exemplar-phy-12-alternating-current

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Ques. 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
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(a) zero
(b) $X_g$
(c) $-X_g$
(d) $R_g$
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Ques. When a voltage measuring device is connected to $A C$ mains, the meter shows the steady input voltage of $220 V$. This means
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(a) input voltage cannot be $AC$ voltage, but a $DC$ voltage
(b) maximum input voltage is $220 V$
(c) the meter reads not $v$ but $<v^{2}>$ and is calibrated to read $\sqrt{<v^{2}>}$
(d) the pointer of the meter is stuck by some mechanical defect
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Ques. Which of the following combinations should be selected for better tuning of an $LCR$ circuit used for communication?
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(a) $R=20 \Omega, L=1.5 H, C=35 \mu F$
(b) $R=25 \Omega, L=2.5 H, C=45 \mu F$
(c) $R=15 \Omega, L=3.5 H, C=30 \mu F$
(d) $R=25 \Omega, L=1.5 H, C=45 \mu F$
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Ans. (c)
Thinking Process
For better tuning of an $LCR$ circuit used for communication, quality factor of the circuit must be as high as possible.
Quality factor $(Q)$ of an $LCR$ circuit is given by,
$Q=\frac{1}{R} \sqrt{\frac{L}{C}}$
where $R$ is resistance, $L$ is inductance and $C$ is capacitance of the circuit. To make $Q$ high,
<Success>Sol. to be continued …</Success>
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Ans. (a, d)
Thinking Process
We can decide the elements, comprising the given circuit by predicting the variation in their reactances with frequency.
Reactance of an inductor of inductance $L$ is, $X_L=2 \pi \nu L$ where $v$ is frequency of the AC circuit.
$X_C $ = Reactance of the capacitive circuit
$ =\frac{1}{2 \pi f C} $
<Success>Sol. to be continued …</Success>
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Ans. (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,
$ X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C} $
Clearly when frequency increases, $X_C$ decreases.
<Success>Sol. to be continued …</Success>
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Ques. Electrical energy is transmitted over large distances at high alternating voltages. Which of the following statements is (are) correct?
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(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
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Ans. (a, b, d)
We have to transmit energy (power) over large distances at high alternating voltages, so current flowing through the wires will be low because for a given power (P).
$ P =E_{rms} I_{rms}, I_{rms} \text{ is low when } E_{rms} \text{ is high. } $
$ \text{ Power loss } =I^{2} _{\text{rms }} R=\text{ low } $
$(\because I_{rms} \text{ is low }) $
Now at the receiving end high voltage is reduced by using step-down transformers.
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Ques. For a $LCR$ circuit, the power transferred from the driving source to the driven oscillator is $P=I^{2} Z \cos \phi$.
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(a) Here, the power factor $\cos \phi \geq 0, P \geq 0$
(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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Ans. (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.
<Success>Sol. to be continued …</Success>
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Ques. Draw the effective equivalent circuit of the circuit shown in figure, at very high frequencies and find the effective impedance.
<img src="https://cdn.mathpix.com/snip/images/7j8DpgIM97G4pyLTY2ca0L7oog4AgasvutYvR-EILtk.original.fullsize.png" height="150px">
Thinking Process
The component with infinite resistance will be considered as open circuit and the component with zero resistance will be considered as short circuited.
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Ans.
We know that inductive reactance $X_L=2 \pi f L$
and capacitive reactance $X_C=\frac{1}{2 \pi f C}$
For very high frequencies $(f \to \infty), X_L \to \infty$ and $X_C \to 0$
When reactance of a circuit is infinite it will be considered as open circuit. When reactance of a circuit is zero it will be considered as short circuited.
<Success>Sol. to be continued …</Success>
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(a) When
$ (I _{rms}) a =(I _{rms}) b $ 
$R =\sqrt{R^{2}+(X_L-X_C)^{2}} $ 
$ \Rightarrow \quad X_L=X_C \text{, resonance condition } $
(b) As $Z \geq R$
$ \Rightarrow \frac{(I _{rms}) a}{(I _{rms}) b} =\frac{\sqrt{R^{2}+(X_L-X_C)^{2}}}{R} $
$ =\frac{Z}{R} \geq 1 $
$ \Rightarrow (I _{rms}) a \geq(I _{rms}) b $
No, the rms current in circuit (b), cannot be larger than that in (a).
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Ques. 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 $X$.
<img src="https://cdn.mathpix.com/snip/images/LAyC9wlyGmScJwcoT2dtDGYGfbeEvu-sdarKXUPfnrI.original.fullsize.png" height="140px">
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Ans.
(a) We know that Power $=P=V I$
that is curve of power will be having maximum amplitude, equals to multiplication of amplitudes of voltage $(V)$ and current $(I)$ curve. So, the curve will be represented by $A$.
(b) As shown by shaded area in the diagram, the full cycle of the graph consists of one positive and one negative symmetrical area.
<Success>Sol. to be continued …</Success>
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Ans.
A capacitor does not allow flow of direct current through it as the resistance across the gap is infinite. When an alternating voltage is applied across the capacitor plates, the plates are alternately charged and discharged. The current through the capacitor is a result of this changing voltage (or charge).
Thus, a capacitor will pass more current through it if the voltage is changing at a faster rate, i.e. if the frequency of supply is higher. This implies that the reactance offered by a capacitor is less with increasing frequency.
Mathematically, the reactance can be written as $X_C=\frac{1}{\omega C}$.
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Ans.
An inductor opposes flow of current through it by developing a back emf according to Lenz’s law. The induced voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice -versa.
Since, the induced emf is proportional to the rate of change of current, it will provide greater reactance to the flow of current if the rate of change is faster, i.e., if the frequency is higher. The reactance of an inductor, therefore, is proportional to the frequency. Mathematically, the reactance offered by the inductor is given by $X_L=\omega L$.
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Ques. An electrical device draws 2 kW power from AC mains (voltage 223 V (rms) $=\sqrt{50000} V$). The current differs (lags) in phase by $\phi \tan \phi=\frac{-3}{4}$ as compared to voltage. Find (a) $R_r$ (b) $X_C-X_L$ and (c) $I_M$. Another device has twice the values for $R, X_C$ and $X_L$. How are the answers affected?
Thinking Process
We have to apply the formula for phase relation, net reactance as well as instantaneous power associate with the circuit in terms of voltage and current.
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From Eq. (ii), we put $X_L-X_C=\frac{3 R}{4}$ in Eq. (i), we get
$ 625=R^{2}+\frac{3 R}4^{2}=R^{2}+\frac{9 R^{2}}{16} $
$or \qquad 625=\frac{25 R^{2}}{16} $
(a) Resistance $R=\sqrt{25 \times 16}=\sqrt{400}=20 \Omega$
(b) $X_L-X_C=\frac{3 R}{4}=\frac{3}{4} \times 20=15 \Omega$
(c) Main current $I_M=\sqrt{2} I=\sqrt{2} \frac{V}{Z}=\frac{223}{25} \times \sqrt{2}=12.6 A$
As $R, X_C, X_L$ are all doubled, $\tan \phi$ does not change. $Z$ is doubled, current is halved. So, power is also halved.
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Ques. 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 220 V. 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.
$(\rho_{c u}=1.7 \times 10^{-8}$ SI unit)
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Ques. Consider the $LCR$ circuit shown in figure. Find the net current $i$ and the phase of $i$. Show that $i=\frac{v}{Z}$. Find the impedance $Z$ for this circuit.
<img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/2c254ccf-2821-486c-2276-7b03f8e3b900/public" height="140px">
Thinking Process
The circuit consists of inductor (L) and capacitor (C) connected in series and the combination is connected parallel with a resistance $R$. Due to this combination there is oscillation of electromagnetic energy.
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$\text{ Let } \therefore \frac{d q_2}{d t} =q_m \omega \cos (\omega t+\phi) $
$\Rightarrow \frac{d^{2} q_2}{d t^{2}} =-q_m \omega^{2} \sin (\omega t+\phi) $
$\Rightarrow \quad \frac{q_2}{C}+\frac{L d_2 q_2}{d t^{2}}=V_m \sin \omega t $
$ \because i_2=\frac{d q_2}{d t}$
Now putting these values in Eq. (ii), we get
$q_m \frac{1}{C}+L(-\omega^{2}) \sin (\omega t+\phi)=V_m \sin \omega t$
<Success>Sol. to be continued …</Success>
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Ques. For a $LCR$ circuit driven at frequency $\omega$ the equation reads
$L \frac{d i}{d t}+R i+\frac{q}{C}=v_i=v_m \sin \omega t$
(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) Intergrate the equation over one cycle to find that the phase difference between $V$ and $i$ must be acute.
<Success>Ques. to be continued …</Success>
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$\frac{q}{C} i=\frac{d}{d t} \frac{q^{2}}{2 C}$ = rate of change of energy stored in the capacitor.
$V i=$ rate at which driving force pours in energy. It goes into (i) ohmic loss and (ii) increase of stored energy.
Hence Eq. (ii) is in the form of conservation of energy statement. Integrating both sides of Eq. (ii) with respect to time over one full cycle $(0 \to T)$ we may write
$\int_0^{T} \frac{d}{d t} \frac{1}{2} L i^{2}+\frac{q^{2}}{2 C} d t+\int_0^{T} R i^{2} d t=\int_0^{T} V i d t $
$\Rightarrow 0+(+v e)=\int_0^{T} V i d t $
$\Rightarrow \int_0^{T} V i d t>0$
if phase difference between $V$ and $i$ is a constant and acute angle.
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Ques. In the $LCR$ circuit, shown in figure the $A C$ driving voltage is $V=V_m$ $\sin \omega t$.
(i) Write down the equation of motion for $q(t)$.
(ii) At $t=t_0$, 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.
<Success>Ques. to be continued …</Success>
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From Eq. (i)
$ \frac{d q}{d t} R+L \frac{d^{2} q}{d t^{2}}+\frac{q}{C}=V_m \sin \omega t $
$ \Rightarrow \quad L \frac{d^{2} q}{d t^{2}}+R \frac{d q}{d t}+\frac{q}{C}=V_m \sin \omega t $
This is the required equation of variation (motion) of charge.
(ii) Let $\quad q=q_m \sin (\omega t+\phi)=-q_m \cos (\omega t+\phi)$
$i =i_m \sin (\omega t+\phi)=q_m \omega \sin (\omega t+\phi) $
<Success>Sol. to be continued …</Success>
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$i_m =\frac{V_m}{Z}=\frac{V_m}{\sqrt{R^{2}+(X_C-X_L)^{2}}} $
$\phi =\tan ^{-1} \frac{X_C-X_L}{R} $
When $R$ is short circuited at $t=t_0$, energy is stored in $L$ and $C$.
and
$U_L =\frac{1}{2} L i^{2}=\frac{1}{2} L \frac{V_m}{\sqrt{(R^{2}+X_C-X_L)^{2}}} \sin ^{2}(\omega t_0+\phi) $
$U_C =\frac{1}{2} \times \frac{q^{2}}{C}=\frac{1}{2 C}[q^{2} m \cos ^{2}(\omega t_0+\phi)] $
<Success>Sol. to be continued …</Success>
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$ =\frac{1}{2 C} \frac{V_m}{\sqrt{R^{2}+(X_C-X_L)^{2}}} $
$ =\frac{1}{2 C} \times \frac{i_m}{\omega} \cos ^{2}(\omega t_0+\phi) $
$ =\frac{i^{2} m}{2 C \omega^{2}} \cos ^{2}(\omega t_0+\phi) $
$ =\frac{1}{2 C} \frac{V_m}{\sqrt{R^{2}+(X_C-X_L)^{2}}} \frac{\cos ^{2}(\omega t_0+\phi)}{\omega^{2}} $
$[\because i_m=q_m \omega] $
<Success>Sol. to be continued …</Success>
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