# Alternating Current (AC) Numericals – Class 12 Physics

### Formulae:

1. The instantaneous value of alternating current is

I  =  Io sin$\omega$t.

Where Io is the maximum (peak) value of current.

And the instantaneous value of alternating voltage is

E  =  Eo sin$\omega$t.

Where Eo is the maximum (peak) value of voltage.

2. For certain phase angle ($\phi$)

E  =  Eo sin ωt and I = Io sin ($\omega$t $\pm$ $\phi$)

where $\phi$ is the phase angle between emf and current.

3. Irms  =  $\frac{{{I}_{o}}}{\sqrt{2}}$ & Erms  =  $\frac{{{E}_{o}}}{\sqrt{2}}$

4. Phase angle ($\phi$) between current and voltage in AC circuit containing

5. Inductive reactance (i.e. resistance offered by inductor),

XL =$\omega$L = 2$\pi$f L

Capacitive reactance (i.e. resistance offered by capacitor),

Xc = $\frac{1}{\omega \text{ }C}$ = $\frac{1}{2\pi f\text{ }C}$

6. Impedance of AC circuit (i.e. total resistance offered by AC circuit)

Z  =  $\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

7. Voltage drop across Resistor, VR = IR

Voltage drop across inductor, VL = I XL

Voltage drop across capacitor, VC = I Xc

8. Circuit Formula E = IZ (in AC) [E = V = I Req , for r = 0 (in D.C)]

9. Electrical Resonance

At resonance, I is maximum and hence Z is minimum as E= IZ

We have, Impedance, Z =  $\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

i.e. at resonance, XL = Xc

Resonating frequency, fo  =  $\frac{1}{2\pi }$  $\sqrt{\frac{1}{LC}}$

10. Quality factor = $\frac{{{V}_{L}}\text{ }or\text{ }{{V}_{C}}}{{{V}_{R}}}$

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

11. Average power consumed in AC circuit

Pavg  =  Irms Erms . cos$\phi$

12. Power factor, cos$\phi$ = $\frac{R}{Z}$ = $\frac{R}{\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}}$

13. A.C. Circuit containing inductor with some internal resistance (rL)

(i) Impedance of AC circuit, Z  =  $\sqrt{{{(R+{{r}_{L}})}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

(ii) Impedance of AC circuit containing L & R, Z  =  $\sqrt{{{(R+{{r}_{L}})}^{2}}+{{X}_{L}}^{2}}$

(iii) Total resistance offered by inductor, ZL = $\sqrt{{{r}_{L}}^{2}+{{X}_{L}}^{2}}$

(iv) Voltage drop across inductor, VL= I ZL

(v) Phase angle ($\phi$), Tan$\phi$ = $\frac{{{Z}_{L}}}{R}$

Type-1:

Q.1. Alternating voltage in an ac circuit is represented by V = 100$\sqrt{\mathbf{2}}$ sin (100$\pi$t) volts. Find the root mean square value and the frequency.

Solution:

Here,

Alternating voltage, V = 100$\sqrt{2}$ sin (100$\pi$t)

(i) Root mean square value of voltage, Vrms = ?

(ii) Frequency, f = ?

Comparing above equation with V = Vo sin$\omega$t, we get

(i) Vo = 100$\sqrt{2}$

Now,

Vrms =  $\frac{{{V}_{o}}}{\sqrt{2}}$ = $\frac{100\sqrt{2}}{\sqrt{2}}$ = 100V

Also,

(ii) $\omega$ = 100$\pi$

Or, 2$\pi$f = 100$\pi$

$\therefore$ Frequency, f  = 50 Hz

Type – 2:

Q. 2. A circuit consists of a capacitor of 2$\mu$F and a resistor of 1000$\Omega$. An alternating emf of 12V and frequency 50Hz is applied. Find the voltage across capacitor and the phase angle between the applied emf and current.

Solution:

Here,

Capacitance, C = 2 $\mu$F = 2×10–6 F

Resistance, R = 1000$\Omega$

Emf, E = 12V,

Frequency, f = 50Hz

(i) Voltage drop across capacitor, VC = ?

(ii) Phase angle, $\phi$ = ?

We have, capacitive reactance,

XC = $\frac{1}{\omega c}$ = $\frac{1}{2\pi f\text{ }C}$ = $\frac{1}{2\pi \times 50\times 2\times {{10}^{-6}}}$ = 1591.55$\Omega$

Using circuit formula,

E = I Z

Or, I = $\frac{E}{Z}$ = $\frac{E}{\sqrt{{{R}^{2}}+{{X}_{C}}^{2}}}$

Or, I = $\frac{12}{\sqrt{{{1000}^{2}}+{{(1591.55)}^{2}}}}$ = 6.38×10–3A

(i) VC = I Xc = 6.38×10–3×1591.55 = 10.16V

(ii) Tan$\phi$ = $\frac{{{X}_{C}}}{R}$

Or, $\phi$ = Tan–1 $\left( \frac{{{X}_{C}}}{R} \right)$

Or, $\phi$ = Tan–1 $\left( \frac{1591.55}{1000} \right)$ = 57.86o

Q.3.  A coil of inductance 0.1 H and negligible resistance is in series with a resistance 40$\Omega$. A supply voltage of 50v (rms) is connected to them. If the voltage across L is equal to that across R, calculate the voltage across the inductor and frequency of the supply.

Solution:

Here,

Inductance, L= 0.1 H

Resistance, R = 40$\Omega$

Emf, Erms = 50V,

Voltage across inductor, VL= Voltage across resistor, VR

(i) Voltage across inductor, VL = ?

(ii) Frequency, f = ?

We have,

VL= VR

Or, I XL= I R

Or, XL = R = 40$\Omega$

Using circuit formula,

E = I Z

Or, I = $\frac{E}{Z}$ = $\frac{E}{\sqrt{R2+{{X}_{L}}^{2}}}$

Or, I = $\frac{50}{\sqrt{{{40}^{2}}+{{(40)}^{2}}}}$ = 0.884A

Now,

(i) VL= I XL = 0.884 ×40 = 35.36 V

Again

(ii) XL =R

Or, $\omega$L=  R

Or, 2$\pi$f L=  R

Or, f =  $\frac{R}{2\pi L}$

Or, f =  $\frac{40}{2\pi \times 0.1}$

$\therefore$ Frequency, f = 63.66 Hz

Q.4. A coil of inductance L = 0.1 H and negligible resistance is in series with a resistance 40$\Omega$. A supply voltage of 40v (rms) is connected to them. If the voltage across L is equal to that across R, calculate the voltage across R and the frequency of the supply.

Solution:

Here,

Inductance, L= 0.1 H

Resistance, R = 40$\Omega$

Emf, Erms = 40 V

Voltage across inductor, VL= Voltage across resistor, VR

(i) Voltage across inductor, VL = ?

(ii) Frequency, f = ?

We have,

VL= VR

Or, I XL= I R

Or, XL = R = 40$\Omega$

Using circuit formula,

E = I Z

Or, I = $\frac{E}{Z}$ = $\frac{E}{\sqrt{{{R}^{2}}+{{X}_{L}}^{2}}}$

Or, I = $\frac{40}{\sqrt{{{40}^{2}}+{{40}^{2}}}}$ = 0.707A

Now,

(i) VR = I R = 0.707 ×40 = 28.28 V

Again

(ii) XL = R

Or, $\omega$L= R

Or, 2$\pi$f L = R

Or, f =$\frac{R}{2\pi L}$

$\therefore$ f =$\frac{40}{2\pi \times 0.1}$ = 63.66 Hz

Q.5. L-C-R alternating current series circuits of L = 1H, C = 1$\mu$F and R = 100$\Omega$ are connected in series with a source of frequency 50 Hz. What is the phase shift between current and voltage?

Solution:

Here,

Inductance, L = 1 H

Capacitance, C = 1 $\mu$F = 1×10–6

Resistance, R = 100$\Omega$

Frequency, f = 50Hz

Phase angle, $\phi$ = ?

We have,

Inductive reactance, XL =$\omega$L = 2$\pi$f L = 2$\pi$×50×1 = 314.16$\Omega$

Capacitive reactance, XC = $\frac{1}{\omega c}$ = $\frac{1}{2\pi f\text{ }C}$= $\frac{1}{2\pi \times 50\times 1\times {{10}^{-6}}}$ = 3183.1$\Omega$

Impedance, Z = $\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

Or, Z = $\sqrt{{{100}^{2}}+{{(314.16-3183.1)}^{2}}}$ = 2870.68$\Omega$

(ii) Tan$\phi$ = $\frac{{{X}_{L}}-{{X}_{C}}}{R}$

Or, $\phi$ = Tan–1 $\left( \frac{{{X}_{L}}-{{X}_{C}}}{R} \right)$

Or, $\phi$ = Tan–1 $\left( \frac{314.16-3183.1}{100} \right)$ = – 88.7o

i.e. current leads by voltage by phase angle 88.7o.

Q.6.  A 100 V, 50 Hz AC source is connected to an LCR circuit containing L = 8.1 mH, c = 12.5$\mu$F and R = 10$\Omega$all connected in series. Find the potential difference across the resistor.

Solution:

Here,

Emf, E = 100V

Frequency, f = 50 Hz

Inductance, L = 8.1 mH = 8.1 ×10–3 H

Capacitance, C = 12.5 $\mu$F = 12.5 ×10–6F

Resistance, R = 10$\Omega$

Voltage drop resistor, VR = ?

We know, VR = IR ……………(i)

Inductive reactance, XL =$\omega$L = 2$\pi$f L = 2$\pi$×50×8.1 ×10–3 = 2.54$\Omega$

Similarly,

We have, XC = $\frac{1}{\omega c}$ = $\frac{1}{2\pi f\text{ }C}$= $\frac{1}{2\pi \times 50\times 12.5\times {{10}^{-6}}}$ = 254.65$\Omega$

Impedance, Z = $\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

Or, Z = $\sqrt{{{10}^{2}}+{{(2.54-254.65)}^{2}}}$ = 252.31$\Omega$

Using circuit formula,

E = I Z

Or, I = $\frac{E}{Z}$ =  $\frac{100}{252.31}$ = 0.396 A

From equation (i),

VR = I R = 0.396 ×10 = 3.96 V

Q.7. An alternating voltage 10v (rms) and 4 KHz frequency is applied to a resistor of resistance 5$\Omega$ in series with a capacitor of capacitance 10$\mu$F. Calculate the r.m.s. potential differences across the resistor and the capacitor.

Solution:

Here,

Emf, Erms = 10V

Frequency, f = 4 KHz = 4×103 Hz

Resistance, R = 5$\Omega$

Capacitance, C = 10 $\mu$F = 10 ×10–6F

(i) VR (rms) = ?

(ii) VC (rms) = ?

We have,

Capacitive reactance, XC = $\frac{1}{\omega c}$

= $\frac{1}{2\pi f\text{ }C}$

= $\frac{1}{2\pi \times 4\times {{10}^{3}}\times 10\times {{10}^{-6}}}$ = 3.979$\Omega$

Impedance, Z =  $\sqrt{{{R}^{2}}+{{X}_{C}}^{2}}$

Or, Z = $\sqrt{{{5}^{2}}+{{3.979}^{2}}}$ = 6.39$\Omega$

Using circuit formula,

Erms = Irms

Or, Irms = $\frac{{{E}_{rms}}}{Z}$ =  $\frac{10}{6.39}$ = 1.565 A

Now

(i) VR (rms) = Irms R = 1.565 ×5 = 7.825 V

(ii) VC (rms) = Irms XC = 1.565 ×3.979 = 6.227 V

Q.8. In a series LCR circuit, R = 25$\Omega$, L = 30 mH and C = 10 $\mu$F and these elements are connected to 240V ac (rms) 50 Hz source. Calculate the current in the circuit and voltmeter reading across a capacitor.

Solution:

Here,

Resistance, R = 25$\Omega$

Inductance, L = 30 mH = 30 ×10–3 H

Capacitance, C = 10 $\mu$F = 10 ×10–6F

Emf, Erms = 240V

Frequency, f = 50 Hz

(i) Current, I = ?

(ii) VC = ?

We have,

Inductive reactance,

XL =$\omega$L = 2$\pi$f

L = 2$\pi$×50×30 ×10–3 = 9.42$\Omega$

Similarly,

We have, XC = $\frac{1}{\omega c}$ = $\frac{1}{2\pi f\text{ }C}$

= $\frac{1}{2\pi \times 50\times 10\times {{10}^{-6}}}$ = 318.31$\Omega$

Impedance, Z =  $\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

Or, Z = $\sqrt{{{25}^{2}}+{{(9.42-318.31)}^{2}}}$ = 309.9$\Omega$

Using circuit formula,

E = I Z

Or, I = $\frac{E}{Z}$ =  $\frac{240}{309.9}$ = 0.774 A

Also,

VC = I Xc = 0.774 ×318.31 = 246.37 V

Q.9. A circuit consists of a capacitor of 10 $\mu$F and a resistor of 1000$\Omega$. An alternating voltage of  12 V (rms) and frequency 50Hz is applied. Calculate the current flowing and voltage across the capacitor.

Solution:

Here,

Capacitance, C = 10 $\mu$F = 10×10–6 F

Resistance, R = 1000$\Omega$

Emf, E = 12 V (rms)

Frequency, f = 50Hz

(i) Current, I = ?

(ii) Voltage across capacitor, VC = ?

We have, XC = $\frac{1}{\omega c}$ = $\frac{1}{2\pi f\text{ }C}$= $\frac{1}{2\pi \times 50\times 10\times {{10}^{-6}}}$ = 318.31$\Omega$

Using circuit formula,

E = I Z

Or, I = $\frac{E}{Z}$ = $\frac{E}{\sqrt{{{R}^{2}}+{{X}_{C}}^{2}}}$

Or, I = $\frac{12}{\sqrt{10002+(318.31)2}}$ = 0.0114 A

(i) VC = I Xc = 0.0114×318.31 = 3.64V

Q. 10. A circuit consists of a capacitor of 2$\mu$F and a resistor of 1000$\Omega$. An alternating emf of 12V (rms) and frequency 50Hz is applied. Find the current flowing, the voltage across capacitor and the phase angle between the applied emf and current.

Solution:

Here,

Capacitance, C = 2$\mu$F = 2×10–6 F

Resistance, R = 1000$\Omega$

Emf, E = 12V

Frequency, f = 50Hz

(i) Current, I = ?

(ii) Voltage across capacitor, VC = ?

(iii) Phase angle,$\phi$ = ?

We have, XC = $\frac{1}{\omega c}$ = $\frac{1}{2\pi f\text{ }C}$ = $\frac{1}{2\pi \times 50\times 2\times {{10}^{-6}}}$ = 1591.55$\Omega$

(i) Using circuit formula,

E = I Z

Or, I = $\frac{E}{Z}$ = $\frac{E}{\sqrt{{{R}^{2}}+{{X}_{C}}^{2}}}$

Or, I = $\frac{12}{\sqrt{{{1000}^{2}}+{{1591.55}^{2}}}}$ = 6.38×10–3A

(ii) VC = I Xc = 6.38×10–3×1591.55 = 10.16V

(iii) Tan$\phi$ = $\frac{{{X}_{C}}}{R}$

Or, $\phi$ = Tan–1 $\left( \frac{{{X}_{C}}}{R} \right)$

$\therefore$ $\phi$ = Tan–1 $\left( \frac{1591.55}{1000} \right)$ = 57.86o

Q.11. A constant A.C. supply is connected to series circuit consisting of a resistance of 300$\Omega$in series with a capacitance of 6.67 MF, the frequency of the supply being $\frac{\mathbf{3000}}{\mathbf{2}\pi }$ Hz. It is desired to reduce the current in the circuit to half of its value. Show how this could be done by placing an additional resistance.

Solution:

Here,

Resistance, R = 300$\Omega$

Capacitance, C = 6.67 MF = 6.67×106 F

Frequency, f = $\frac{3000}{2\pi }$ Hz

Let current in the first case, I1 = I

Then current in the first case, I2 = $\frac{I}{2}$

We have,

Capacitive reactance, XC = $\frac{1}{\omega c}$ = $\frac{1}{2\pi f\text{ }C}$= $\frac{1}{2\pi \times \frac{3000}{2\pi }\times 6.67\times {{10}^{6}}}$ = 5×10–11 $\Omega$

Case-1

Using circuit formula,

Emf, E = I1 Z1

Or, I =  = $\frac{E}{\sqrt{{{R}^{2}}+{{X}_{C}}^{2}}}$    ….(i)

Case-2

Emf, E = I2 Z2

Or, $\frac{I}{2}$ = $\frac{E}{{{Z}_{2}}}$ = $\frac{E}{\sqrt{{{(R+{{R}_{A}})}^{2}}+{{X}_{C}}^{2}}}$ ….(ii)

Dividing (i) by (ii), we get

2 = $\sqrt{\frac{{{(R+{{R}_{A}})}^{2}}+{{X}_{C}}^{2}}{{{R}^{2}}+{{X}_{C}}^{2}}}$

Or, 4 R2+4Xc2 = (R+RA)2 + Xc2

Or, (R+RA)2   = 4 R2+ 3 Xc

Or, RA = $\sqrt{4{{R}^{2}}+3{{X}_{C}}^{2}}$– R

Or, RA = $\sqrt{4\times {{300}^{2}}+3\times {{(5\times {{10}^{-11}})}^{2}}}$– 300

$\therefore$ RA = 300$\Omega$

Q.12. A coil having inductance and resistance is connected to an oscillator giving a fixed sinusoidal output voltage of 5 V rms. With the oscillator set at a frequency of 50 Hz, the rms current in the coil is 1 ampere and a frequency of 100 Hz, the rms current is 0.625 A. Determine the inductance of the coil.

Solution:

Here,

Emf, Erms = 5V

Frequency, f1 = 50 Hz Current, I1 = 1 A

Inductance, L = ?

Frequency, f2 = 100 Hz Current, I2 = 0.625 A

We have,

E = I1 Z1

Or, Z1 = $\frac{E}{{{I}_{1}}}$= $\frac{5}{1}$= 5$\Omega$

Again,

E = I2 Z2

Or, Z2 = $\frac{E}{{{I}_{2}}}$= $\frac{5}{0.625}$= 8$\Omega$

Also,

Z1 = $\sqrt{{{R}^{2}}+{{X}_{L}}{{_{1}}^{2}}}$

Or, 5 = $\sqrt{{{R}^{2}}+{{(2\pi {{f}_{1}}L)}^{2}}}$ ………..(i)

Similarly,

Z2 = $\sqrt{{{R}^{2}}+{{X}_{L2}}^{2}}$

Or, 8 = $\sqrt{{{R}^{2}}+{{(2\pi {{f}_{2}}L)}^{2}}}$ ………..(ii)

Squaring and then subtracting above equations

Or, 64 – 25 = R2+ (2$\pi$f2L)2 – R2– (2$\pi$f1L)2

Or, 39 = L2 [(2$\pi$f2)2– (2$\pi$f1)2]

Or, L = $\sqrt{\frac{39}{{{(2\pi \times 100)}^{2}}-{{(2\pi \times 50)}^{2}}}}$

$\therefore$ Inductance, L = 0.0115 H

Type – 3:

Q. 13. An ac source of 220V, 50Hz is connected to a series circuit containing a resistor R and inductor L and capacitor C. If R = 200 $\Omega$, L = 0.5 H and C = 10$\mu$F, (i) calculate the current in the circuit, (ii) the phase angle, and (iii) the power consumed in the circuit.

Solution:

Here,

Emf, E = 220V

Frequency, f = 50Hz

Resistance, R = 200$\Omega$

Inductance, L = 0.5 H

Capacitance, C = 10$\mu$F = 10×10–6 F,

(i) Current, I = ?

(ii) Phase angle,$\phi$= ?

(iii) Power, P = ?

We have,

Inductive reactance, XL =$\omega$

L = 2$\pi$f L

= 2$\pi$×50×0.5 = 157.08$\Omega$

Capacitive reactance, XC = $\frac{1}{\omega c}$

= $\frac{1}{2\pi f\text{ }C}$

= $\frac{1}{2\pi \times 50\times 10\times {{10}^{-6}}}$ = 318.31$\Omega$

Impedance, Z =  $\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

Or, Z = $\sqrt{{{200}^{2}}+{{(157.08-318.31)}^{2}}}$ = 256.90$\Omega$

Using circuit formula,

E = I Z

Or, I = $\frac{E}{Z}$ =  $\frac{220}{256.90}$ = 0.856 A

(ii) Tan$\phi$= $\frac{{{X}_{L}}-{{X}_{C}}}{R}$

Or, $\phi$ = Tan–1 $\left( \frac{{{X}_{L}}-{{X}_{C}}}{R} \right)$

Or, $\phi$ = Tan–1 $\left( \frac{157.08-318.31}{200} \right)$ = –38.87o

i.e. current leads by voltage by phase angle 38.87o.

(iii) P =  I E  cos$\phi$ [or Pavg  =  Irms Erms . cos φ]

Or, P =  I E  cos$\phi$= 0.856 ×220×cos (38.87o) = 146.62 W

Alternative Method

P =  I E $\frac{R}{Z}$= 0.856 ×220×$\frac{200}{256.9}$= 146.61 W

Q. 14. A.C. mains of 200 volts and 50 Hz are joined to a circuit containing an inductance of 100 mH and a resistance of 20$\Omega$in series. Calculate the power consumed.

Solution:

Here,

Emf, E = 200V

Frequency, f = 50Hz

Resistance, R = 20$\Omega$

Inductance, L = 100 mH = 100×10–3 H

Power, P = ?

We have,

Inductive reactance, XL =$\omega$L = 2$\pi$f L = 2$\pi$×50×100×10–3 = 31.42$\Omega$

Impedance, Z =  $\sqrt{{{R}^{2}}+{{X}_{L}}^{2}}$

Or, Z = $\sqrt{{{20}^{2}}+{{31.42}^{2}}}$ = 37.24$\Omega$

Using circuit formula,

E = I Z

Or, I =$\frac{E}{Z}$ =  $\frac{200}{37.24}$ = 5.37 A

Now power consumed,

P = I E  cos$\phi$

Or, P = I E$\frac{R}{Z}$ [As power factor, cos$\phi$= $\frac{R}{Z}$]

Or, P = 5.37 ×200×$\frac{20}{37.24}$= 576.86 watt

Q. 15. A circuit consists of an inductor of 200 $\mu$H and resistance of 10Ω in series with a variable capacitor and a 0.10 V (r.m.s.), 1.0 MHz supply. Calculate (i) the capacitance to give resonance (ii) the quality factor of the circuit at resonance.

Solution:

Here,

Inductance, L = 200 $\mu$H = 200×10–6 H

Resistance, R = 10$\Omega$

Emf, Erms = 0.1V

Frequency, f = 1 MHz = 1×106 Hz

(i) Capacitance, C = ?

(ii) Quality factor = ?

We have,

Inductive reactance, XL =$\omega$L = 2$\pi$f L = 2$\pi$×1×106 ×200×10–6 = 1256.64$\Omega$

(i) Resonating frequency, fo =  $\frac{1}{2\pi }$  $\sqrt{\frac{1}{LC}}$

$\therefore$  Capacitance, C = $\frac{1}{L{{(2\pi f)}^{2}}}$ = $\frac{1}{200\times {{10}^{-6}}{{(2\pi \times 1\times {{10}^{6}})}^{2}}}$ = 1.266×10–10 F

Alternative method

XL = XC

Or, XL = $\frac{1}{\omega C}$

Or, C = $\frac{1}{2\pi f\text{ }{{X}_{L}}}$

Or, C = $\frac{1}{2\pi \times 1\times {{10}^{6}}\times 1256.64}$

$\therefore$ C = 1.266×10–10 F

(ii) Quality factor = $\frac{{{V}_{L}}}{{{V}_{R}}}$

= $\frac{I{{X}_{L}}}{IR}$= $\frac{{{X}_{L}}}{R}$

=$\frac{1256.64}{10}$ = 125.66

Alternative Method:

Quality factor = $\frac{{{V}_{C}}}{{{V}_{R}}}$

= $\frac{I{{X}_{C}}}{IR}$

= $\frac{1}{\omega C\times R}$

= $\frac{1}{2\pi f\text{ }C\times R}$

= $\frac{1}{2\pi \times 1\times {{10}^{6}}\times 1.266\times {{10}^{-10}}\times 10}$

= 125.71

Alternative Method:

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

or, Q. factor =  $\frac{1}{10}$  $\sqrt{\frac{200\times {{10}^{-6}}}{1.266\times {{10}^{-10}}}}$ = 125.69

Q. 16. An inductor, a resistor and a capacitor are connected in series across an a.c. circuit. A voltmeter reads 60 V when connected across the inductor, 16 V across the resistor and 30V across the capacitor

(i) What will the voltmeter read when placed across the series circuit?

(ii) What is the power factor of the circuit?

Solution:

Here,

Voltage drop across inductor, VL = 60 V

Voltage drop across resistor, VR = 16 V

Voltage drop across capacitor, VC = 30 V

(i) Emf, E (i.e. VLCR) = ?

(ii) Power factor = ?

We have,

E = IZ

Or, E = I× $\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

Or, E =  $\sqrt{{{I}^{2}}{{R}^{2}}+{{(I{{X}_{L}}-I{{X}_{C}})}^{2}}}$

Or, E = $\sqrt{{{V}_{R}}^{2}+{{({{V}_{L}}-V{}_{C})}^{2}}}$

Or, E = $\sqrt{{{16}^{2}}+{{(60-30)}^{2}}}$

$\therefore$ E = 34 V

(ii) Power factor, (cos$\phi$) = $\frac{R}{Z}$

Or, Power factor = $\frac{IR}{IZ}$

Or, Power factor = $\frac{{{V}_{R}}}{E}$

Or, Power factor = $\frac{16}{34}$

$\therefore$ Power factor = 0.47

Q. 17. The maximum capacitance of a variable capacitor is 33 pF. What should be the self-inductance to be connected to this capacitor for the natural frequency of the LC circuit to be 810 KHz corresponding to A.m. broadcast band of Radio Nepal.

Solution:

Here,

Capacitance, C = 33 pF = 33×10–12 F

Resistance, R = 10$\Omega$

Inductance, L = ?

Frequency, f = 810 KHz = 810×103 Hz

This is the condition of resonance,

Resonating frequency, fo =  $\frac{1}{2\pi }$  $\sqrt{\frac{1}{LC}}$

Or, LC = $\frac{1}{{{(2\pi f)}^{2}}}$

Or, L = $\frac{1}{{{(2\pi f)}^{2}}\times C}$

Or, L = $\frac{1}{33\times {{10}^{-12}}\times {{(2\pi \times 810\times {{10}^{3}})}^{2}}}$

$\therefore$ L = 1.17×10–3 H

Alternative Method:

At resonance,

XL = XC

Or, $\omega$L = $\frac{1}{\omega C}$

Or, L = $\frac{1}{{{\omega }^{2}}\times C}$

Or, L = $\frac{1}{{{(2\pi f)}^{2}}\times C}$

Or, L = $\frac{1}{{{(2\pi \times 810\times {{10}^{3}})}^{2}}\times 33\times {{10}^{-12}}}$

$\therefore$ L = 1.17×10–3 H

Type – 4:

Q. 18. An iron cored coil of inductance 2 H and resistance 50$\Omega$is connected in series with a resistor of 950 $\Omega$ A 220 V, 50 Hz ac supply is connected across the arrangement. Find the current flowing in the circuit and the voltage drop across the coil.

Solution:

Here,

Inductance, L = 2 H

Resistance of wire of solenoid, rL = 50$\Omega$

Resistance, R = 950$\Omega$

Emf, E = 220 V

Frequency, f = 50 Hz

(i) Current, I = ?

(ii) Voltage drop across inductor, VL = ?

We have,

Inductive reactance, XL =$\omega$L = 2$\pi$f L = 2$\pi$×50 × 2 = 200$\pi $$\Omega Impedance of AC circuit, Z = \sqrt{{{(R+{{r}_{L}})}^{2}}+{{X}_{L}}^{2}} = \sqrt{{{(950+50)}^{2}}+{{(200\pi )}^{2}}} = 1181.01\Omega Using circuit formula, E = I Z Or, I = \frac{E}{Z} = \frac{220}{1181.01} = 0.186 A Total resistance offered by inductor i.e. impedance of the coil ZL = \sqrt{{{r}_{L}}^{2}+{{X}_{L}}^{2}} Or, ZL = \sqrt{{{50}^{2}}+{{(200\pi )}^{2}}} = 630.3\Omega \therefore VL = I ZL = 0.186×630.3 = 117.24 V Q. 19. An iron cored coil of inductance 2 H and 50\Omega resistance placed in series with a resistor of 450\Omega and 220 V, 50 Hz ac supply is connected across the arrangement. Find and the voltage drop across the coil. (i) the current flowing in the coil (ii) its phase angle relative to the voltage supply (iii) the voltage across the coil. Solution: Here, Inductance, L = 2 H Resistance of wire of solenoid, rL = 50\Omega Resistance, R = 450\Omega Emf, E = 200 V Frequency, f = 50 Hz (i) Current, I = ? (ii) Phase angle,\phi = ? (iii) VL = ? Inductive reactance, XL =\omega L = 2\pi f L = 2\pi ×50 × 2 = 200\pi$$\Omega$

Impedance of AC circuit, Z = $\sqrt{{{(R+{{r}_{L}})}^{2}}+{{X}_{L}}^{2}}$

= $\sqrt{{{(450+50)}^{2}}+{{(200\pi )}^{2}}}$ = 802.98$\Omega$

Using circuit formula,

E = I Z

Or, I = $\frac{E}{Z}$ =  $\frac{200}{802.98}$ = 0.25 A

Total resistance offered by inductor i.e. impedance of the coil

ZL = $\sqrt{{{r}_{L}}^{2}+{{X}_{L}}^{2}}$

Or, ZL = $\sqrt{{{50}^{2}}+{{(200\pi )}^{2}}}$ = 630.3$\Omega$

(ii) Tan$\phi$= $\frac{{{Z}_{L}}}{R}$ [instead of Tan$\phi$= $\frac{{{X}_{L}}}{R}$ ]

Or, $\phi$ = Tan–1 $\left( \frac{630.3}{450} \right)$ = 54.48o

(iii) VL = I ZL = 0.25×630.3 = 157.58 V

Q. 20. An iron cored coil of inductance 3 H and 50$\Omega$resistance is placed in series with a resistor of 550$\Omega$ and a 100 V, 50 Hz ac supply is connected across the arrangements. Find the current flowing in the coil and the voltage drop across the coil.

Solution:

Here,

Inductance, L = 3 H

Resistance of wire of solenoid, rL = 50 Ω

Resistance, R = 550$\Omega$

Emf, E = 100 V

Frequency, f = 50 Hz

(i) Current, I = ?

(ii) Voltage drop across inductor, VL = ?

Inductive reactance, XL =$\omega$L = 2$\pi$f L = 2$\pi$×50 × 3 = 300$\pi $$\Omega Impedance of AC circuit, Z = \sqrt{{{(R+{{r}_{L}})}^{2}}+{{X}_{L}}^{2}} = \sqrt{{{(550+50)}^{2}}+{{(300\pi )}^{2}}} = 1117.26\Omega Using circuit formula, E = I Z Or, I = \frac{E}{Z} = \frac{100}{1117.26} = 0.09 A Total resistance offered by inductor i.e. impedance of the coil ZL = \sqrt{{{r}_{L}}^{2}+{{X}_{L}}^{2}} Or, ZL = \sqrt{{{50}^{2}}+{{(300\pi )}^{2}}} = 943.80\Omega (ii) VL = I ZL = 0.09×943.80 = 84.89 V Q. 21. A 50 V ac supply is connected to a resistor having resistance 50\Omega , in series with a solenoid whose inductance is 0.25H. The potential difference between the ends of the resistor is 25 V. Find the resistance of the wire of the solenoid. Assume the frequency of the ac source is 50 Hz. Solution: Here, Emf, E = 50 V Resistance, R = 50\Omega Inductance, L = 0.25 H Voltage drop across resistor, VR = 25 V Resistance of wire of solenoid, rL = ? Frequency, f = 50 Hz We have, Inductive reactance, XL =\omega L = 2\pi f L = 2\pi ×50 × 0.25 = 25\pi \Omega Given, VR = 25 IR = 25 Or, I = \frac{25}{R} = \frac{25}{50} = 0.5 A Or, Z = \frac{E}{I} = \frac{50}{0.5}= 100 \Omega Again Z = \sqrt{{{(R+{{r}_{L}})}^{2}}+{{X}_{L}}^{2}} Or, (R+rL)2 = Z2– XL2 Or, rL = \sqrt{{{Z}^{2}}-{{X}_{L}}^{2}} – R Or, rL =\sqrt{{{100}^{2}}-{{(25\pi )}^{2}}} – 50 \therefore Resistance of wire of solenoid, rL = 11.9\Omega Q. 22. An iron cored coil of inductance 2 H and of resistance 50\Omega is connected in series with a resistor of 950 \Omega and a 220 V, 50 Hz ac supply. Find the current flowing through the circuit and the voltage drop across the coil. Solution: Here, Inductance, L = 2 H Resistance of wire of solenoid, rL = 50\Omega Resistance, R = 950\Omega Emf, E = 220 V Frequency, f = 50 Hz (i) Current, I = ? (ii) Voltage drop across inductor, VL = ? We have, Inductive reactance, XL =\omega L = 2\pi f L = 2\pi ×50 × 2 = 200\pi$$\Omega$

Impedance of AC circuit, Z = $\sqrt{{{(R+{{r}_{L}})}^{2}}+{{X}_{L}}^{2}}$

= $\sqrt{{{(950+50)}^{2}}+{{(200\pi )}^{2}}}$ = 1181.01$\Omega$

Using circuit formula,

E = I Z

Or, I = $\frac{E}{Z}$ =  $\frac{220}{1181.01}$ = 0.186 A

Total resistance offered by inductor i.e. impedance of the coil

ZL = $\sqrt{{{r}_{L}}^{2}+{{X}_{L}}^{2}}$

Or, ZL = $\sqrt{{{50}^{2}}+{{(200\pi )}^{2}}}$ = 630.3$\Omega$

$\therefore$ VL = I ZL = 0.186×630.3 = 117.24 V

Q. 23. A 50V, 50Hz, a.c. supply is connected to a resistor, of resistance 40$\Omega$, in series with a solenoid whose inductance is 0.20H. The p.d. between the ends of the resistor is found to be 20 V. What is the resistance of the wire of the solenoid? (Assume $\pi$2 = 10)

Solution:

Here,

Emf, E = 50 V

Frequency, f = 50 Hz

Resistance, R = 40$\Omega$

Inductance, L = 0.2 H

Voltage drop across resistor, VR = 20 V

Resistance of wire of solenoid, rL = ?

We have,

Inductive reactance, XL = $\omega$L = 2$\pi$f L = 2$\pi$×50 × 0.2 = 20$\pi $$\Omega Given, VR = 20 IR = 20 Or, I = \frac{20}{R} = \frac{20}{40} = 0.5 A Using circuit formula, E = I Z Or, Z = \frac{E}{I} = \frac{50}{0.5}= 100\Omega Again Z =\sqrt{{{(R+{{r}_{L}})}^{2}}+{{X}_{L}}^{2}} Or, (R+rL)2 = Z2– XL2 Or, rL = \sqrt{{{Z}^{2}}-{{X}_{L}}^{2}} – R Or, rL =\sqrt{{{100}^{2}}-{{(20\pi )}^{2}}} – 40 \therefore Resistance of wire of solenoid, rL = 37.8 \Omega Q.24. A 50V, 50Hz, a.c. supply is connected to a resistor of resistance 40\Omega in series with a solenoid having inductance is 200 mH with same resistance. The potential difference across the ends of the 40 \Omega resistor is found to be 20 V. Find the resistance of the wire of the solenoid. Solution: Here, Emf, E = 50 V Frequency, f = 50 Hz Resistance, R = 40\Omega Inductance, L = 200 mH = 200×10–3 H Voltage drop across resistor, VR = 20 V Resistance of wire of solenoid, rL = ? We have, Inductive reactance, XL =\omega L = 2\pi f L = 2\pi ×50 × 200×10–3 = 20\pi$$\Omega$

Given, VR = 20

IR = 20

Or, I = $\frac{20}{R}$ =  $\frac{20}{40}$ = 0.5 A

Using circuit formula,

E = I Z

Or, Z = $\frac{E}{I}$ = $\frac{50}{0.5}$= 100$\Omega$

Again Z =$\sqrt{{{(R+{{r}_{L}})}^{2}}+{{X}_{L}}^{2}}$

Or, (R+rL)2 = Z2– XL2

Or, rL = $\sqrt{{{Z}^{2}}-{{X}_{L}}^{2}}$ – R

Or, rL =$\sqrt{{{100}^{2}}-{{(20\pi )}^{2}}}$  – 40

$\therefore$ Resistance of wire of solenoid, rL = 37.8 $\Omega$