negative?
Ans. Let the applied e.m.f. =E=E₀sin (ωt)
I = I₀ sin (ωt - ϕ)
Instantaneous power output of ac source
P = EI
=E₀ sin ωt I₀ sin (ωt - ϕ )
= E₀I₀ sin ωt [sin ωt cos ϕ - cosωt sinϕ]
= E₀I₀ [sin² ωt cos ϕ - sinωt cosωt sinϕ]


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.

Q.18. In series LCR circuit, the plot of I_max vs ω is shown in Figure. Find the bandwidth and mark in the figure.
Ans. We know that band width=(ω₂-ω₁).
Where ω₁ and ω₂ are two frequencies where the current amplitude of LCR circuit becomes 1/√2 times (i.e., I_rms) the value of current is maximum at resonant frequency.


From graph ω₁ and ω₂ at 0.707A current is 0.8 and 1.2 rad/sec.
So Bandwidth ω₂- ω₁=1.2-0.8=0.4 rad/sec.

Q.19. The alternating current in a circuit is described by the graph shown in Figure. Show rms current in this graph.
Ans. Graph or rms current is shown below by dotted line



Q.20. How does the sign of the phase angle φ, by which the supply voltage leads the current in an LCR series circuit, change as the supply frequency is gradually increased from very low to very high values.
Ans. The phase angle 'ϕ' by which voltage leads the current in LCR series circuit where X_L>X_C.

If v is small X_C>X_L so  is negative, so tan ϕ < 0.
For v is large, X_L>X_C
So X_L-X_C is positive or tan ϕ>0
For X_L=X_C i.e. at resonant frequency
X_L - X_C=0 so tan ϕ=0.
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.

Q.23. A coil of 0.01 henry inductance and 1 ohm resistance is connected to 200 volt, 50 Hz ac supply. Find the impedance of the circuit and time lag between max. alternating voltage and current.
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^-1 13.14= 72⁰
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_S=V_SI_S
60=V_S×0.54
60/0.54 = V_s
V_S= 111.10 Volt
In multiple of 11
V_S  ≅ 110 Volt
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_SV_S=I_PV_P

I_P = 0.27 ampere

Q.25. Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency.
Ans. When AC current is applied across a capacitor plate, the plates of the capacitor get charge and discharge alternately. Current through the capacitor is a result of this charging and discharging by AC voltage or current.
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_C = 1/ωC.

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_L=2πvL=ωL.

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?
Ans. P= 2000W Current lags the voltage so

V² = 50,000 V  X_C>X_L

I_m =  I₀
P = V²/Z
2000 = 50000/Z
Z = 50000/2000 = 25 Ω

R² + (X_C - X_L)²= 25²  ...(I)
R² + (X_C - X_L)² = 625
tan ϕ = -3/4

Put the value of (X_C-X_L)² in (I)

R² = 16×25
R=4×5 = 20Ω

X_C-X_L = -15Ω

I₀ = 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_Cu= 1.7 ×10^-8 SI unit)
Ans. When power is transmitted at 220 V.
Power lost in transmission P = I²R
P = VI
P=1 MW=10⁶W
V=220 Volt


⇒
l=10 km × 2 = 20,000 m
∴ A = πr²
r = 0.5 cm = 0.5 cm×10^-2 = 5×10^-3 m
ρ_Cu = 1.7×10^-8Ω-m




∴ Power loss = I²R

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 = 10⁶ W
VI = 1000000
11000 I = 1000000

∴ R_cu = 4Ω [as already calculated in part (i)]
∴ Power loss  = P = I²R

P = 3.3×10⁴ Watt
Fractional power loss
Power loss in % = 3.3%

Q.29. 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.
Ans.
Total current i from the source V_m sin ωt is divided at B in two part, i₁ through
capacitor and inductor and part i₂ through resistance.
Potential across R= Potential of source
P.D. across R=V_m sin ωt
i²R = V_msin ωt

q₁ 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 q₁=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=i₂+i₁ 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)
A²+B² = C² cos²ϕ + C² sin²ϕ
=C²[cos² ϕ + sin² ϕ)
A² + B² = C²







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 acute
Ans. Consider L-C-R series circuit with AC supply
V=V_m sin ωt
Applying voltage Kirchhoff’s law over the circuit
∴ V_L+V_C+V_R = V_m sin ωt

(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) i²R 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 i²RT is +ve [ ∵ i², 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.
Ans. (i) Consider series LCR circuit and tapping key K to short circuit R. Let i be the current in circuit. Then by Kirchhoff’s voltage law, when key K is open,
V_R + V_L + V_C=V_m sin ωt

⇒ 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_m cos(ωt + ϕ)

At t = t₀, R is short circuited, then energy stored in L and C, when K is closed will be,

At t= t₀,
i=i_m sin (ωt₀ + ϕ) ...(IV)
From (II),



Comparing (IV) and (I) I_m=q_mω



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.