Alternating Current
Chapter 7
ALTERNATING CURRENT
MCQ 1
7.1 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) $5 \sqrt{2} A$
(b) $5 \sqrt{3 / 2} A$
(c) $5 / 6 A$
(d) $5 / \sqrt{2} A$
Show Answer
Answer: (b)7.2 An alternating current generator has an internal resistance $Rg$ and an internal reactance $Xg$. 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) $R_g$.
Show Answer
Answer: (c)7.3 When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of $220 V$. This means
(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 $\langle v^{2}\rangle$ and is calibrated to read $\sqrt{\langle v^{2}\rangle}$.
(d) the pointer of the meter is stuck by some mechanical defect.
Show Answer
Answer: (c)7.4 To reduce the reasonant 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.
Show Answer
Answer: (b)7.5 Which of the following combinations should be selected for better tuning of an LCR circuit used for communication?
(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) $R=25 \Omega, L=1.5 H, C=45 \mu F$.
Show Answer
Answer: (c)7.6 An inductor of reactance $1 \Omega$ and a resistor of $2 \Omega$ are connected in series to the terminals of a $6 V(rms)$ a.c. source. The power dissipated in the circuit is
(a) $8 W$.
(b) $12 W$.
(c) $14.4 W$.
(d) $18 W$.
Show Answer
Answer: (c)7.7 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) $1 / \sqrt{2} A$.
(b) $\sqrt{2} A$.
(c) $2 A$.
(d) $2 \sqrt{2} A$.
MCQ II
Show Answer
Answer: (a)7.8 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.
Show Answer
Answer: (a), (d)7.9 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.
Show Answer
Answer: (c), (d)7.10 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.
Show Answer
Answer: (a), (b), (d)7.11 For an $L C R$ circuit, the power transferred from the driving source to the driven oscillator is $P=I^{2} Z \cos \phi$.
(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.
Show Answer
Answer: (a), (b), (c)7.12 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.
Show Answer
Answer: (c), (d)7.13 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 $90^{\circ}$.
(d) voltage and current possibly differing in phase $\phi$ such that $|\phi|<\frac{\pi}{2}$.
Show Answer
Answer: (a), (d)7.14 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?
Show Answer
Answer: Magnetic energy analogous to kinetic energy and electrical energy analogous to potential energy.7.15 Draw the effective equivalent circuit of the circuit shown in Fig 7.1, at very high frequencies and find the effective impedance.
Fig. 7.1
Show Answer
Answer: At high frequencies, capacitor $\approx$ short circuit (low reactance) and inductor $\approx$ open circuit (high reactance). Therefore, the equivalent circuit $Z \approx R_1+R_3$ as shown in the Fig.
7.16 Study the circuits (a) and (b) shown in Fig 7.2 and answer the following questions.
(a) 
(a) 
Fig. 7.2
(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)?
Show Answer
Answer: (a) Yes, if rms voltage in the two circuits are same then at resonance, the rms current in $L C R$ will be same as that in $R$ circuit.
(d) No, because $R \leq Z$, so $I_a \geq I_b$.
7.17 Can the instantaneous power output of an ac source ever be negative? Can the average power output be negative?
Show Answer
Answer: Yes, No.7.18 In series LCR circuit, the plot of $I _{\max }$ vs $\omega$ is shown in Fig 7.3. Find the bandwidth and mark in the figure.
Fig. 7.3
Show Answer
Answer: Bandwidth corresponds to frequencies at which $I_m=\frac{1}{\sqrt{2}} I _{\max }$ $\approx 0.7 I _{\max }$.
It is shown in the Fig.
$\Delta \omega=1.2-0.8=0.4 rad / s$
7.19 The alternating current in a circuit is described by the graph shown in Fig 7.4. Show rms current in this graph.
Fig. 7.4
Show Answer
Answer: $I _{\text{rms }}=1.6 A$ (shown in Fig. by dotted line)
7.20 How does the sign of the phase angle $\phi$, by which the supply voltage leads the current in an $L C R$ series circuit, change as the supply frequency is gradually increased from very low to very high values.
SA
Show Answer
Answer: From negative to zero to positive; zero at resonant frequency.7.21 A device ’ $X$ ’ is connected to an a.c source. The variation of voltage, current and power in one complete cycle is shown in Fig 7.5.
(a) Which curve shows power consumption over a full cycle?
(b) What is the average power consumption over a cycle?
(c) Identify the device ’ $X$ ‘.
Fig. 7.5
Show Answer
Answer: (a) A
(b) Zero
(c) $L$ or $C$ or $L C$
7.22 Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current?
Show Answer
Answer: An a.c current changes direction with the source frequency and the attractive force would average to zero. Thus, the a.c 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 a.c.7.23 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.
Show Answer
Answer: $ X_L=\omega L=2 p f L$
$=3.14 \Omega$
$Z=\sqrt{R^{2}+L^{2}}$
$=\sqrt{(3.14)^{2}+(1)^{2}}=\sqrt{10.86}$
$\simeq 3.3 \Omega$
$\tan \phi=\frac{\omega L}{R}=3.14$
$\phi=\tan ^{-1}(3.14)$
$\simeq 72^{\circ}$
$\simeq \frac{72 \times \pi}{180} rad$.
Timelag $\Delta t=\frac{\phi}{\omega}=\frac{72 \times \pi}{180 \times 2 \pi \times 50}=\frac{1}{250} s$
7.24 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.
Show Answer
Answer: $P_L=60 W, I_L=0.54 A$
$V_L=\frac{60}{0.54}=110 V$
The transformer is step-down and have $\frac{1}{2}$ input voltage. Hence
$i_p=\frac{1}{2} \times I_2=0.27 A$.
7.25 Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency.
Show Answer
Answer: 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; it is given by $1 / \omega C$.
Exemplar Problems-Physics
7.26 Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.
LA
Show Answer
Answer: 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, being given by $\omega L$.7.27 An electrical device draws $2 kW$ power from AC mains (voltage $223 V$ (rms) $=\sqrt{50,000} V$ ). The current differs (lags) in phase by $\phi(\tan \phi=\frac{-3}{4})$ 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 ans affected?
Show Answer
Answer: Power $P=\frac{V^{2}}{Z} \Rightarrow \frac{50,000}{2000}=25=Z$
$Z^{2}=R^{2}+(X_C-X_L)^{2}=625$
$\tan \phi=\frac{X_C-X_L}{R}=-\frac{3}{4}$
$625=R^{2}+(-\frac{3}{4} R)^{2}=\frac{25}{16}$
$R^{2}=400 \Rightarrow R=20 \Omega$
$X_C-X_L=-15 \Omega$
$I=\frac{V}{Z}=\frac{223}{25} \simeq 9 A$.
$I_M=\sqrt{2} \times 9=12.6 A$.
If $R, X_C, X_L$ are all doubled, $\tan \phi$ does not change.
$Z$ is doubled, current is halfed.
Power drawn is halfed.
7.28 1MW 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 transfomer is used to bring voltage to $220 V$.
$ (\rho _{C u}=1.7 \times 10^{-8} \text{ SI unit }) $
Show Answer
Answer: (i) Resistance of $Cu$ wires, $R$
$=\rho \frac{l}{A}=\frac{1.7 \times 10^{-8} \times 20000}{\pi \times(\frac{1}{2})^{2} \times 10^{-4}}=4 \Omega$
$I$ at $220 V: V I=10^{6} W ; I=\frac{10^{6}}{220}=0.45 \times 10^{4} A$
$R I^{2}=$ Power loss
$ \begin{aligned} & =4 \times(0.45)^{2} \times 10^{8} W \\ & >10^{6} W \end{aligned} $
This method cannot be used for transmission
(ii) $V^{\prime} I^{\prime}=10^{6} W=11000 I^{\prime}$
$I^{\prime}=\frac{1}{1.1} \times 10^{2}$
$R I^{\prime 2}=\frac{1}{1.21} \times 4 \times 10^{4}=3.3 \times 10^{4} W$
Fraction of power loss $=\frac{3.3 \times 10^{4}}{10^{6}}=3.3 %$
7.29 Consider the $L C R$ circuit shown in Fig 7.6. Find the net current $i$ and the phase of $i$. Show that $i=\frac{v}{Z}$. Find the impedence $Z$ for this circuit.
Fig. 7.6
Show Answer
Answer: $R i_1=v_m \sin \omega t i_1=\frac{v_m \sin \omega t}{R}$
$\frac{q_2}{C}+L \frac{d q_2^{2}}{d t^{2}}=v_m \sin \omega t$
Let $q_2=q_m \sin (\omega t+\phi)$
$q_m(\frac{q_m}{C}-L \omega^{2}) \sin (\omega t+\phi)=v_m \sin \omega t$
$q_m=\frac{v_m}{\frac{1}{C}-L \omega^{2}}, \phi=0 ; \frac{1}{C}-\omega^{2} L>0$
$v_R=\frac{v_m}{L w^{2}-\frac{1}{C}}, \phi=\pi L \omega^{2}-\frac{1}{C}>0$
$i_2=\frac{d q_2}{d t}=\omega q_m \cos (\omega t+\phi)$
$i_1$ and $i_2$ are out of phase. Let us assume $\frac{1}{C}-\omega^{2} L>0$
$i_1+i_2=\frac{v_m \sin \omega t}{R}+\frac{v_m}{L \omega-\frac{1}{c \omega}} \cos \omega t$
Now $A \sin \omega t+B \cos \omega t=C \sin (\omega t+\phi)$
$C \cos \phi=A, C \sin \phi=B ; C=\sqrt{A^{2}+B^{2}}$
Therefore, $i_1+i_2=[\frac{v_m{ }^{2}}{R^{2}}+\frac{v_m{ }^{2}}{[\omega l-1 / \omega C]^{2}}]^{\frac{1}{2}} \sin (\omega t+\phi)$
$\phi=\tan ^{-1} \frac{R}{X_L-X_C}$
$\frac{1}{Z}={\frac{1}{R^{2}}+\frac{1}{(L \omega-1 / \omega C)^{2}}}^{1 / 2}$
7.30 For an $L C R$ 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.
Show Answer
Answer: $L i \frac{d i}{d t}+R i^{2}+\frac{q i}{c}=v i ; L i \frac{d i}{d t}=\frac{d}{d t}(\frac{1}{2} L i^{2})=$ rate of change of energy stored in an inductor.
$R i^{2}=$ joule heating loss
$\frac{q}{C} i=\frac{d}{d t}(\frac{q^{2}}{2 C})=$ 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.
$\int_0^{T} d t \frac{d}{d t}(\frac{1}{2} i^{2}+\frac{q^{2}}{C})+\int_0^{T} R i^{2} d t=\int_0^{T} v i d t$ $0+(+v e)=\int_0^{T} v i d t$
$\int_0^{T}$ vidt $>0$ if phase difference, a constant is acute.
7.31 In the $L C R$ circuit shown in Fig 7.7, the ac 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.
Fig. 7.7
Show Answer
Answer: (i) $L \frac{d^{2} q}{d t^{2}}+R \frac{d q}{d t}+\frac{q}{C}=v_m \sin \omega t$
Let $q=q_m \sin (\omega t+\phi)=-q_m \cos (\omega t+\phi)$
$i=i_m \sin (\omega t+\phi)=q_m \omega \sin (w t+\phi)$
$i_m=\frac{v_m}{Z}=\frac{v_m}{\sqrt{R^{2}+(X_C-X_L)^{2}}} ; \phi=\tan ^{-1}(\frac{X_C-X_L}{R})$
(ii) $U_L=\frac{1}{2} L i^{2}=\frac{1}{2} L[\frac{v_m}{\sqrt{.R^{2}+X_C-X_L)^{2}}}]^{2} \sin ^{2}(\omega t_0+\phi)$
$U_C=\frac{1}{2} \frac{q^{2}}{C}=\frac{1}{2 C}[\frac{v_m}{\sqrt{R^{2}+(X_C-X_L)^{2}}}]^{2} \frac{1}{\omega^{2}} \cos ^{2}(\omega t_0+\phi)$
(iii) Left to itself, it is an $L C$ oscillator. The capacitor will go on discharging and all energy will go to $L$ and back and forth.
