Short Answer type questions
Question 1 Why it is not possible to have electrolysis by A.C?
Solution
An alternating current reverses its direction after each half cycle. Therefore on passing A.C through a solution, the motion of the positive and negative ions become vibratory .So ions are not separated. So that is the reason electrolysis does not happen in A.C. Also Batteries cannot be charged through A.C for this reason only.
Question 2 What is the value of the inductive reactance X
L in a DC circuit?
Solution
inductive reactance XL is given by
XL=ωL
For DC ,ω=0, So XL=0
Multiple Choice Questions
Question 3 Which of the these works on the principle of self-induction?
(a) Transformer
(b) Transistor
(c) choke
(d) None of these
Solution
Answer is (c)
Question 4If the velocity of the object is zero,the acceleration would also be zero
(a) True
(b) False
Solution
False.
Question 5 What is true of any AC circuit?
(a) None of these
(b) The power dissipation happens in resistance only in AC circuit
(c) The algebraic sum of the rms voltage across the different element of the circuit is equal to the applied rms voltage
(d) The algebraic sum of the instantaneous voltage across the different element of the circuit is equal to the applied instantaneous voltage
Solution
Answer is (b)
C and d are incorrect as the voltage across different elements are not in phase.
Question 6 Transformers are used
(a) To step up or step down ac voltage
(b) To convert a.c to d.c
(c) To step up d.c voltage
(d) To convert d.c to a.c
Solution
(a)
Question 7 What is true of AC GENERATOR?
(a) It works on the principle of electromagnetic induction
(b) The mechanical energy of the rotation is converted into electrical energy
(c) A pair of slip rings are used in the AC generator
(d) All the above
Solution
Answer is (d)
Question 8 An alternating voltage V=200sin300t is applied across a series combination of R=10Ω and L=800mH.
(a) The impedence of the circuit is 240.2Ω
(b) Power factor is given by .05
(c) The peak current in the circuit is .8326 A
(d) None of these
Solution
(a) and (c)
Formula used
$Z=\sqrt {R^2 + (\omega L)^2}$
$i_0=\frac {V}{Z}$
$cos \phi =\frac {R}{Z}$
Question 9 In an series LC circuit, the voltage across the L and C are
(a) 180 out of phase
(b) 90 out of phase
(c) are in same phase
(d) 45 out of phase
Solution
(a)
Question 10 Let A and B are the heat production of the electric heater in AC and DC of same voltage.
(a) A=B
(b) A> B
(c) A <B
(d) none of the above
Solution
(c)
Question 11 An inductor having a reactance of 25 and a resistance R gives off heat at the rate of 10W when it carries a current of 0.50A (rms). What is the impedance of the inductor?
Question 12 At the moment depicted in the LC circuit the current is non-zero and the capacitor plates are charged (as shown in the figure below). The energy in the circuit is stored
(a) only in the electric field and is decreasing.
(b) only in the electric field and is constant.
(c) only in the magnetic field and is decreasing.
(d) only in the magnetic field and is constant.
(e) in both the electric and magnetic field and is constant.
(f) in both the electric and magnetic field and is decreasing.
Solution
(e)
Question 13
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.
Solution
(c) and (d)
Question 14
A resistor of 50 ohm ,an inductor of 20/π H and a capacitor of 5/π μ F are connected in series to a voltage supply of 230 V-50 Hz, The impedance of the circuit is
(a) 10 ohm.
(b) 50 ohm.
(c) 100 ohm.
(d) 1000 ohm.
Solution
$X_L = 2 \pi f L =2 \pi \times 50 \times \frac {20}{\pi} = 2000 \Omega$
$X_C= \frac {1}{2 \pi f C} = 2000 \Omega$
Now
$Z= \sqrt {R^2 + (X_L - X_C)^2} = \sqrt { 50^2 + (2000 -2000)^2} = 50\Omega$
Hence (b) is the correct option
Question 15
Dimension of the impedance is
(a) $[M L^2 T^{-3} A^{-2}]$
(b) $[M L^2 T^{-3} A^{-1}]$
(c) $[M L^2 T^{-2} A^{-2}]$
(d) $[M L^2 T^{-1} A^{-2}]$
Solution
(a) is the correct option
Question 16
In pure inductive circuit, the curves between frequency f and reciprocal of inductive reactance $\frac {1}{X_L}$ is
Solution
$X_L = 2 \pi f L$
or
$\frac {1}{X_L} = \frac {1}{2 \pi f L}$
Therefore $\frac {1}{X_L}$ is inversely proportional to f
Hence (b) is the correct option
Question 17
if C and R denote Capacitance and Resistance, then dimension of CR is
(a) $[M^0 L^0 T^{-1} A^{0}]$
(b) $[M^0 L^0 T^{1} A^{0}]$
(c) $[M^0 L^0 T^{1} A^{-2}]$
(d) $[M^0 L^1 T^{1} A^{0}]$
Solution
CR is the time constant. Hence dimension of CR is $[M^0 L^0 T^{1} A^{0}]$. Therefore (b) is the correct option
Question 18
In a AC circuit ,the current is expressed as
$I= 200 \ sin 100 \pi t$.
In this circuit , the current rises from zero to peak value in time
(a) 1/200 sec
(b) 1/100 sec
(c) 1/300 sec
(d) 1/400 sec
Solution
$I= 200 \ sin 100 \pi t$
comparing with
$I=I_0 \sin 2\pi f t$
2f=100
f=50 hz
or
T= 1/f = 1/50 sec
Now Time taken by the current to rise from zero to peak value
=T/4 = 1/200 sec
Hence (a) is correct answer
Question 19
In LCR series circuit connected with an AC voltage source of frequency f, the current leads the voltage by 45°, The value of C is
(a) $\frac {1}{\pi f (2 \pi fL -R)}$
(b) $\frac {1}{2\pi f (2 \pi fL -R)}$
(c) $\frac {1}{\pi f (2 \pi fL +R)}$
(d) $\frac {1}{2\pi f (2 \pi fL +R)}$
Solution
$\tan \phi = \frac {X_L - X_C}{R}$
Now $\tan \phi= \tan 45 =1$
Therefore
$1 = \frac {X_L - X_C}{R}$
$X_C = X_L+ R$
$\frac {1}{2 \pi f C}= 2\pi f L + R$
or
$C=\frac {1}{2\pi f (2 \pi fL +R)}$
Hence (d) is correct answer
Question 20
Quantity that remains unchanged in a transformer is
(a) Voltage
(b) Current
(c) Frequency
(d) none of these
Solution
(c)
Text Based Questions
Question 1
Show that in the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the inductor is constant in time.
Question 2
What is the principle of transformer? . Explain how laminating the core of a transformer helps to reduce eddy current losses in it. Why the primary and secondary coils of a transformer are preferably wound on the same core
Solution
i. Transformer is based on the principle of Mutual induction
ii. Laminations are thing making the resistance higher,Eddy current are confined within each thin lamination and this reduces the net Eddy current
iii. For maximum sharing of magnetic flux , magnetic flux per turn should be same in both primary and secondary coil
Question 3
A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced?
Solution
When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (1/ω C) and the current flows in the circuit. Consequently, the lamp will shine.Reducing C will increase reactance and the lamp will shine less brightly than before.
Question 4
Prove that an ideal capacitor in an A.C circuit does not dissipate power?
Question 5
(i) Explain briefy with the help if labelled diagram , the basic principle of the working of an AC generator
(ii) In an A.C generator , coil of N turns and Area A is rotataed at 'v' revolutions per sec in a uniform magnetic field B. Write the expression for the EMF produced
Question 6
(i) Derive the expression for the average power consumed in a series LCR circuit connected to AC source where phase difference between voltage and current is φ
(ii) Define the quality factor in an AC circuit?
Question 7
Why is choke coil neeed in the use of fluorescent tubles with AC mains?
Numerical Questions
Question 1
A 100 μF capacitor in series with a 40 &Omega resistance is connected to 100 V 60Hz supply
Calculate the following
a. Reactance
b. Impedance
c. Maximum Current in the circuit
d. rms voltage across the resistor and capacitor.Is the algebraic sum of these voltages more than the source voltage?. if Yes resolve the paradox
Solution
a. $X_C= \frac {1}{2 \pi f C} = \frac {1}{2 \times 3.14 \times 60 \times 100 \times 10^{-6}}=26.5 \Omega$
b. Now
$Z= \sqrt {R^2 + X_C^2} = \sqrt { 40^2 + (26.5)^2} = 48 \Omega$
c. $V_{rms}=100$V
$I_{max} = \frac {V_{max}}{Z} = \frac { V_{rms} \times \sqrt {2}}{Z} = \frac {1.414 \times 100}{48}=2.95$ A
d. $I_{rms} = \frac {V_{max}}{Z}= \frac {100}{48} = 2.08$ A
$V_R= i_{rms} R = 2.08 \times 40 =83.2$ V
$V_C= i_{rms} X_C = 2.08 \times 26.5 =55.12$ V
Now
$V_R + V_C= 83.2 + 55.12=138.32 > 100$
This is because these voltages are not in same phase and cannot be added like these
Infact
$V = \sqrt { V_R^2 +V_C^2} = \sqrt { 6922.24 + 3038.21} = 99.80$ V
Question 2
An A.C voltage given by $V= 70 \sin 100 \pi t$ is connected across a pure resistor of 25 Ω. Find the following
a. Frequency of the source
b. rms current through the resistor
Solution
a. $V= 70 \sin 100 \pi t$
Comparing with
$V=V_0 \sin 2 \pi f t$
we get
$V_0=70$ V
$2 \pi f = 100 \pi $
f=50 hz
b. $V_{rms} = \frac {V_0}{\sqrt {2}}= 35 \sqrt {2}$
$I_{rms} = \frac {V_{rms}}{R} = 1.98$ A
Question 3
An inductor L of inductance $X_L$ is connected in series with a bulb B and an ac source.
How would brightness of the bulb change when (i) number of turn in the inductor is reduced, (ii) an iron rod is inserted in the inductor and (iii) a capacitor of reactance $X_C = X_L$ is inserted in series in the circuit. Justify your answer in each case.
Solution
Net resistance in the circuit is given by
$Z= \sqrt {R^2 + X_L^2}$
Inductance is given by,
$X_L = \omega L$
(i) As number of turns decreases, L decreases. Hence $X_L$ decreases and Hence net Resistance of the circuit is decreased
Thus, current increases and brightness of the bulb is increased.
(ii) When a soft iron rod is inserted in the circuit, L increases. Therefore, inductive reactance also increases. Net resistance increases and flow of current in the circuit decreases. Thus, the brightness of the bulb will decrease.
(iii) When a capacitor of reactance $X_L = X_C$ is connected in series with the circuit net resistance becomes,
$Z= \sqrt {R^2 + (X_L - X_C)^2}$
hence at $X_L = X_C$
We have
Z= R
which is the condition of resonance.
At resonance, maximum current will flow through the circuit. Therefore, the brightness of the bulb will increase.
Question 4
The given circuit diagram shows a series LCR circuit connected to a variable frequency 230 V source. Here L=5.0 H, C=80 μ F, R=40 Ω
(a) Determine the source frequency, which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit.
(d) How do you explain the observation that the algebraic sum of the voltage of the three elements obtained in (c) is greater than the supplied voltage?
Solution
Here L=5.0 H, C=80 μ F, R=40 Ω
and $V_{rms}=230$V
(a) source frequency at resonance
$\omega _r = \frac {1}{\sqrt {LC}} = 50$ rad/sec
(b) $Z= \sqrt {R^2 + (X_L - X_C)^2}$
At resonance $X_L = X_C$
$Z= R = 40$ Ω
Current Amplitude is given by
$I_0 = \frac {V_0}{Z} = \frac {\sqrt {2} V_{rms}}{R} = 8.1$ A
(c) $I_{rms} = \frac {V_{rms}}{Z} = \frac {230}{40} = 5.75 $A
$V_L = I_{rms} X_L = I_{rms} \times \omega L = 1437.5 $V
$V_C = I_{rms} X_C = \frac {I_{rms}}{\omega C} = 1437.5 $V
$V_R= I_{rms} R= 230$ V
The algebraic sum of the three voltages is more than the source voltage of 230V. These voltages are not in the same phase and cannot be added like ordinary numbers. The voltage across L and C are out of phase and get added to zero. So,
$V_R$= Applied rms voltage.
Question 5
A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.
Solution
The inductive reactance,
$X_L= 2\pi f L= 2 \times 3.13 \times 50 \times 25 \times 10^{-3} = 7.85 \Omega$
The rms current in the circuit is
$I= \frac {V}{X_L} = 28A$
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