NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current

• An alternating voltage υ = υm sin ωt applied to a resistor R drives a current i = im sin ωt in the resistor, $i_m=\frac{{\nu }_m}{R}.$ The current is in phase with the applied voltage.

• For an alternating current i = im sin ωt passing through a resistor R, the average power loss P (averaged over a cycle) due to joule heating is $\frac{1}{2}(i_m{)}^{2}R.$ To express it in the same form as the dc power (P = I2R), a special value of current is used. It is called root mean square (rms) current and is denoted by I: 

$I=\frac{i_m}{\sqrt{2}}=0.707i_m$

Similarly, the rms voltage is defined by

$V=\frac{{\nu }_m}{\sqrt{2}}=7.07{\nu }_m$

We have P = IV = I2 R

• An ac voltage υ = υm sin ωt applied to a pure inductor L, drives a current in the inductor i = im sin (ωt – π/2), where im = υm/XL. XL = ωL is called inductive reactance. The current in the inductor lags the voltage by π/2. The average power supplied to an inductor over one complete cycle is zero.

• An ac voltage υ = υm sin ωt applied to a capacitor drives a current in the capacitor: i = im sin (ωt + π/2). Here, 

$i_m=\frac{{\nu }_m}{X_C},X_C=\frac{1}{\omega C}$ is called capacitive reactance. 

The current through the capacitor is π/2 ahead of the applied voltage. As in the case of inductor, the average power supplied to a capacitor over one complete cycle is zero.

• For a series RLC circuit driven by voltage υ = υm sin ωt, the current is given by i = im sin (ωt + φ)

where $i_m=\frac{{\nu }_m}{\sqrt{R^2+(X_L{)}^2}}$.

and $\varphi ={\tan }^{-1}\frac{X_C-X_L}{R}$

$Z=\sqrt{R^2+(X_C-X_L{)}^2}$ is called the impedance of the circuit.

• In a purely inductive or capacitive circuit, cosφ = 0 and no power is dissipated even though a current is flowing in the circuit. In such cases, current is referred to as a wattless current.

• An interesting characteristic of a series RLC circuit is the phenomenon of resonance. The circuit exhibits resonance, i.e., the amplitude of the current is maximum at the resonant frequency, ${\omega }_0=\frac{1}{\sqrt{LC}}.$ 

The quality factor Q defined by $Q=\frac{{\omega }_{0}L}{R}=\frac{1}{{\omega }_{0}CR}$  is an indicator of the sharpness of the resonance, the higher value of Q indicating sharper peak in the current.

• A circuit containing an inductor L and a capacitor C (initially charged) with no ac source and no resistors exhibits free oscillations. The charge q of the capacitor satisfies the equation of simple harmonic motion: 

$\frac{d^{2}q}{d{t}^2}+\frac{1}{LC}q=0$

and therefore, the frequency ω of free oscillation is ${\omega }_0=\frac{1}{\sqrt{LC}}.$ The energy in the system oscillates between the capacitor and the inductor but their sum or the total energy is constant in time.

• A transformer consists of an iron core on which are bound a primary coil of Np turns and a secondary coil of Ns turns. If the primary coil is connected to an ac source, the primary and secondary voltages and the currents are related by 

$V_s=\left(\frac{N_s}{N_p}\right)V_p$  and $I_s=\left(\frac{N_p}{N_s}\right)I_p$

If the secondary coil has a greater number of turns than the primary, the voltage is stepped-up (Vs > Vp). This type of arrangement is called a step- up transformer. If the secondary coil has turns less than the primary, we have a step-down transformer.

• When a value is given for ac voltage or current, it is ordinarily the rms value. The voltage across the terminals of an outlet in your room is normally 240 V. This refers to the rms value of the voltage. The amplitude of this voltage is ${\upsilon }_m=\sqrt{2}=\sqrt{2}(240)=340V$

• The power rating of an element used in ac circuits refers to its average power rating.

• The power consumed in an ac circuit is never negative.

• In an ac circuit, while adding voltages across different elements, one should take care of their phases properly. For example, if VR and VC are voltages across R and C, respectively in an RC circuit, then the total voltage across RC combination is $V_{RC}=\sqrt{V_R^2+V_C^2}$  and not VR + VC since Vc is π/2 out of phase of VR.

• In a RLC circuit, resonance phenomenon occur when XL = XC or ${\omega }_0=\frac{1}{\sqrt{LC}}.$ For resonance to occur, the presence of both L and C elements in the circuit is a must. With only one of these (L or C) elements, there is no possibility of voltage cancellation and hence, no resonance is possible.

• The power factor in a RLC circuit is a measure of how close the circuit is to expending the maximum power.

• Displacement current is that current which comes into play in the region in which the electric field and the electric flux is changing with time. $I_D={\in }_0=\frac{d{\phi }_E}{dt}$

Maxwell modified Ampere’s circuital law in order to make the same logically consistent. 

$\int \overrightarrow{B}\cdot d\overrightarrow{l}=\mu 0(I+I_0)={\mu }_0(I+{\epsilon }_0{\displaystyle \frac{d{\phi }_E}{dt}})$

• The orderly distribution of electromagnetic radiations according to their wavelength or frequency is called the electromagnetic spectrum.

• Radio waves are the electromagnetic wave of frequency range from 5 × 105 Hz to 109 Hz. These waves are produced by oscillating electric circuits having an inductor and capacitor. 

• Microwaves are the electromagnetic waves of frequency range 1 GHz to 300 GHz. They are produced by special vacuum tubes. namely; klystrons, magnetrons and Gunn diodes etc. 

• Infrared waves were discovered by Herschell. These are the electromagnetic waves of frequency range 3 × 1011 Hz to 4 × 1014 Hz. Infrared waves are sometimes called heat waves. Infrared waves are produced by hot bodies and molecules. These waves are not detected by the human eye but snakes can detect them. 

• It is the narrow region of the electromagnetic spectrum, which is detected by the human eye. Its frequency ranges from 4×1014 Hz to 8×1014 Hz. It is produced due to atomic excitation. 

• The ultraviolet rays were discovered by Ritter in 1801. The frequency range of ultraviolet rays is 8 × 1014 Hz to 5 × 1016 Hz. The ultraviolet rays are produced by sun, special lamps and very hot bodies. 

• The X–rays were discovered by German Physicist W. Roentgen. Their frequency range is 1016 Hz to 3 × 1021 Hz. These are produced when high energy electrons are stopped suddenly on a metal of high atomic number. X–rays have high penetrating power. 

• γ–rays are the electromagnetic waves of frequency range 3 × 1018 Hz to 5 × 1022 Hz. γ–rays have nuclear origin. These rays are highly energetic and are produced by the nucleus of the radioactive substances.

• Intensity of an electromagnetic wave at a point is defined as the energy crossing per second per unit area normally around that point during the propagation of electromagnetic waves. 

$I=\frac{1}{2}\frac{B_0^2}{{\mu }_0}c=\frac{1}{{\mu }_0}{B}^2{\text{rms}}^C$

Alternating Current Class 12 NCERT Solutions PDF

1. A $100\mathrm{\Omega }$ resistor is connected to a $220V$, $50Hz$ ac supply.

a) What is the rms value of current in the circuit?

Ans: It is given that,

Resistance, $R=100\mathrm{\Omega }$

Voltage, $V=220V$  

Frequency, $f=50Hz$ 

It is known that,

$I_{rms}=\frac{V_{rms}}{R}$ 

$\Rightarrow I_{rms}=\frac{220}{100}=2.2A$ 

Therefore, the rms value of current in the circuit is$I_{rms}=2.2A$.

b) What is the net power consumed over a full cycle?

Ans: It is known that,

$Power=V\times I$ 

$\Rightarrow Power=220\times 2.2$

$\Rightarrow Power=484W$

Therefore, the net power consumed over a full cycle is$484W$.

2.

a) The peak voltage of an ac supply is $300V$. What is the rms voltage?

Ans: It is given that,

Peak voltage of the ac supply, $V_0=300V$ 

It is known that,

$V_{rms}=\frac{V_0}{\sqrt{2}}$ 

$\Rightarrow V_{rms}=\frac{300}{\sqrt{2}}$

$\Rightarrow V_{rms}=212.1V$

Therefore, the rms voltage is $212.1V$.

c) The rms value of current in an ac circuit is $10A$. What is the peak current?

Ans: It is given that,

Rms value of current in an ac circuit, $I_{rms}=10A$

It is known that,

$I_0=\sqrt{2}\times I_{rms}$ 

$\Rightarrow I_0=1.414\times 10$ 

$\Rightarrow I_0=14.14A$

Therefore, the peak current is $14.14A$.

3. A $44mH$ inductor is connected to $220V$,$50Hz$ ac supply. Determine the rms value of the current in the circuit.

Ans: It is known that,

Inductance, $L=44mH=44\times {10}^{-3}H$ 

Voltage,$V=220V$

Frequency, $f_L=50Hz$ 

Angular frequency, ${\omega }_L=2\pi f_L$ 

It is known that,

Inductive reactance, $X_L={\omega }_{L}L=2\pi f_{L}L$ 

$\Rightarrow X_L=2\times 3.14\times 50\times 44\times {10}^{-3}\mathrm{\Omega }$ 

$\Rightarrow X_L=13.8\mathrm{\Omega }$

$I_{rms}=\frac{V}{X_L}$ 

$\Rightarrow I_{rms}=\frac{220}{13.82}$ 

$\Rightarrow I_{rms}=15.92A$

Therefore, the rms value of the current in the circuit is $15.92A$.

4. A $60\mu F$ capacitor is connected to a $110V$,$60Hz$ ac supply. Determine the rms value of the current in the circuit.

Ans: It is given that,

Capacitance, $C=60\mu F=60\times {10}^{-6}F$ 

Voltage, $V=110V$ 

Frequency, $f_C=60Hz$ 

It is known that,

$I_{rms}=\frac{V}{X_C}$ 

$X_C=\frac{1}{{\omega }_{C}C}=\frac{1}{2\pi f_{C}C}$ 

$\Rightarrow X_C=\frac{1}{2\times 3.14\times 60\times 60\times {10}^{-6}}$ 

$\Rightarrow X_C=44.248\mathrm{\Omega }$ 

$\Rightarrow I_{rms}=\frac{110}{44.28}$ 

$\Rightarrow I_{rms}=2.488A$

Therefore, the rms value of the current in the circuit is $2.488A$.

5. In exercises $4$ and $5$ What is the net power absorbed by each circuit over a complete cycle? Explain your answer.

Ans: From the inductive circuit,

Rms value of current, $I_{rms}=15.92A$ 

Rms value of voltage, $V_{rms}=220V$ 

It is known that,

Net power absorbed, $P=V_{rms}\times I_{rms}\cos \varphi$ 

Where,

$\varphi$ is the phase difference between voltage and current

For a pure inductive circuit, the phase difference between alternating voltage and current is ${90}^0$i.e., $\varphi ={90}^0$

$\Rightarrow P=220\times 15.92\cos {90}^0=0$ 

Therefore, net power absorbed is zero in a pure inductive circuit.

In a capacitive circuit,

Rms value of current, $I_{rms}=2.49A$ 

Rms value of voltage, $V_{rms}=110V$ 

It is known that,

Net power absorbed, $P=V_{rms}\times I_{rms}\cos \varphi$ 

Where,

$\varphi$ is the phase difference between voltage and current

For a pure capacitive circuit, the phase difference between alternating voltage and current is ${90}^0$i.e., $\varphi ={90}^0$

$\Rightarrow P=110\times 2.49\cos {90}^0=0$ 

Therefore, net power absorbed is zero in a pure capacitive circuit.

6. Obtain the resonant frequency ${\omega }_r$ of a series LCR circuit with $L=2.0H$, $C=32\mu F$ and $R=10\mathrm{\Omega }$ . What is the Q-value of this current?

Ans: It is given that,

Inductance, $L=2H$ 

Capacitance, $$C=32\mu F=32\times {10}^{-6}F$$ 

$R=10\mathrm{\Omega }$ 

It is known that,

Resonant frequency, ${\omega }_r=\frac{1}{\sqrt{LC}}$ 

$\Rightarrow {\omega }_r=\frac{1}{\sqrt{2\times 32\times {10}^{-6}}}$ 

$\Rightarrow {\omega }_r=\frac{1}{8\times {10}^{-3}}$

$\Rightarrow {\omega }_r=125rad/s$

$Q-value=\frac{{\omega }_{r}L}{R}$ 

$\Rightarrow Q=\frac{1}{R}\sqrt{\frac{L}{C}}$ 

$\Rightarrow Q=\frac{1}{10}\sqrt{\frac{2}{32\times {10}^{-6}}}$

$\Rightarrow Q=\frac{1}{10\times 4\times {10}^{-3}}$

$\Rightarrow Q=25$ 

Therefore, the resonant frequency is $125rad/s$ and Q-value is $25$.

7. A charged $30\mu F$ capacitor is connected to a $27mH$ inductor. What is the angular frequency of free oscillations of the circuit?

Ans: It is given that,

Capacitance, $C=30\mu F=30\times {10}^{-6}F$

Inductance, $L=27mH=27\times {10}^{-3}H$ 

It is known that,

Angular frequency of free oscillations, ${\omega }_r=\frac{1}{\sqrt{LC}}$ 

$$\Rightarrow {\omega }_r=\frac{1}{\sqrt{27\times {10}^{-3}\times 30\times {10}^{-6}}}$$

$$\Rightarrow {\omega }_r=\frac{1}{9\times {10}^{-4}}$$

$$\Rightarrow {\omega }_r=1.11\times {10}^{3}rad/s$$

Therefore, the angular frequency of free oscillations of the circuit is $$1.11\times {10}^{3}rad/s$$.

8. Suppose the initial charge on the capacitor in exercise $7$ is $6mC$ . What is the total energy stored in the circuit initially? What is the total energy at a later time?

Ans: It is known that,

Capacitance of the capacitor, $C=30\mu F=30\times {10}^{-6}F$ 

Inductance of the capacitor, $L=27mH=27\times {10}^{-3}H$ 

Charge on the capacitor, $Q=6mC=6\times {10}^{-3}C$ 

It is known that,

Energy,$E=\frac{1}{2}\frac{Q^2}{C}$ 

$\Rightarrow E=\frac{1}{2}\frac{{(6\times {10}^{-3})}^2}{30\times {10}^{-6}}$

$\Rightarrow E=\frac{6}{10}=0.6J$

Therefore, the energy stored in the circuit initially is $E=0.6J$.

Total energy at later time will remain same as the initially stored i.e., $0.6J$ because energy is shared between the capacitor and the inductor.

9. A series LCR circuit with $R=20\mathrm{\Omega }$, $L=1.5H$ and $C=35\mu F$ is connected to a variable frequency $200V$ ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Ans: It is known that,

Resistance, $R=20\mathrm{\Omega }$ 

Inductance, $L=1.5H$ 

Capacitance, $C=35\mu F=35\times {10}^{-6}F$ 

Voltage, $V=200V$ 

It is known that,

Impedance,$Z=\sqrt{R^2+{(X_L-X_C)}^2}$ 

At resonance, $X_L=X_C$ 

$\Rightarrow Z=R=20\mathrm{\Omega }$ 

$I=\frac{V}{Z}=\frac{200}{20}$ 

$\Rightarrow I=10A$ 

Average power, $P=I^{2}R$ 

$\Rightarrow P={10}^2\times 20$ 

$\Rightarrow P=2000W$ 

Therefore, the average power transferred is $2000W$.

10. A radio can tune over the frequency range of a portion of $MW$ broadcast band: ($800kHz$ to $1200kHz$). If its LC circuit has an effective inductance of $200\mu H$, what must be the range of its variable capacitor?

(Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radio wave.)

Ans: It is given that,

The range of frequency$(f)$ of a radio is $800kHz$ to $1200kHz$.

Effective inductance of the circuit, $L=200\mu H=200\times {10}^{-6}H$ 

It is known that,

Capacitance of variable capacitor for $f_1$ is $C_1=\frac{1}{{{\omega }_1}^{2}L}$ 

Where,

${\omega }_1$ is the angular frequency for capacitor for $f_1=2\pi f_1$ 

$$\Rightarrow {\omega }_1=2\times 3.14\times 800\times {10}^{3}rad/s$$ 

$\Rightarrow C_1=\frac{1}{{(2\times 3.14\times 800\times {10}^3)}^2\times 200\times {10}^{-6}}$ 

$\Rightarrow C_1=1.9809\times {10}^{-10}F$ 

$\Rightarrow C_1=198.1pF$

$C_2=\frac{1}{{{\omega }_2}^{2}L}$

$\Rightarrow C_2=\frac{1}{{(2\times 3.14\times 1200\times {10}^3)}^2\times 200\times {10}^{-6}}$

$\Rightarrow C_2=0.8804\times {10}^{-10}F$

$\Rightarrow C_2=88.04pF$

Therefore, the range of the variable capacitor is from $88.04pF$ to $198.1pF$.

11. Figure shows a series LCR circuit connected to a variable frequency $230V$ source. $L=5.0H$, $C=80\mu F$, $R=40\mathrm{\Omega }$ .

(Image will be Uploaded Soon)

a) Determine the source frequency which drives the circuit in resonance.

Ans: It is given that,

Voltage, $V=230V$ 

Inductance, $L=5.0H$ 

Capacitance, $C=80\mu F=80\times {10}^{-6}F$ 

Resistance, $R=40\mathrm{\Omega }$ 

It is known that,

Source frequency at resonance$=\frac{1}{\sqrt{LC}}$ 

$\Rightarrow \frac{1}{\sqrt{5\times 80\times {10}^{-6}}}=50rad/s$ 

Therefore, the source frequency of the circuit in resonance is $50rad/s$.

b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

Ans: It is known that,

At resonance, Impedance,$Z=$ Resistance,$R$ 

$\Rightarrow Z=R=40\mathrm{\Omega }$ 

$I=\frac{V}{Z}$ 

$\Rightarrow I=\frac{230}{40}=5.75A$ 

Amplitude, $I_0=1.414\times I$

$\Rightarrow I_0=1.414\times 5.75$

$\Rightarrow I_0=8.13A$

Therefore, the impedance of the circuit is $40\mathrm{\Omega }$ and the amplitude of current at resonating frequency is $8.13A$.

c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

Ans: It is known that,

Potential drop, $V=IR$ 

Across resistor, $V_R=IR$

$\Rightarrow V_R=5.75\times 40=230V$ 

Across capacitor, $V_C=I{X}_C=\frac{I}{\omega C}$

$\Rightarrow V_C=5.75\times \frac{1}{50\times 80\times {10}^{-6}}$ 

$\Rightarrow V_C=1437.5V$ 

Across Inductor, $V_L=I{X}_L=I\omega L$ 

$\Rightarrow V_L=5.75\times 50\times 5$ 

$\Rightarrow V_L=1437.5V$

Across LC combination, $V_{LC}=I(X_L-X_C)$ 

At resonance, $X_L=X_C$ 

$\Rightarrow V_{LC}=0$ 

Therefore, the rms potential drop across Resistor is $230V$, Capacitor is $1437.5V$, Inductor is $1437.5V$ and the potential drop across LC combination is zero at resonating frequency.

12. An LC circuit contains a $20mH$ inductor and a $50\mu F$ capacitor with an initial charge of $10mC$. The resistance of the circuit is negligible. Let the instant the circuit is closed be $t=0$.

a) What is the total energy stored initially? Is it conserved during LC oscillations?

Ans: It is given that,

Inductance of the inductor, $L=20mH=20\times {10}^{-3}H$ 

Capacitance of the capacitor, $C=50\mu F=50\times {10}^{-6}F$ 

Initial charge on the capacitor, $Q=10mC=10\times {10}^{-3}C$ 

It is known that,

Total energy stored initially in the circuit, $E=\frac{1}{2}\frac{Q^2}{C}$ 

$\Rightarrow E=\frac{{(10\times {10}^{-3})}^2}{2\times 50\times {10}^{-6}}=1J$ 

Therefore, the total energy stored in the LC circuit will be conserved because there is no resistor $(R=0)$ connected in the circuit.

b) What is the natural frequency of the circuit?

Ans: It is known that,

Natural frequency of the circuit, $\nu =\frac{1}{2\pi \sqrt{LC}}$ 

$\Rightarrow \nu =\frac{1}{2\pi \sqrt{20\times {10}^{-3}\times 50\times {10}^{-6}}}$ 

$\Rightarrow \nu =\frac{{10}^3}{2\pi }=159.24Hz$

Natural angular frequency, ${\omega }_r=\frac{1}{\sqrt{LC}}$ 

$\Rightarrow {\omega }_r=\frac{1}{\sqrt{20\times {10}^{-3}\times 50\times {10}^{-6}}}$ 

$\Rightarrow {\omega }_r=\frac{1}{\sqrt{{10}^{-6}}}={10}^{3}rad/s$

Therefore, the natural frequency is $159.24Hz$ and the natural angular frequency is ${10}^{3}rad/s$.

c) At what time is the energy stored $(i)$ completely electrical (i.e., stored in the capacitor)? $(ii)$ completely magnetic (i.e., stored in the inductor)?

Ans:

(i) Completely electrical:

It is known that,

Time period for LC oscillations, $T=\frac{1}{\nu }$ 

$\Rightarrow T=\frac{1}{159.24}=6.28ms$

Total charge on the capacitor at time $t$, $Q^{\prime }=Q\cos \left(\frac{2\pi }{T}t\right)$  

If energy stored is electrical, $Q^{\prime }=\pm Q$ 

Therefore, it can be inferred that the energy stored in the capacitor is completely electrical at time, $t=0,\frac{T}{2},T,\frac{3T}{2},..............$ where, $T=6.3ms$.

(ii) Completely magnetic:

Magnetic energy is maximum, when electrical energy $Q^{\prime }$ is equal to $0$.

Therefore, it can be inferred that the energy stored is completely magnetic at time, $t=\frac{T}{4},\frac{3T}{4},\frac{5T}{4},...........$ where, $T=6.3ms$.

d) At what times is the total energy shared equally between the inductor and the capacitor?

Ans: Consider, $Q^{\prime }$ be the charge on capacitor when total energy is equally shared between the capacitor and the inductor at time $t$.

When total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor$={}^1/{}_2($maximum energy$)$.

$\Rightarrow \frac{1}{2}\frac{{\left(Q^{\prime }\right)}^2}{C}=\frac{1}{2}\left(\frac{1}{2}\frac{Q^2}{C}\right)$ 

$\Rightarrow \frac{1}{2}\frac{{\left(Q^{\prime }\right)}^2}{C}=\frac{1}{4}\frac{Q^2}{C}$

$\Rightarrow Q^{\prime }=\frac{Q}{\sqrt{2}}$ 

It is known that, $Q^{\prime }=Q\cos \frac{2\pi }{T}t$ 

$\Rightarrow \frac{Q}{\sqrt{2}}=Q\cos \frac{2\pi }{T}t$ 

$\Rightarrow \cos \frac{2\pi }{T}t=\frac{1}{\sqrt{2}}=\cos (2n+1)\frac{\pi }{4};$ $n=0,1,2,3.....$ 

$\Rightarrow t=(2n+1)\frac{T}{8}$ 

Therefore, the total energy is equally shared between the inductor and the capacitor at the time,$t=\frac{T}{8},\frac{3T}{8},\frac{5T}{8}...........$.

e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

Ans: If a resistor is included in the circuit, then the total initial energy gets dissipated as heat energy in the circuit. The LC oscillation gets damped due to the resistance.

13. A coil of inductance $0.5H$ and resistance $100\mathrm{\Omega }$ is connected to a $240V,50Hz$ ac supply.

a) What is the maximum current in the coil?

Ans: It is given that,

Inductance of the inductor, $L=0.5H$ 

Resistance of the resistor, $R=100\mathrm{\Omega }$ 

Potential of the supply voltage, $V=240V$ 

Frequency of the supply, $\nu =50Hz$ 

It is known that,

Peak voltage, $V_0=\sqrt{2}V$ 

$\Rightarrow V_0=\sqrt{2}\times 240$

$\Rightarrow V_0=339.41V$

Angular frequency of the supply, $\omega =2\pi \nu$ 

$\Rightarrow \omega =2\pi \times 50=100\pi rad/s$ 

Maximum current in the circuit, $I_0=\frac{V_0}{\sqrt{R^2+{\omega }^2{L}^2}}$ 

$\Rightarrow I_0=\frac{339.41}{\sqrt{{(100)}^2+{(100\pi )}^2{(0.50)}^2}}=1.82A$

Therefore, the maximum current in the coil is $1.82A$. 

b) What is the time lag between the voltage maximum and the current maximum?

Ans: It is known that,

Equation for voltage, $V=V_0\cos \omega t$ 

Equation for current, $I=I_0\cos (\omega t-\varphi )$ 

Where,

$\varphi$ is the phase difference between voltage and current.

At time $t=0$, $V=V_0$ (voltage is maximum)

If $\omega t-\varphi =0$ i.e., at $t=\frac{\varphi }{\omega }$ , $I=I_0$ (current is maximum)

Therefore, the time lag between maximum voltage and maximum current is $\frac{\varphi }{\omega }$ .

$\Rightarrow \tan \varphi =\frac{\omega L}{R}$ 

$\Rightarrow \tan \varphi =\frac{2\pi \times 50\times 0.5}{100}=1.57$

$\Rightarrow \varphi ={\tan }^{-1}(1.57)$ 

$\Rightarrow \varphi ={57.5}^{\circ }=\frac{57.5\pi }{180}rad$ 

Time lag, $t=\frac{\varphi }{\omega }$ 

$\Rightarrow t=\frac{57.5\pi }{180\times 2\pi \times 50}$ 

$\Rightarrow t=3.19\times {10}^{-3}s$

$\Rightarrow t=3.2ms$

Therefore, the time lag between the maximum voltage and maximum current is $3.2ms$.

14. Obtain the answers $(a)$ to $(b)$ in Exercise $13$ if the circuit is connected to a high frequency supply $(240V,10kHz)$. Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?

Ans: It is given that,

Inductance of the inductor, $L=0.5H$ 

Resistance of the resistor, $R=100\mathrm{\Omega }$ 

Potential of the supply voltage, $V=240V$ 

Frequency of the supply, $\nu =10kHz={10}^{4}Hz$

Angular frequency, $\omega =2\pi \nu =2\pi \times {10}^{4}rad/s$ 

Peak Voltage, $V_0=V\sqrt{2}=110\sqrt{2}V$ 

1. Maximum current, $I_0=\frac{V_0}{\sqrt{R^2+{\omega }^2{C}^2}}$  

$\Rightarrow I_0=\frac{240\sqrt{2}}{\sqrt{{(100)}^2+{(2\pi \times {10}^4)}^2\times {(0.5)}^2}}=1.1\times {10}^{-2}A$

Therefore, the maximum current in the coil is $1.1\times {10}^{-2}A$.  

2. The time lag between maximum voltage and maximum current is $\frac{\varphi }{\omega }$ .

For phase difference $\varphi$:$\tan \varphi =\frac{\omega L}{R}$ 

$\Rightarrow \tan \varphi =\frac{2\pi \times {10}^4\times 0.5}{100}=100\pi$

$\Rightarrow \varphi ={\tan }^{-1}(100\pi )$ 

$\Rightarrow \varphi ={89.82}^{\circ }=\frac{89.82\pi }{180}rad$ 

Time lag, $t=\frac{\varphi }{\omega }$ 

$\Rightarrow t=\frac{89.82\pi }{180\times 2\pi \times {10}^4}$ 

$\Rightarrow t=25\times {10}^{-6}s$

$\Rightarrow t=25\mu s$

Therefore, the time lag between the maximum voltage and maximum current is $25\mu s$.

It can be observed that $I_0$ is very small in this case.

Thus, at high frequencies, the inductor amounts to an open circuit.

In a dc circuit, after a steady state is achieved, $\omega =0$. Thus, inductor $L$ behaves like a pure conducting object.

15. A $100\mu F$ capacitor in series with a $40\mathrm{\Omega }$ resistance is connected to a $110V,60Hz$ supply.

a) What is the maximum current in the circuit?

Ans: It is given that,

Capacitance of the capacitor, $C=100\mu F=100\times {10}^{-6}F$ 

Resistance of the resistor, $R=40\mathrm{\Omega }$ 

Supply voltage, $V=110V$ 

Frequency oscillations, $\nu =60Hz$ 

Angular frequency, $\omega =2\pi \nu =2\pi \times 60rad/s$ 

It is known that,

For a RC circuit, Impedance: $Z=\sqrt{R^2+\frac{1}{{\omega }^2{C}^2}}$ 

Peak Voltage, $V_0=V\sqrt{2}=110\sqrt{2}V$ 

Maximum current; $I_0=\frac{V_0}{Z}$ 

$\Rightarrow I_0=\frac{V_0}{\sqrt{R^2+\frac{1}{{\omega }^2{C}^2}}}$ 

$\Rightarrow I_0=\frac{110\sqrt{2}}{\sqrt{{(40)}^2+\frac{1}{{(120\pi )}^2{({10}^{-4})}^2}}}$

$$\Rightarrow I_0=\frac{110\sqrt{2}}{\sqrt{1600+\frac{1}{{(120\pi )}^2{({10}^{-4})}^2}}}=3.24A$$

Therefore, the maximum current in the circuit is $3.24A$.

b) What is the time lag between the current maximum and the voltage maximum?

Ans: It is known that,

In a capacitor circuit, the voltage lags behind the current by a phase angle of $\varphi$.

$\tan \varphi =\frac{\frac{1}{\omega C}}{R}=\frac{1}{\omega CR}$ 

$\Rightarrow \tan \varphi =\frac{1}{120\pi \times {10}^{-4}\times 40}=0.6635$

$\Rightarrow \varphi ={\tan }^{-1}(0.6635)$ 

$\Rightarrow \varphi ={33.56}^{\circ }=\frac{33.56\pi }{180}rad$ 

It is known that,

Time lag, $t=\frac{\varphi }{\omega }$ 

$\Rightarrow t=\frac{33.56\pi }{180\times 120\pi }$ 

$\Rightarrow t=1.55\times {10}^{-3}s$

$\Rightarrow t=1.55ms$

Therefore, the time lag between maximum current and maximum voltage is $1.55ms$.

16. Obtain the answers to $(a)$ and $(b)$ in Exercise $15$ if the circuit is connected to a $$\mathbf{110}\mathbf{V},\mathbf{12}\mathbf{kHz}$$ supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.

Ans: It is given that,

Capacitance of the capacitor, $C=100\mu F=100\times {10}^{-6}F$ 

Resistance of the resistor, $R=40\mathrm{\Omega }$ 

Supply voltage, $V=110V$ 

Frequency oscillations, $\nu =12kHz=12\times {10}^{3}Hz$ 

Angular frequency, $\omega =2\pi \nu =2\pi \times 12\times {10}^{3}rad/s=24\pi \times {10}^{3}rad/s$ 

Peak Voltage, $V_0=V\sqrt{2}=110\sqrt{2}V$ 

1. It is known that,

For a RC circuit, Impedance: $Z=\sqrt{R^2+\frac{1}{{\omega }^2{C}^2}}$ 

Maximum current; $I_0=\frac{V_0}{Z}$ 

$\Rightarrow I_0=\frac{V_0}{\sqrt{R^2+\frac{1}{{\omega }^2{C}^2}}}$ 

$\Rightarrow I_0=\frac{110\sqrt{2}}{\sqrt{{(40)}^2+\frac{1}{{(24\pi \times {10}^3)}^2{({10}^{-4})}^2}}}$

$$\Rightarrow I_0=\frac{110\sqrt{2}}{\sqrt{1600+{\left(\frac{10}{24\pi }\right)}^2}}=3.9A$$

Therefore, the maximum current in the circuit is $3.9A$.

2. It is known that,

In a capacitor circuit, the voltage lags behind the current by a phase angle of $\varphi$.

$\tan \varphi =\frac{\frac{1}{\omega C}}{R}=\frac{1}{\omega CR}$ 

$\Rightarrow \tan \varphi =\frac{1}{24\pi \times {10}^3\times {10}^{-4}\times 40}=\frac{1}{96\pi }$

$\Rightarrow \varphi ={\tan }^{-1}(\frac{1}{96\pi })$ 

$\Rightarrow \varphi ={0.2}^{\circ }=\frac{0.2\pi }{180}rad$ 

It is known that,

Time lag, $t=\frac{\varphi }{\omega }$ 

$\Rightarrow t=\frac{0.2\pi }{180\times 24\pi \times {10}^3}$ 

$\Rightarrow t=0.04\times {10}^{-6}s$

$\Rightarrow t=0.04\mu s$

Therefore, the time lag between maximum current and maximum voltage is $0.04\mu s$.

It can be concluded that $\varphi$ tends to become zero at high frequencies. At a high frequency, capacitor $C$ acts as a conductor.

In a dc circuit, after the steady state is achieved, $\omega =0$. Therefore, capacitor $C$ amounts to an open circuit.

17. Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if three elements, L,C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 11 for this frequency.

Ans: It is given that,

An inductor $(L)$, a capacitor $(C)$ and a resistor $(R)$ is connected in parallel with each other in a circuit where,

Inductance, $L=5.0H$ 

Capacitance, $C=80\mu F=80\times {10}^{-6}F$ 

Resistance, $R=40\mathrm{\Omega }$ 

Potential of the voltage source, $V=230V$ 

It is known that,

Impedance $(Z)$ of the given LCR circuit is given as:

$\frac{1}{Z}=\sqrt{\frac{1}{R^2}+{(\frac{1}{\omega L}-\omega C)}^2}$ 

Where,

$\omega$ is the angular frequency

At resonance: $\frac{1}{\omega L}-\omega C=0$ 

$\Rightarrow \omega =\frac{1}{\sqrt{LC}}$ 

$\Rightarrow \omega =\frac{1}{\sqrt{5\times 80\times {10}^{-6}}}=50rad/s$

Therefore, the magnitude of $Z$ is maximum at $50rad/s$ and the total current is minimum. 

Rms current flowing through inductor $L$: $I_L=\frac{V}{\omega L}$ 

$\Rightarrow I_L=\frac{230}{50\times 5}=0.92A$ 

Rms current flowing through capacitor $C$: $I_C=\frac{V}{\frac{1}{\omega C}}=\omega CV$ 

 $\Rightarrow I_C=50\times 80\times {10}^{-6}\times 230=0.92A$

Rms current flowing through resistor $R$: $I_R=\frac{V}{R}$ 

$\Rightarrow I_R=\frac{230}{40}=5.75A$ 

The current RMS value in the inductor is $0.92A$, in the capacitor is $0.92A$ and in the resistor is $5.75A$. 

18. A circuit containing an $80mH$ inductor and a $60\mu F$ capacitor in series is connected to a $$\mathbf{230}\mathbf{V},\mathbf{50}\mathbf{Hz}$$ supply. The resistance of the circuit is negligible. 

a) Obtain the current amplitude and rms values. 

Ans: It is given that,

Inductance, $L=80mH=80\times {10}^{-3}H$ 

Capacitance, $C=60\mu F=60\times {10}^{-6}F$ 

Supply voltage, $V=230V$ 

Frequency, $\nu =50Hz$ 

Angular frequency, $\omega =2\pi \nu =100\pi rad/s$ 

Peak voltage, $V_0=V\sqrt{2}=230\sqrt{2}V$ 

It is known that, 

Maximum current: $I_0=\frac{V_0}{(\omega L-\frac{1}{\omega C})}$ 

$$\Rightarrow I_0=\frac{230\sqrt{3}}{(100\pi \times 80\times {10}^{-3}-\frac{1}{100\pi \times 60\times {10}^{-6}})}$$

$$\Rightarrow I_0=\frac{230\sqrt{3}}{(8\pi -\frac{1000}{6\pi })}=-11.63A$$

The negative sign is because $\omega L<\frac{1}{\omega C}$ 

Amplitude of maximum current, $\left|I_0\right|=11.63A$ 

$\Rightarrow I=\frac{I_0}{\sqrt{2}}=\frac{-11.63}{\sqrt{2}}$ 

$\Rightarrow I=-8.22A$, which is the rms value of current.

b) Obtain the rms values of potential drops across each element.

Ans: It is known that,

Potential difference across the inductor, $V_L=I\times \omega L$ 

$\Rightarrow V_L=8.22\times 100\pi \times 80\times {10}^{-3}$

$\Rightarrow V_L=206.61V$

Potential difference across the capacitor, $V_C=I\times \frac{1}{\omega C}$ 

$\Rightarrow V_C=8.22\times \frac{1}{100\pi \times 60\times {10}^{-6}}$

$\Rightarrow V_C=436.3V$, which is the rms value of potential drop.

c) What is the average power transferred to the inductor? 

Ans: Average power transferred to the inductor is zero as actual voltage leads the current by $\frac{\pi }{2}$.

d) What is the average power transferred to the capacitor? 

Ans: Average power transferred to the capacitor is zero as actual voltage lags the current by $\frac{\pi }{2}$.

e) What is the total average power absorbed by the circuit? (‘Average’ implies ‘averaged over one cycle’.)

Ans: The total average power absorbed (averaged over one cycle) is zero.

19. Suppose the circuit in Exercise $18$  has a resistance of $15\mathrm{\Omega }$ . Obtain the average power transferred to each element of the circuit, and the total power absorbed.

Ans:  It is given that,

Average power transferred to the resistor $$=788.44W$$ 

Average power transferred to the capacitor $=0W$ 

Total power absorbed by the circuit $$=788.44W$$

Inductance of inductor, $$L=80mH=80\times {10}^{-3}H$$ 

Capacitance of capacitor, $$C=60\mu F=60\times {10}^{-6}F$$ 

Resistance of resistor, $$R=15\mathrm{\Omega }$$ 

Potential of voltage supply, $$V=230V$$ 

Frequency of signal, $$\nu =50Hz$$ 

Angular frequency of signal, $$\omega =2\pi \nu =2\pi \times \left(50\right)=100\pi rad/s$$ 

It is known that,

Impedance, $Z=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}$

$\Rightarrow Z=\sqrt{{(15)}^2+{(100\pi (80\times {10}^{-3})-\frac{1}{(100\pi \times 60\times {10}^{-6})})}^2}$ 

$$\Rightarrow Z=\sqrt{{(15)}^2+{(25.12-53.08)}^2}=31.728\mathrm{\Omega }$$

Now,

$I=\frac{V}{Z}$ 

$\Rightarrow I=\frac{230}{31.728}=7.25A$ 

The elements are connected in series to each other. Therefore, impedance of the circuit is given as current flowing in the circuit,

Average power transferred to resistance is given as: $P_R=I^{2}R$